Solutions-Manual-for-Complex-Analysis-by-T-W-Gamelin.pdf

IV 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 191 X X X X X X X X X X X 2 X X X X X X X X 3 X X X X X X X 4 X X X X 5 X X X X 6 X X X X 7 X X X X X X X X X X X 8 X X X X X X X X 1 I.1.1 Identify and sketch the set of points satisfying. (a) jz 1 ij = 1 (f) 0 < Im z < (b) 1 < j2z 6j < 2 (g) < Re z < (c) jz 1j2 + jz + 1j2 < 8 (h) jRe zj < jzj (d) jz 1j + jz + 1j 2 (i) Re (iz + 2) > 0 (e) jz 1j < jzj (j) jz ij2 + jz + ij2 < 2 Solution Let z = x + iy, where x; y 2 R. (a) Circle, centre 1 + i, radius 1. jz 1 ij = 1 , j(x 1) + i (y 1)j = 1 , (x 1)2 + (y 1)2 = 12 (b) Annulus with centre 3, inner radius 1=2, outer radius 1. 1 < j2z 6j < 2 , 1 < 2 jz 3j < 2 , , 1=2 < jz 3j < 1 , (1=2)2 < (x 3)2 + y 2 < 12 p (c) Disk, centre 0, radius 3. jx + iy 1j2 + jx + iy + 1j2 < 8 , p 2 , (x 1)2 + y 2 + (x + 1)2 + y 2 < 8 , x2 + y 2 < 3 (d) Interval [ 1; 1]. q q 2 jz 1j + jz + 1j 2 , (x 1) + y2 2 (x + 1)2 + y 2 , q 2 q 2 2 2 , (x 1) + y2 2 (x + 1) + y 2 , q q 2 , (x + 1)2 + y 2 x+1, (x + 1) +2 y2 (x + 1)2 , y = 0 Now, take y = 0 in the inequality, and compute the three intervals 2 x < 1; then jx 1j + jx + 1j = (x 1) (x + 1) = 2x 2; 1 x 1 then jx 1j + jx + 1j = (x 1) + (x + 1) = 2 2 x > 1; then jx 1j + jx + 1j = (x 1) + (x + 1) = 2x 2: (e) Half–plane x > 1=2. jz 1j < jzj , jz 1j2 < jzj2 , jx + iy 1j2 < jx + iyj2 , , (x 1)2 + y 2 < x2 + y 2 , x > 1=2 (f) Horizontal strip, 0 < y < . (g) Vertical strip, <x< . (h) CnR. jRe zj < jzj , jRe (x + iy)j2 < jx + iyj2 , x2 < x2 + y 2 , jyj > 0 (i) Half plane y < 2. Re (iz + 2) > 0 , Re (i (x + iy) + 2) > 0 , y+2>0,y <2 (j) Empty set. jz ij2 + jz + ij2 < 2 , jx + iy ij2 + jx + iy + ij2 < 2 , , x2 + (y 1)2 + x2 + (y + 1)2 < 2 , x2 + y 2 < 0 3 I.1.2 Verify from the de…nitions each of the identities (a) z + w = z + w (b) zw = z w (c) jzj = jzj (d) jzj2 = z z Draw sketches to illustrate (a) and (c). Solution Substitute z = x + iy and w = u + iv, and use the de…nitions. (a) z + w = (x + iy) + (u + iv) = (x + u) + (y + v) i = = (x + u) (y + v) i = (x iy) + (u iv) = z + w: (b) zw = (x + iy) (u + iv) = (xu yv) + (xv + yu) i = = (xu yv) (xv + yu) i = (x iy) (u iv) = z w: (c) q p jzj = x + iy = jx iyj = x2 + ( y)2 = x2 + y 2 = jx + iyj = jzj : (d) p 2 jzj2 = jx + iyj2 = x2 + y 2 = x2 + y 2 = = x2 i2 y 2 = (x + iy) (x iy) = z z: 4 I.1.3 Show that the equation jzj2 2 Re (az) + jaj2 = 2 represents a circle centered at a with radius . Solution Let z = x + iy and a = + i , we have jzj2 2 Re (az) + jaj2 = 2 = x2 + y 2 2 Re (( i ) (x + iy)) + 2 + = 2 2 2 2 =x +y 2 Re (( x + y) + i ( y x)) + + = 2 2 2 2 =x +y 2 ( x + y) + + = = (x )2 + (y )2 : Thus the equation jzj2 2 Re (az) + jaj2 = 2 , becomes (x )2 + (y )2 = 2 ; which is the equation for a circle of radius centered at ( ; ), which in complex notation is the point a = + i . 5 I.1.4 Show that jzj jRe zj + jIm zj, and sketch the set of points for which equality holds. Solution Apply triangle inequality to z = Re z + i Im z, to obtain jzj 6 jRe zj + jIm zj. Now set z = x + iy, and see then equality holds p p p p 2 p p 2 jzj = jRe zj+jIm zj , x2 + y 2 = x2 + y 2 , x2 + y 2 = x2 + y 2 , p 2 p 2 p p , x2 +y 2 = x2 + x2 +2 x2 y 2 , x2 y 2 = 0 , x = 0 or y = 0: Equality holds only when z is real or pure imaginary, which are all the points on the real and imaginary axis. 6 I.1.5 Show that jRe zj jzj ; jIm zj jzj : Show that jz + wj2 = jzj2 + jwj2 + 2 Re (z w) : Use this to prove the triangle inequality jz + wj jzj + jwj. Solution Let z = x + iy. Then since the square root function is monotone, we have p p jRe zj = jxj = x2 6 x2 + y 2 = jzj ; p p jIm zj = jyj = y 2 6 x2 + y 2 = jzj : Now for z; w we have jz + wj2 = = (z + w) (z + w) = (z + w) (z + w) = z z + z w + wz + ww = = jzj2 + 2 Re (z w) + jwj2 ; where we have used the fact that 2 Re (z w) = z w +z w = z w + wz = z w +wz. We use both of the above facts and the trivial identities jzj = jzj and jzwj = jzj jwj to prove the triangle inequality for z; w. We have jz + wj2 = jzj2 + 2 Re (z w) + jwj2 jzj2 + 2 jRe (z w)j + jwj2 jzj2 + 2 jz wj + jwj2 = jzj2 + 2 jzj jwj + jwj2 = (jzj + jwj)2 : The desired inequality now follows by taking square root of both sides. 7 I.1.6 For …xed a 2 C, show that jz aj = j1 azj = 1 if jzj = 1 and 1 az 6= 0. Solution If jzj = 1, then jzj = 1 and z z = 1. Use this and get jz aj = jz aj jzj = jz z azj = j1 azj = j1 azj : We have jz aj = 1; j1 azj as was to be shown. 8 I.1.7 Fix > 0, = 6 1, and …x z0 ; z1 2 C. Show that the set of z satisfying jz z0 j = jz z1 j is a circle. Sketch it for = 12 and = 2, with z0 = 0 and z1 = 1. What happens when = 1? Solution Recall that a circle in R2 centered at (a; b) with radius r is given by the equation (x a)2 + (y b)2 = r2 : We manipulate the equation jz z0 j = jz z1 j : The solutions set of the equation above remains the same if we square both sides, jz z0 j2 = 2 jz z1 j2 : Let z = x + iy, z0 = x0 + iy0 and z1 = x1 + iy1 . Thus our equation becomes (x x0 )2 + (y y0 )2 = 2 (x x1 )2 + (y y1 )2 : Expanding the squares, and grouping terms, we have 2 1 x2 2 x0 2 x1 x+ x20 2 2 x1 + 1 2 y 2 2 y0 2 y1 y+ y02 2 2 y1 =0 2 Dividing both sides by (1 ), we have 2 (x0 x1 ) (x20 2 2 x1 ) (y0 2 y1 ) (y02 2 2 y1 ) x2 2 2 x+ 2 + y2 2 2 y+ 2 =0 1 1 1 1 Now complete the squares for both the x and y terms. Recall that x2 2ax + b = (x a)2 a2 + b: So we have 9 2 2 2 2 x0 x1 (1 ) (x20 2 2 x1 ) (x0 2 x1 ) x + + 1 2 (1 2 )2 2 2 2 2 y0 y1 (1 ) (y02 2 2 y1 ) (y0 2 y1 ) + y + = 0: 1 2 (1 2 )2 This becomes 2 2 2 2 x0 x1 y0 y1 x 2 + y 2 = 1 1 2 2 2 2 2 (x0 x1 ) + (y0 y1 ) (1 ) (x20 + y02 2 (x21 + y12 )) = = (1 2 )2 2 (x1 x0 )2 + (y1 y0 )2 = ; (1 2 )2 which is the equation for a circle. If z0 = 0, and z1 = 1, we have 2 2 2 2 x 2 +y = 2 : 1 1 When = 21 we have a circle of radius 23 centered at 1 3 ; 0 , and when = 2, 2 4 we have a circle of radius 3 centered at 3 ; 0 . When = 1, we have the equation jz z0 j = jz z1 j ; which is the line bisecting the two points. When z0 = 0; z1 = 1, this is the line Re z = 21 . 10 I.1.7 I.1.7 I.1.7 2 2 2 Im z Im z Im z 1 1 1 -1 1 2 -1 1 2 -1 1 2 -1 Re z -1 Re z -1 Re z -2 -2 -2 1 = 2 =1 =2 11 I.1.8 Let p (z) be a polynomial of degree n 1 and let z0 2 C. Show that there is a polynomial h (z) of degree n 1 such that p (z) = (z z0 ) h (z)+p (z0 ). In particular, if p (z0 ) = 0, then p (z) = (z z0 ) h (z). Solution Set p (z) = an z n + an 1 z n 1 + an 2 z n 2 + + a2 z 2 + a1 z + a0 : and h (z) = bn 1 z n 1 + bn 2 z n 2 + bn 3 z n 3 + : : : + b2 z 2 + b1 z + b0 : We equate coe¢ cients in the polynomial identity p (z) = (z z0 ) h (z) + p (z0 ), and get an z n + an 1 z n 1 + an 2 z n 2 + + a2 z 2 + a1 z + a0 = = (z z0 ) bn 1 z n 1 + bn 2 z n 2 + bn 3 z n 3 + : : : + b2 z 2 + b1 z + b0 +p (z0 ) = = bn 1 z n + bn 2 z n 1 + bn 3 z n 2 + : : : + b2 z 3 + b1 z 2 + b0 z + p (z0 ) bn 1 z0 z n 1 bn 2 z0 z n 2 bn 3 z0 z n 3 : : : b2 z0 z 2 b1 z0 z b0 z0 = bn 1 z n +(bn 2 bn 1 z0 ) z n 1 +(bn 3 bn 2 z0 ) z n 2 +(bn 4 bn 3 z0 ) z n 3 + + (b2 b3 z0 ) z 3 + (b1 b2 z0 ) z 2 + (b0 b1 z0 ) z + p (z0 ) b0 z0 : Equate and solve for the bj ´s in terms of aj ´s. 8 8 > > b n 1 = an > > nP k 1 < an = b n 1 < bk = ak+1+i z0i ; 0 k n 2 ak = bk 1 bk z0 ; 0 k n 1 ) i=0 : > > Pn a0 = p (z0 ) b0 z0 > > : p (z0 ) = ai z0i i=0 Proof by induction on degree n of p (z), set p (z) = an z n + an 1 z n 1 : : : + a0 ; where an 6= 0. 12 Fix z0 and write p (z) = an (z z0 ) z n 1 + r (z) ; where deg r (z) n 1. By using the induction hypothesis, we can assume that r (z) = q (z) (z z0 ) + c; where deg q (z) deg r (z). Then p (z) = an z n 1 + q (z) (z z0 ) + c = h (z) (z z0 ) + c Since deg q (z) n 2, deg r (z) n 1. Plug in z0 , get p (z0 ) = c. 13 I.1.9 Find the polynomial h (z) in the preceding exercise for the following choices of p (z) and z0 (a) p (z) = z 2 and z0 = i (b) p (z) = z 3 + z 2 + z and z0 = 1 (c) p (z) = 1 + z + z 2 + + z m and z0 = 1 Solution From the preceding exercise we have p (z) = (z z0 ) h (z) + p (z0 ) : We solve for h (z) then p (z) p (z0 ) h (z) = : z z0 (a) We have that p (z) = z 2 and z0 = i, thus p (z0 ) = p (i) = 1. Thus p (z) p (z0 ) z2 + 1 h (z) = = = z + i; z z0 z i and z 2 = (z i) (z + i) 1: (b) We have that p (z) = z 3 + z 2 + z and z0 = 1, thus p (z0 ) = p ( 1) = 1. Thus p (z) p (z0 ) z3 + z2 + z + 1 (z + 1) (z 2 + 1) h (z) = = = = z 2 + 1; z z0 z+1 z+1 and z 3 + z 2 + z = (z + 1) z 2 + 1 1: (c) We have that p (z) = 1 + z + z 2 + + z m and z0 = 1, thus 14 m odd p (z0 ) = p ( 1) = 1. 0. m even Thus p (z) p (z0 ) h (z) = = 8 z z0 < (zm 1 +z m 3 +z 2 +1)(z+1) = zm 1 + zm 3 + + z2 + 1 if m is odd = z+1 : (zm 1 +z m 3 +z 3 +z )(z+1)+1 = zm 1 + zm 3 + : : : + z 3 + z if m is even z+1 and zm + zm 1 + z2 + z + 1 = (z + 1) (z m 1 + zm 3 + + z 2 + 1) if m is odd = (z + 1) (z m 1 + zm 3 + 3 + z + z) + 1 if m is even 15 . the polynomials are h (z) = 0 and r (z) = 0. bm 6= 0 and m n. h (z) q (z)+r (z) = 0 implies that h (z) = 0. where h (z) and r (z) are polynomials and the degree of the remain- der r (z) is strictly less than m.) The degree of r (z) must be less than the degree of q (z). So. So. suppose that the degree of q (z) is less than or equal to the degree of p (z).) If the degree of q (z) is greater than the degree of p (z). So.1.I. This proves both existence and uniqueness of h (z) and r (z). it follows that the degree of h (z) q (z) is greater than the degree of r (z). and thus r (z) = 0. where an 6= 0. Now. The resulting method is called the division algorithm. Proceed by induction on the degree of p (z). This then implies that h (z) q (z)+r (z) 6= 0. if the degree of q (z) is greater than the degree of p (z). and h (z) is nonzero.10 Let q (z) be a polynomial of degree m 1. Let an n m p1 (z) = z q (z) p (z) . then h (z) = 0 and thus r (z) = p (z). Solution First suppose p (z) is the zero polynomial. Now assume that the division algorithm is true for all polynomials p (z) of degree less than n. (Where n 0. If h (z) 6= 0. and these polynomials are the only ones that satisfy both conditions. then h (z) q (z) + r (z) has degree greater than p (z). Set p (z) = an z n + an 1 z n 1 + + a1 z + a0 and q (z) = bm z m + bm 1 z m 1 + + b1 z + b0 . the degree of p (z) is 1. Hint. bm 16 . in this case. Show that any polynomial p (z) can be expressed in the form p (z) = h (z) q (z) + r (z) . (So. bm and let r (z) = r1 (z) : Then h (z) q (z) + r (z) = an n m = z h1 (z) q (z) r1 (z) = bm an n m = z q (z) (h1 (z) q (z) + r1 (z)) = bm an n m = z q (z) p1 (z) = bm = p (z) : Thus given p (z) and q (z). that p1 (z) = h1 (z) q (z)+r1 (z). there exist polynomials h (z) and r (z) satisfying the above two conditions. by assumption. and therefore p1 (z) is a polynomial of de- gree at most n 1. where h1 (z) and r1 (z) are the unique polynomials satisfying the conditions above. Let an n m h (z) = z h1 (z) . It follows. 17 .then an n m p1 (z) = z bm z m + bm 1 z m 1 + + b1 z + b0 an z n + an 1 z n 1 + + a1 z + a0 : bm The monomials of degree n cancel. and 1.11 Find the polynomials h (z) and r (z) in the preceding exercise for p (z) = z n and q (z) = z 2 1. if n is even. zn 2 + zn 4 + zn 6 + + z 3 + z. if n is odd. if n is odd.I. if n is even.1. Solution Require z n = h (z) z 2 1 + r (z) . h (z) = . deg r (z) 1: If n is even zn = zn 2 + zn 4 + zn 6 + + z2 + 1 z2 1 + 1: If n is odd zn = zn 2 + zn 4 + zn 6 + + z3 + z z2 1 + z: Thus zn 2 + zn 4 + zn 6 + + z 2 + 1. 18 . r (z) = : z. 1 = ei =4 i5 =4 . 1. 1 i 3 : (f) 19 .I. k = 0. ei13 =8 = = f (cos ( =8) + i sin ( =8)) .1 Express all values of the following expressions in both polar and Cartesian p coordinates. (a) i (c) 4 1 (e) ( 8)1=3 (g) (1 + i)8 p p 25 (b) i 1 (d) 4 i (f) (3 4i)1=8 (h) 1+i p 2 Solution (a) p n o n p o i( =2+2k ) 1=2 i( =4+k ) i= e =e . ei5 =8 . 2 = n p p o = 2ei =3 . p and plot them. 1 = = 21=4 ei3 =8 . k = 0. 2. 3 = n p p o i =4 i3 i=4 i5 =4 i7 =4 = e .e = (1 i)= 2. 1. 21=4 ei11 =8 = 21=4 (cos (3 =8) + i sin (3 =8)) : (c) p n o +2k ) 1=4 4 1 = ei( = ei( =4+k =2) . 3 = ei =8 . k = 0. 2ei . k = 0.e = (1 + i) = 2 : (b) p p 1=2 i 1 = 2ei(3 =4+2k ) = 21=4 ei(3 =8+k ) . 1. 2ei5 =3 = 1 + i 3. 2.e . ei9 =8 . 2. ( 1 i) = 2 : (d) p n o 4 =2+2k ) 1=4 i( =8+k =2) i= ei( e .2. k = 0.e . (cos (5 =8) + i sin (5 =8))g : (e) n o +2k ) 1=3 ( 8)1=3 = 23 ei( = 2ei( =3+2k =3) . 1f 2 1 1 -1 1 -2 2 -1 1 -1 -1 -2 20 . 51=8 ei( 0 =8+ =4) .2. 51=8 (cos ( 0 =8 + 3 =4) + i sin ( 0 =8 + 3 =4)) : (g) p 8 (1 + i)8 = 2ei( =4+2k ) = 16ei(2 +16k ) = 16ei2 = 16: (h) 25 1+i =4+2k ) 25 1+i p = ei( = ei(25 =4+50k ) = ei( =4+6 +50k ) = ei =4 = p : 2 2 I. 51=8 (cos ( 0 =8 + =2) + i sin ( 0 =8 + =2)) .1c 1 1 1 -1 1 -1 1 -1 1 -1 -1 -1 I. 1. 1 4 Draw …gure and get tan 0 = 4=3 ) 0 = tan 3 . 7 = = 51=8 ei( 0 =8) . n o ) 1=8 (3 4i)1=8 = 5ei( 0 +2k = 51=8 ei( 0 =8+k =4) .1a I. 51=8 ei( 0 =8+3 =4) = 1=8 1=8 5 (cos ( 0 =8) + i sin ( 0 =8)) . 51=8 ei( 0 =8+ =2) .2. k = 0. 5 (cos ( 0 =8 + =4) + i sin ( 0 =8 + =4)) .2. : : : .1e I.2.1d I.2.2.1b I. 2.1g I.2.1h 20 1 -20 20 -1 1 -1 -20 21 .I. 0) -5 5 -5 5 (−3π/2.2.2. b) Sector. c) Is a spiral curve starting at 0. and to 1.0) -5 -5 I. I. spiraling to 1.2d (-30 < x < 30) I.2b I.2a I.2 Sketch the following sets (a) jarg zj < =4 (c) jzj = arg z (b) 0 < arg (z 1 i) < =3 (d) log jzj = 2 arg z Solution a) Sector.2.2d (-800 < x < 800) (1.1.I.2.1.2c 5 5 π/3 (π/2. spiraling to 0.0) π/4 π/4 (−π.2d (-3 < x < 3) I. (d) Is a spiral curve.0) (2π.0) 22 .2. sketch the curve ei + be i : 0 2 . an ellipse traversed in negative direction. an ellipse x2 = (1 + b)2 + y 2 = (1 b)2 = 1. jbj = 1 and jbj < 1.2. and then reduce the general case to this case by a rotation. traversed in positive direction with increasing .2. For b = ei' . an interval [ 2. For b = 1. For 1 < b < +1. Solution For 0 < b < 1.7 (b = 1. I. express equation as ei'=2 ei( '=2) + e i( '=2) to see that curve is rotate of ellipse or interval by '=2. Di¤erentiate between the cases jbj < 1.7 (b = 0. Hint.7 (b = 1) I.5) I.5) 3 3 3 y y y 2 2 2 1 1 1 -3 -2 -1 1 2 3 -3 -2 -1 1 2 3 -3 -2 -1 1 2 3 -1 x -1 x -1 x -2 -2 -2 -3 -3 -3 23 .2. 2].2.3 For a …xed complex number b. First consider the case b > 0.I. i is an nth root of unity for in = 1.2.4 For which n is i an n-th root of unity? Solution.I. k = 1. 2. 3. n = 4k. : : : 24 . 5 For n 1. z 6= 1 1 sin n+ 1 2 (b) 1 + cos + cos 2 + + cos n = 2 + 2 sin =2 . 25 . 1ei(n+1) 1 + ei + ei2 + + ein = . and multiply sn with z and have zsn = z + z 2 + z 3 + + z n+1 : Now subtract zsn from the sum sn . 1 z n+1 sn = : 1 z (b) Apply (a) to z = ei and to z = e i . 1 ei i i2 in 1 e i(n+1) 1+e +e + +e = : 1 e i Add the identities. and use the de…nitions of sine and cosine.2. show that (a) 1 + z + z 2 + + z n = (1 z n+1 ) = (1 z). Solution (a) Set sn = 1 + z + z 2 + + zn. and have sn (1 z) = 1 z n+1 : If z 6= 1 we have after division by 1 z.I. 2 (1 + cos + cos 2 + + cos n ) = 1ei(n+1) 1 e i(n+1) = + = 1 ei 1 e i 1 ei(n+1) e i =2 1 e i(n+1) ei =2 = + = e i =2 ei =2 ei =2 e i =2 1 e i =2 1 ei(n+1) 1 ei =2 1 e i(n+1) = + = 2i sin ( =2) 2i sin ( =2) " 1 1 # 1 ei =2 e i =2 + ei(n+ 2 ) e i(n+ 2 ) sin ( =2) + sin n + 1 2 = = = 2i sin ( =2) sin ( =2) sin n + 12 =1+ : sin ( =2) Divide both sides with 2 and get 1 sin n + 21 1 + cos + cos 2 + + cos n = + : 2 2 sin =2 26 . and wj are a roots of z n 1 since ji=n n wjn 1 = e2 1 = e2 ji 1 = 0: By the fundamental theorem of algebra. k = n: Solution (a) Let w0 . and since the root are simple and the coe¢ cient for the z n term is 1 it follows that zn 1 = (z w0 ) (z w1 ) (z wn 1 ) : (b) Using exercise (a) and multiply the factors in the product we get zn 1 = zn (w0 + w1 + wn 1 ) z n 1 +c2 z n 2 + +( 1)n w0 w1 wn 1 : By identi…cation of the coe¢ cients we see that w0 + w1 + + wn 1 =0 if n 2: (c) From (b) follows that 1 = ( 1)n w0 w1 wn 1 . Show that the n th roots of unity w0 . 1 k n 1. w0 w1 wn 1 = ( 1)n 1 : (d) If 1 k n 1 n X n 1 X n 1 X n 1 1 e2 ki=n 1 e2 ki 2 ji=n k 2 ki=n j wjk = e = e = = = 0: j=0 j=0 j=0 1 e2 ki=n 1 e2 ki=n 27 . wn 1 be the n th roots of unity then wj = e2 ji=n .I. : : : . : : : . (d) wjk = j=0 n.2.6 Fix n 1. wn 1 satisfy (a) (z w0 ) (z w1 ) : : : (z wn 1 ) = z n 1: (b) w0 + + wn 1 = 0 if n 2: (c) w0 wn 1 = ( 1)n 1 : nP1 0. If k = n X n 1 X n 1 X n 1 X n 1 2 ji=n n wjk = e = 2 ji e = 1 = n: j=0 j=0 j=0 j=0 28 . 7 (n = 5) 1 1 1 -1 1 -1 1 -1 1 -1 -1 -1 We use that jz wj jzj jwj see (1. Show that zm Rm .1) p. and have that jz n + 1j > jz n j 1 = Rn 1.7 (n = 2) 1 1 a+b b -1 1 -1 1 a -1 -1 I. jz n + 1j (Rn 1) 29 .2.2.1) on page 2 in CA.7 Fix R > 1 and n 1. where the last equaltity is due to that jzj = R. Solution I.2. jzj = R: zn + 1 Rn 1 Sketch the set where equality holds. Hint. See (1.7 I.I.2.2. Some rearrangement gives 1 1 . m 0.2.7 (n = 3) I.7 (n = 1) I.2.7 (n = 4) I.2. 2. and zm Rm = : zn + 1 Rn 1 30 . b lie on the same ray. we must have jz n + 1j = Rn 1. jzj = R: zn + 1 Rn 1 For equality. see …rst …gure I. we require that ein = 1 thus i=n z = wk Re . then zn + 1 = Rn + 1. If i=n z = wk Re . (We have used the fact that jaj+jbj = ja + bj implies that a. where wk is an n th root of unity. We rearrarange this equality to j 1j + jz n + 1j = jz n j and we conclude that 1 and z n +1 must lie on the same ray.7) If z = Rei then z n = Rn ein : Since R > 1. because jzjm = Rm .and mulitplication by jz m j = Rm gives zm Rm . by setting the real and imaginary parts equal to each other we obtain cos 4 = cos4 6 cos2 sin2 + sin4 sin 4 = 4 cos3 sin 4 cos sin3 : as required. by setting the real and imaginary parts equal to each other we obtain cos 2 = cos2 sin2 sin 2 = 2 cos sin : Similarly. applying de Moivre’s formulae for n = 4 we get (using the Binomial Theorem) cos 4 + i sin 4 = (cos + i sin )4 = = cos4 + i4 cos3 sin 6 cos2 sin2 i4 cos sin3 + sin4 = = cos4 6 cos2 sin2 + sin4 + i 4 cos3 sin 4 cos sin3 . then. 31 .8 Show that cos 2 = cos2 sin2 and sin 2 = 2 cos sin using de Moivre’s formulae. then. Then by de Moivre’s formulae (on page 8 in CA) we have for all n 2 Z cos n + i sin n = (cos + i sin )n Hence for n = 2 we get cos 2 + i sin 2 = = (cos + i sin )2 = cos2 + i2 cos sin sin2 = = cos2 sin2 + i (2 cos sin ) .2. Solution Let 2 R be given. Find formulae for cos 4 and sin 4 in terms of cos and sin .I. 1c 2 4 2 y y y 1 2 1 -2 -1 1 2 -4 -2 2 4 -2 -1 1 2 -1 -2 -1 -2 -4 -2 I.3. for …xed A. according to various ranges of A.1 Sketch the image under the spherical projection of the following sets on the sphere (a) the lower hemisphere Z 0 (b) the polar cap 43 Z 1 p p (c) lines of lattitude X = 1 Z 2 cos . for Z …xed and 0 2 p p (d) lines of longitude X = 1 Z 2 cos .1d I. for …xed and 1 Z 1 (e) the spherical cap A X 1.I.1a I.1e (A = 0) 2 4 2 y y y 1 2 1 θ -2 -1 1 2 -6 -4 -2 2 -2 -1 1 2 -1 -2 -1 -2 -4 -2 32 .3. with center lying on the equator. Y = 1 Z 2 sin .3. Separate into cases.1e (A = -1/2) I.3.3.3. Y = 1 Z 2 sin . Solution I.3.1b I. thus x = 7. 33 . p p (c) Image p is the disk. I. (e) Case 1: 1 < A < 0 p Image is the exterior of the disk centered at 1=A with radius 1 A2 = jAj. (b) p Set Y = 0 and Z = 3=4 in X 2 + Y 2 + Zp 2 = 1.3. We have the formula x = X= (1 p Z). Case 2: A = 0 Image is the left half–plane.1e (A = 1/2) 4 y 2 -2 2 4 6 -2 -4 (a) Image is the unit disk. thus r = 1 Z = p1+Z 1 Z . Set Y = 0 then Z goes from 1 A2 top 1 A2 . we have X = 7=4. We have that 2 X = 1 Z cos and 2 Y = p 1 Z sin . Case 3: 0 < A < 1 p Image is a disk centered at 1=A p with radius p 1 A2 = jAj. we can take the radius to image p 1 Z2 for = 0. Image is the exterior of a disk. (d) Image is lines of longitude issuing from 0. We have p the formula x = X= (1 Z) thus x goes from A= 1 + 1 A = 1 2 1 A2 =A to p p A= 1 1 A2 = 1 + 1 A2 =A. with centre p 0 and radius 1 + z= 1 z. with centre 0 and radius 7. Z) on the sphere corresponds to z 7 ! 1=z of C. 1 2 2 jzj + 1 z + 1 1 2 0 z 1 jzj2 1 Z = = = Z: 1 2 jzj2 + 1 z + 1 Hence. Y 0 . jzj + 1 jzj2 1 Z = : jzj2 + 1 For z 6= 0. the map (X. jzj2 + 1 2y Y = 2 .I. Z 0 ) where 2x 0 jzj2 2x X = = = X. Z) 7 ! ( X. we have 1 1 x + iy x + iy x y = = = 2 = i 2 z x iy (x iy) (x + iy) x2 + y 2 jzj jzj which corresponds to the point on the sphere given by (X 0 . show that the antipodal point P on the sphere corresponds to 1=z. Y.2 If the point P on the sphere corresponds to z under the stereo- graphic projection. Y.3. 1 2 2 jzj + 1 z + 1 2y jzj2 2y Y0 = = = Y. Y. Solution For z = x + iy the corresponding point on the sphere under transformation given on p. Z) where 2x X = . 34 . 12 in CA is given by (X. 35 .3. and seeing that the image circles moves continuously.) Solution Draw the picture. the corresponding point P on the sphere traverses a small circle in the negative (clockwise) direction with respect to someone standing at the center of the circle and with body outside the sphere.I. (Thus the stereographic projection is orientation reversing. as a map from the sphere with orientation determined by the unit outer normal vector to the com- plex plane with the usual orientation. And the image circle is the unity and positive direction of the unit circle ( ) is the converse. Or argue as follows. Thus we need to shrink it only for the equator of the sphere oriented as indicated ( ).3 Show that as z travers a small circle in the complex plane in the positive (counterclockwise) direction. if the South Pole is inside it. This can be seen by moving one circle continuously to the other. So the orientation of the image is clockwise (negative). The orientation of the image circle is the same for all circles on the sphere orientated so that N is outside the circle. Y 0 . jzj + 1 jzj2 1 Z = : jzj2 + 1 For z 6= 0. Z) on the sphere corresponds to inversion z 7 ! 1=z of C. Y. 1 2 2 jzj + 1 z + 1 1 jzj2 1 jzj2 1 Z0 = 1 = = Z: jzj 2 + 1 jzj2 + 1 Hence. the map (X. Z) where 2x X = . Y. Solution For z = x + iy the corresponding point on the sphere under transformation given on p. the map (X. 1 2 jzj2 + 1 z + 1 2y jzj2 2y Y0 = = = Y. Z 0 ) where 2x 0 jzj2 2x X = = = X. Y.I. Z) is given by rotation of the sphere by 180 about the X-axis. 12 in CA is given by (X. Z) 7 ! (X. Z) 7 ! (X. we have 1 1 x iy x iy x y = = = 2 = i z x + iy (x + iy) (x iy) x + y2 jzj2 jzj2 which corresponds to the point on the sphere given by (X 0 .4 Show that a rotation of the sphere of 180 about the X-axis corre- sponds under stereographic projection to the inversion z 7 ! 1=z of C. 36 . Y.3. Moreover. jzj2 + 1 2y Y = 2 . Y. 0) is the spherical projection of (X. X and the distance from the north pole N = (0. Y. By the Pythagorean Law. 0) = . Z) lie on the equator. 37 . Y. What is the situation when (X. y.I. 1) to (x. Z) lies on the equator on the sphere? Solution The distance from from the north pole N = (0. Z). this is 1 + 1 = 2.5 Suppose (x. Y. Y . 0. Show that the product of the distances from the north pole N to (X.0 1 Z 1 Z is s X2 Y2 + + 1: (1 Z)2 (1 Z)2 The product of distances is 1=2 1=2 X2 Y2 X 2 + Y 2 + (Z 1)2 X 2 + Y 2 + (Z 1)2 + +1 = = 2: (1 Z)2 (1 Z)2 1 Z When (X. Y. 1) to (X. Z) and from N to (x. Z) is q X 2 + Y 2 + (Z 1)2 . 0) is 2.3. the product is simply the square of the distance from N to a point on the equator. 0. y. y. Y. it is sym- metric. z. we have jz wj2 = = (z w) (z w) = (z w) (z w) = 2 2 = zz zw zw + ww = jzj + jwj zw zw = 2 2 = jzj + jwj (x1 + iy1 ) (x2 iy2 ) (x1 iy1 ) (x2 + iy2 ) = = jzj + jwj2 2 2x1 x2 2y1 y2 : (1) 38 . 1 + jzj2 where ds = jdzj is the usual Euclidean in…nitesimal arc length. w 2 C: 2 2 1 + jzj 1 + jwj (c) What is d (z. The expression for d (z.6 We de…ne the chordal distance d (z. satis…es the triangle inequality d (z. w) shows that in…nitesimal arc length corresponding to the chordal metric is given by 2ds d (z) = . respectively. w).3. w). the spher- ical metric (z. z). w) between two points z. d (z.I. w 2 C to be the length of the straight line segment joining to the points P and Q on the unit sphere whose stereographic projections are z and w. and d (z. ) + d ( . 1)? Remark. that is. (a) Show that the chordal distance is a metric. w) = d (w. w) = q q . on the extended complex plane.3. (b) Show that the chordal distance from z to w is given by 2 jz wj d (z. w) = 0 if and only if z = w. w) d (z. The in…nitesimal arc length d (z) determines another metric. w = x2 + iy2 . See Section IX. Solution (a) Follows from fact that Euclidean distance in R3 is a metric on the sphere. (b) Set z = x1 + iy1 . 1) = lim q q = lim q q =q : w!1 2 2 w!1 2 2 2 1 + jzj 1 + jwj 1 + jzj 1= jwj + 1 1 + jzj 39 . w 2 C: 1 + jzj2 1 + jwj2 (c) 2 jz wj 2 jz=w 1j 2 d (z. w) = q q . z. w)2 = (X X 0 ) + (Y Y 0 ) + (Z Z 0) = = X 2 + Y 2 + Z 2 + X 02 + Y 02 + Z 02 2 (XX 0 + Y Y 0 + ZZ 0 ) = ! 4x1 x2 + 4y1 y2 + jzj2 1 jwj2 1 =2 2 = jzj2 + 1 jwj2 + 1 ! jzj2 + 1 jwj2 + 1 4x1 x2 4y1 y2 jzj2 1 jwj2 1 =2 = jzj2 + 1 jwj2 + 1 ! 2jzj2 + 2 jwj2 4x1 x2 4y1 y2 (1) 4 jz wj2 =2 = : jzj2 + 1 2 jwj + 1 2 jzj + 1 jwj2 + 1 Taking the positive square root we have 2 jz wj d (z.We take square of distance between P and Q 2 2 2 d (z. 1). and the equations for X. 0) plane. Z) (0. Y. Y. Y. 1 + t (Z 1)) : By simultaneous equations we have. Y and Z 8 8 < x = tX < X = xt y = tY ) Y = yt : : 1 + t (Z 1) = 0 Z = 1 1t 1 2 1 We solve for the t that is a point on the sphere X 2 + Y 2 + Z 2 = 4 we have 2 2 x2 y 2 1 1 1 t t2 + 2 + = . 0. y. Y.3. jzj2 t + 1 = 0 . t = jzj2 + 1 thus we have 8 x > X = 1+jzj < y 2 Y = 1+jzj2 > : Z = jzj2 : 1+jzj2 40 . 0) lie on the same line through the north pole. Y. tY. jzj2 + 1 = . We de…ne a stereographic projection of the sphere onto the complex plane as before. Y. 0. so that corresponding points (X. 1) is on the form N + t (P N ).7 Consider the sphere of radius 21 in (X. 0. 0. 0) = (0. the equation of the sphere is 2 X 2 + Y 2 + Z 12 = 41 .I. Y. Z in terms of z. Z) and z (x. Z) and N = (0. Find the equations for z = x + iy in terms of X. 0) and north pole at (0. In this case. and it meats the xy plane when (x. 1)) = (tX. resting on the (X. t2 t 2 t 4 2 4 . 1) + t ((X. Z) space. Solution The line through P = (X. Z. 0. y. with south pole at the origin (0. What is the corresponding formula for the chordal distance? Note. and solve for X. Set z = x1 + iy1 . w = x2 + iy2 . we have jz wj2 = = (z w) (z w) = (z w) (z w) = 2 2 = zz zw zw + ww = jzj + jwj zw zw = 2 2 = jzj + jwj (x1 + iy1 ) (x2 iy2 ) (x1 iy1 ) (x2 + iy2 ) = = jzj2 + jwj2 2 (x1 x2 + y1 y2 ) : (1) And the coordical distance 41 . d (z. w)2 = (X1 X 2 )2 + (Y1 Y 2 )2 + (Z1 Z 2 )2 = !2 2 2 x1 x2 y1 y2 jz1 j2 jz2 j2 = + + = 1 + jzj2 1 + jwj2 1 + jzj2 1 + jwj2 1 + jzj2 1 + jwj2 2 2 2 x1 1 + jwj2 x2 1 + jzj2 + y1 1 + jwj2 y2 1 + jzj2 + jzj2 jwj2 = 2 2 = 1 + jzj2 1 + jwj2 2 2 (x21 + y12 ) 1 + jwj2 + (x22 + y22 ) 1 + jzj2 + jzj4 2 jzj2 jwj2 + jwj4 = 2 2 + 1 + jzj2 1 + jwj2 2x1 x2 1 + jwj2 1 + jzj2 2y1 y2 1 + jwj2 1 + jzj2 + 2 2 = 1 + jzj2 1 + jwj2 2 2 jzj2 1 + jwj2 + jwj2 1 + jzj2 + jzj4 2 jzj2 jwj2 + jwj4 = 2 2 + 1 + jzj2 1 + jwj2 2 (x1 x2 + y1 y2 ) 1 + jwj2 1 + jzj2 + 2 2 = 1 + jzj2 1 + jwj2 jzj2 + 2 jzj2 jwj2 + jzj2 jwj4 + jwj2 + 2 jwj2 jzj2 + jwj2 jzj4 + jzj4 2 jzj2 jwj2 + jwj4 = 2 2 + 1 + jzj2 1 + jwj2 2 (x1 x2 + y1 y2 ) 1 + jwj2 1 + jzj2 + 2 2 = 1 + jzj2 1 + jwj2 jzj2 + jwj2 1 + jwj2 1 + jzj2 2 (x1 x2 + y1 y2 ) 1 + jwj2 1 + jzj2 = 2 2 = 1 + jzj2 1 + jwj2 jzj2 + jwj2 2 (x1 x2 + y1 y2 ) 1 + jwj2 1 + jzj2 = 2 2 = 1 + jzj2 1 + jwj2 jzj2 + jwj2 2 (x1 x2 + y1 y2 ) (1) = 2 2 = 1 + jzj 1 + jwj (1) jz wj2 = : jzj2 + 1 jwj2 + 1 This gives 42 . jzwj d (z. w) = q q : 2 2 jzj + 1 jwj + 1 43 . 4. x 6= 0 Solution I. x > 0 (b) x = 1 (d) y = x + 1 (f) y = 1=x.4.1c z-plane y 4 y 4 y 4 2 2 2 -4 -2 2 4 -4 -2 2 4 -4 -2 2 4 -2 x -2 x -2 x -4 -4 -4 I.4.4.1c w-plane y 4 y 4 y 4 2 2 2 -4 -2 2 4 -4 -2 2 4 -4 -2 2 4 -2 x -2 x -2 x -4 -4 -4 I.1d z-plane I.4.1f z-plane y 4 y 4 y 4 2 2 2 -4 -2 2 4 -4 -2 2 4 -4 -2 2 4 -2 x -2 x -2 x -4 -4 -4 44 .I.1b z-plane I.4.4.1e z-plane I.1 Sketch each curve and its image under w = z 2 .1b w-plane I.4.1a w-plane I. (a) jz 1j = 1 (c) y = 1 (e) y 2 = x2 1.4.1a z-plane I.4. 1d w-plane I.4.I.1e w-plane I.4.1f w-plane y 4 y 4 y 4 2 2 2 -4 -2 2 4 -4 -2 2 4 -4 -2 2 4 -2 x -2 x -2 x -4 -4 -4 45 .4. 2 Sketch the image of eachpcurve in the preceding problem under the principal branch of w = z.4. on the same grid but in apdi¤erent color.4.2a z-plane I. Solution 1=2 n 1=2 p w = z 1=2 = jzj ei argjzj = jzj ei Argjzj+i2 = jzjei Arg z=2 I.4. and also sketch.2a w-plane I.4. the image of each curve under the other branch of z.2c w-plane y 4 y 4 y 4 2 2 2 -4 -2 2 4 -4 -2 2 4 -4 -2 2 4 -2 x -2 x -2 x -4 -4 -4 46 .2b w-plane I.I.4.2c z-plane y 4 y 4 y 4 2 2 2 -4 -2 2 4 -4 -2 2 4 -4 -2 2 4 -2 x -2 x -2 x -4 -4 -4 I.4.2b z-plane I.4. 2f z-plane y 4 y 4 y 4 2 2 2 -4 -2 2 4 -4 -2 2 4 -4 -2 2 4 -2 x -2 x -2 x -4 -4 -4 I.2f w-plane y 4 y 4 y 4 2 2 2 -4 -2 2 4 -4 -2 2 4 -4 -2 2 4 -2 x -2 x -2 x -4 -4 -4 47 .2d w-plane I.4.2e w-plane I.4.4.4.4.I.4.2e z-plane I.2d z-plane I. Circles. (b) We make branch cuts at ( 1.4. centered at the origin of radius r. (Describe what happens to w as z traverses a ray emanating from the origin. Sheet 3 : Take f3 (z) = e4 i=3 g (z). 0]. Sheet 2 : Take f2 (z) = e2 i=3 g (z). considered as a mapping from the z plane to the w. (c) Describe the construction of the Riemann surface of z 1=3 . Top edge of cut on sheet 3 to bottom edge of cut on sheet 1. Top edge of cut on sheet 2 to bottom edge of cut on sheet 3. (c) Top edge of cut on sheet 1 to bottom edge of cut on sheet 2.plane.) (b) Make branch cuts and de…ne explicitly three branches of the inverse mapping. Solution (a) The function w = f (z) = z 3 . The radical rays at angle are mapped to rays at angle 3 . arg w = 3 arg z. are mapped to cocentric circles with radius r3 .I. The endpoints for f (z) is continuous on surface Spara 48 . 1=3 n 1=3 p z = w1=3 = jwj ei arg w = jwj ei Arg w+i2 = ei2 n=3 jwjei Arg w=3 we choose g (z) = z 1=3 = r1=3 ei =3 . The magnitude of a complex number is cubed. that is. Sheet 1 : Take f1 (z) = g (z).3 (a) Give a brief description of the function z 7 ! w = z 3 . jwj = jzj3 . and as z traverses a ray a circle centered at the origin. < < . For z = rei . we have z 3 = r3 ei3 . I.3a ± I.1c ± ± ± ±± 1 1 ± ± ± ± ± ± ±± ± ± ± ± ±±±±±±± ± ± -1 1 -1 1 ±± ±± ± -1 -1 ± Spara 49 .1b I.2.4.2. I. To describe the Riemann surface of a multivalued function. can make branch cuts along horizontal line from 1 + i to i. begin with one sheet for each branch of the function.4. and identify each edge of a cut on one sheet to another appropriate edge so that the function values match up continuously. (b) w = z i. 50 . Solution (a) Use four sheets. make branch cuts so that the branches are de…ned continuously on each sheet. (c) Use …ve sheets. can make branch cuts along real axis from 1 to 1.4 Describe how to construct thep Riemann surfaces for the 2=5 following functions (a) w = z 1=4 . (c) w = (z 1) : Remark. (b) Use two sheets. can make branch cuts along real axis from 1 to 0. 2. 3.2 -1 -0. 2. m = 1. (a) 0 (c) (i 1) =3 (e) i=m. : : : Solution I.1 Calculate and plot for ez for the following points z.1e I.1f 1 0. m = 1.2 (a) e0 = 1: (b) i+1 e = e i e1 = e: (c) p ! (i 1)3 =3 i=3 =3 1 3 e =e e =e + i = 0:351 + 0:006i: 2 2 (d) e37 i = e36 i e i = 1: (e) i=m We take the limit for the sequence e .5.2 -1 1 -0.I.5.2 0. 51 .4 -0.5. 3. : : : as m ! 1. : : : (b) i + 1 (d) 37 i (f) m (i 1) m = 1. 3. 2. and have i=m e ! 1. 52 . Because je e j=e the sequence spiraling to its limit in origo. : : : as m ! 1.as m ! 1. (f) We take the limit for the sequence em(i 1) = e m emi . m mi m as m ! 1. Because e i=m = 1 the sequence approaches its limit along a circle with radius 1. m = 1. 3. and have m mi e e ! 0. 2. 5. 0 < y < =4. (f) the disk jzj 3 =2. (d) the disk jzj =2.5.I. (b) the horizontal strip 5 =3 < Im z < 8 =3.2a w-plane I.2b z-plane I.5.5.5. Solution I.5.2a z-plane I.2c w-plane 4 4 3 2 2 2 -4 -2 2 4 -4 -2 2 4 1 -2 -2 -4 -4 1 2 3 53 . Indicate the images of horizontal and vertical lines in your sketch.2 Sketch each of the following …gures and its image under the expo- nential map w = ez .2b w-plane I.2c z-plane 4 3 8 2 6 2 4 -4 -2 2 4 1 -2 2 -4 -4 -2 0 2 4 1 2 3 I. (c) the rectangle 0 < x < 1. (a) the vertical strip 0 < Re z < 1. (e) the disk jzj .5. .2e z-plane I.2e w-plane I.5.5.2d w-plane I.5.5.2d z-plane I.5..2f z-plane 4 4 4 2 2 2 -4 -2 2 4 -4 -2 2 4 -4 -2 2 4 -2 -2 -2 -4 -4 -4 I.I.2f w-plane 20 100 4 2 10 -4 -2 2 4 -10 10 20 30 100 -2 -10 -4 -20 -100 54 .5. Solution ez = ex iy = ex e iy = ex (cos ( y) + i sin ( y)) = ex (cos y i sin y) = = ex (cos y + i sin y) = ex (cos y + i sin y) = ex eiy = ex+iy = ez : 55 .5.3 Show that ez = ez .I. 4 Show that the only periods of ez are the integral multiples of 2 i. if ez+ = ez for all z. Solution Show that the only periods of ez are the integral muliples of 2 i that is. then is an integer times 2 i.5. that is. ez = ez+ = ez e ) e = 1 ) = 2 mi 56 . then is an integer times 2 i.I. if ez+ = ez for all z. 2. 2. 1. 2. p (a) 2 (b) i (c) 1 + i (d) 1 + i 3 =2 Solution (a) Suppose that n = 0. : : : p 1+i 3 log 2 = p p 1+i 3 1+i 3 = log 2 + i Arg 2 + 2 ni = = i=3 + i2 n: 57 . 1.I. 1. : : : log 2 = log j2j + i Arg 2 + 2 ni = log 2 + i2 n: (b) Suppose that n = 0. Specify the principal value. : : : log (1 + i) = p = log j1 + ij + i Arg (1 + i) + 2 ni = log 2 + i =4 + i2 n = = log 2=2 + i =4 + i2 n: (d) Suppose that n = 0. 2. 1.1 Find and plot log z for the following complex numbers z.6. : : : log i = log jij + i Arg i + 2 ni = i =2 + i2 n: (c) Suppose that n = 0. 6.1a I.1c 20 20 20 10 10 10 -1 1 2 -1 1 2 -1 1 2 -10 -10 -10 -20 -20 -20 I.6.I.1d 20 10 -1 1 2 -10 -20 58 .1b I.6.6. 2 Sketch the image under the map w = Log z of each of the following …gures. bounded by x = log j ej = 12 and x = log je2 j = 2. it goes from 0 to 1. p (d) the slit annulus e < jzj < e2 . the polar angle is from to . Solution (b) We have a disk with radius less then 1.2a z-plane I. the log function is the inverse of the exponential map. (b) the half-disk jzj < 1. (f) the vertical line x = e. Re z > 0. thus log jzj < 0 and is unbounded. (e) the horizontal line y = e. the image p of this annulus under the log function is the rectangle. I. therefore. we have an annulus. thus the argu- ment is in this range.2b z-plane I.6.6.6. (d) Here.2c z-plane 3 4 2 2 2 1 -3 -2 -1 1 2 3 -2 2 -4 -2 2 4 -1 -2 -2 -2 -3 -4 59 . circles in z plane are stright lines in the w plane (w = log z). (c) the unit circle jzjp= 1.I.6. Since Re (z) > 0. e). Since. (a) the right halv-plane Re z > 0. Therefore. the polar angle is between 2 and 2 . this means jzj < 1. z 62 ( e2 . 2f w-plane 4 4 4 2 2 2 -4 -2 2 4 -4 -2 2 4 -4 -2 2 4 -2 -2 -2 -4 -4 -4 60 .6.2c w-plane 3 π 2 π/2 1 -3 -2 -1 1 2 3 -1 −π/2 -2 −π -3 I.2a w-plane I.2d w-plane I.2b w-plane I.2d z-plane I.6.2e z-plane I.6.I.6.6.2f z-plane 8 4 4 6 4 2 2 2 -8 -6 -4 -2 -2 2 4 6 8 -4 -2 2 4 -4 -2 2 4 -4 -2 -2 -6 -8 -4 -4 I.6.6.6.2e w-plane I.6. i1).6.I. Cn [0.6. To avoid the negative imaginary axis we chose =2 < < 3 =2.3 De…ne explicitly a continuous branch of log z in the complex plane slit along the negative imaginary axis. We use the branch f rei = log r + i .3 We have log z = log rei = log r + i . of log z. Solution I. 61 . =2 < < 3 =2. I.4 How would you make a branch cut to de…ne a single-valued branch of the function log (z + i 1)? How about log (z z0 )? Solution Any straight line cut from z0 to 1. in any direction. 62 . will do.6. 1. 2. : : : (1 + i)i = p p = ei log(1+i) = ei(logj1+ij+i Arg(1+i)+i2 m) = ei(log 2+i =4+i2 m) =e 2 m e =4 i log e 2 = [m = n] = p = e2 n e =4 i log e 2 : (b) Suppose that m. p (1 i) (a) (1 + i)i (b) ( i)1+i (c) 2 1=2 (d) 1+i 3 Solution (a) Suppose that m. 2. 2. : : : 63 . 1. n = 0. n = 0.I. : : : 1=2 2 = log 2=2 (logj2j+i Arg 2+i2 m)=2 (log 2+i2 m)=2 log 2=2 i m =e =e =e =e e = p = 1= 2: (d) Suppose that n = 0. 2. : : : ( i)1+i = = e(1+i) log( i) = e(1+i)(logj ij+i Arg( i)+i2 m) = e(1+i)( i =2+i2 m) = 2 m =2 i( =2+2 m) =e e e = [m = n] = e2 n e =2 i( e =2 2 n) = = ie2 n e =2 : (c) Suppose that m = 0. 1. 1.1 Find all values and plot them.7. p 1 i 1+i 3 = p p p i) log(1+i 3) i)(logj1+i 3j+i Arg(1+i 3)+i2 n) = e(1 = e(1 = = e(1 i)(log 2+i =3+i2 n) = e2 n elog 2+ =3 i( e log 2+ =3+2 n) = = e2 n elog 2+ =3 i( e log 2+ =3) : 64 . n = 0. I.2 h i 2i Compute and plot log (1 + i) .7. m. 2.2 20 10 -10 10 20 -10 -20 65 . Solution We rewrite the expression. : : : h i log (1 + i)2i = h i log(1+i) 2i = log e = log e2i log(1+i) = log e2i(logj1+ij+i Arg(1+i)+i2 k) = h p i = log e2i(log 2+i =4+i2 k) = log e 4 k e =2 ei log 2 = [k = m] = log e4 m e =2 i log 2 e = = log e4 m e =2 i log 2 e +i Arg e4 m e =2 i log 2 e +i2 n = 4 m =2+i log 2+i2 n = = =2 + i log 2 + 4 m + 2 in: Thus the set is a square lattice.7.I. suppose that k. 1. 7. 0) too.I.3 Sketch the image of the sector f0 < arg z < =6g under the map w = z a for (a) a = 32 (b) a = i (c) a = i + 2 Use only the principal branch of z a . Solution I.7.7.7.3a z-plane I.3c z-plane 5 5 5 -5 5 -5 5 -5 5 -5 -5 -5 I. and the point (1.7.3a w-plane I. 66 .7.3b w-plane I.3c w-plane 5 2 1 -5 5 -2 2 1 -5 -2 Some points on the y-axis is marked.7.3b z-plane I. 1. Solution Let 2 (zw)a . where on the right we take all possible products. m. 2.7. and k. say = = ea log z ea log w = ea(logjzj+i Arg z+i2 m) a(logjwj+i Arg w+i2 n) e = = ea(logjzj+logjwj+i Arg z+i Arg w+i2 (m+n)) = ea(logjzwj+i Arg(zw)+i2 k+i2 (m+n)) = a(logjzwj+i Arg(zw)+i2 (k+m+n)) =e = = ea log(zw) 2 (zw)a We have that (zw)a = z a wa as sets 67 . n = 0.4 Show that (zw)a = z a wa .I. : : : = = ea log(zw) = ea(logjzwj+i Arg(zw)+i2 n) = ea(logjzj+logjwj+i Arg z+i Arg w+i2 m+i2 n) = a(logjzj+i Arg z+i2 (m+n)) a(logjwj+i Arg w) a a =e e 2z w : Conversely. if 2 z a wa . : : : i ii = = ii i = ei log(i ) = ei(log(e i 2 k =2 )) = ei(logje 2 k =2 j+i Arg(e 2 k =2 )+i2 m) = i( 2 k =2+i2 m) 2 m i( 2 k =2) 2 m =e =e e = ie = = ie2 m : Which not coincide with ii i = i 1 = 1=i = i which is smaller than f ie2 m .7. suppose that k = 0. m = 0. 2. 1. : : : ii = = ei log i = ei(logjij+i Arg(i)+i2 k) = ei(i =2+i2 k) = 2 k =2 =e : i Now we can …nd ii we use that we have ii . Show that is does not coincide with ii i = i 1 . 2. Solution We rewrite the expression ii . suppose that k.I.5 i Find ii . 68 . 2. m = 0. : : :g. 1. 1. 1]. 1]? Solution Phase factors is e2 ia at 0 and e2 ib at 1.7. where n 2 Z . 69 . Require e2 ia e2 ib = 1.6 Determine the phase factors of the function z a (1 z)b at the branch points z = 0 and z = 1. or a + b = n.I. What conditions on a and b guarantee that z a (1 z)b can be de…ned as a (continuous) single-valued function on Cn [0. to have a continuous single–valued determination of z a (1 z)b on Cn [0. its Riemann surface will be a sphere with n 2 1 holes. +1) . and where do they come from?) Solution p p For n = 1 and n = 2. and its Riemann surface will be a sphere with n2 1 holes. the functions f1 p= z x1 and f2 = (z x1 ) (z x2 ). If n = 3 or n = 4. x2 ] [ [x3 . For a general even n. 70 . For a given n. fn (x) will be continuous on Cn ([x1 . +1]). x4 ] . and for a general n we have fn (x) = (z x1 ) (z x2 ) : : : (z xn ). x4 ] [ : : : [ [xn 1 . What is it when n = 3 or n = 4? Can you say anything for general n? (Any compact Riemann surface is topologically a sphere with handles. we also need a slit [xn . x4 ] [ : : : [ [xn . Identify top edge of slit on one sheet with bottom edge of slit on other sheet.I. If n = 1 and n = 2 the surface is topologically a sphere with certain punctures corresponding to the branch points and 1. xn ]). : : :. [x3 . For a general odd n though. x2 ] . fn (x) will be continuous on Cn ([x1 . Show that for n = 1 and n = 2 the surface is topologically a sphere with certain punctures corresponding to the branch points and 1. x2 ] [ [x3 . Describe the Riemann surface of (z x1 ) (z xn ). surface is a torus. To con- stuct the Riemann Suface we use two sheets with slits [x1 .7 Let x1 < x2 < < xn be n consecutive p points on the real axis.7. If n is odd. how many handles are there. Thus a torus is topologically a sphere with one handle. connect 0 to 1 by a straight line. 1. for jzj > 1. Describe the Riemann surface of the function. that is at 0. If jzj > 1. Draw branch cuts so that the function can be de…ned continuous o¤ the branch cuts. Make two branch cuts by connecting any two pairs of point by curves. can use the principal value of the square root to de…ne a branch of the function. that is.7. e2 i=3 and e 2 i=3 .I. 71 .8 p Show that z 2 1=z can be de…ned as a (single-valued) continuous function outside the unit disk. Solution p The function is z 1 1=z 3 . for instance. There are branch points at z = 0 and z 3 = 1. and the other cube roots of unity by a straight line or arc of unit circle. The resulting two-sheeted surface with identi…cation of cuts and with points at in…nity is a torus. 7.9 q Consider the branch of the function z (z 3 1) (z + 1)3 that is posi- tive at z = 2. the second square root can be de…ned to be single-valued for jzj > 1. To what value at z = 2 does this branch return if it is continued continuously once counterclockwise around the circle fjzj = 2g? Solution We have that the function is p p (z + 1) z 2 z (1 1=z 3 ) (1 + 1=z): If jzj > 1.I. Describe the Rie- mann surface of the function. e 2 i=3 and 1. because z does. We can construct a Riemann Surface by making cuts at 1. e2 i=3 . 0. The value of the function at z = 2 returns to thepnegative of their initial value then we travers the circle jzj = 2 . Draw branch cuts so that this branch of the function can be de…ned continuously o¤ the branch cuts. The function for the Riemann Surface is q (z + 1) z (z 1) (z + 1) (z e2 i=3 ) (z e 2 i=3 ): 72 . 1. it is de…ned continuously for jzj > 1.10 q Consider the branch of the function z (z 3 1) (z + 1)3 (z 1) that is positive at z = 2.I.7. To what value at z = 2 does this branch return if it is continued continuously once counterclockwise around the circle fjzj = 2g? Solution We have that the function is q z (1 1=z 3 ) (1 + 1=z)3 (1 4 1=z) The branch that is positive for z = 2 is. The function for the Riemann Surface is q (z 1) (z + 1) z (z + 1) (z e2 i=3 ) (z e 2 i=3 ): 73 . Describe the Riemann surface of the function. Draw branch cut so that this branch of the func- tion can be de…ned continuously o¤ the branch cuts. Branch return to original value around circle jzj = 2. e2 i=3 and e 2 i=3 . 1. We can construct a Riemann Surface by making cuts at 0. which are the cube root of unity. we end up with a one hole torus.7.7. So the phase factor is e2 i=3 . from 1 to e 2 i=3 .11b I. Solution I. The Riemann surface is obtained ty pasting three sheets with the corresponding branch cuts.7. Make two cuts. In this case the cuts share common endpoint.11c 1 1 1 -1 1 -1 1 -1 1 -1 -1 -1 We rewrite as follows p q 3 3 z3 1= (z 1) (z e2 i=3 ) (z e 2 i=3 ): This equation will have 3 branch points. We nee three shets where f0 (z). on each sheet.11 p Find the branch points of 3 z 3 1 and describe the Riemann surface of the function. f1 (z) = e2 i=3 f0 (z) and f2 (z) = e 2 i=3 f0 (z) Note: Can use Riemann–formula to see that surface is a torus. 74 .7. Cheek by going around each little tip what phase change to. which of for your sheet.11a I.I. (b) sin (z + w) = sin z cos w + cos z sin w. (d) sinh (z + w) = sinh z cosh w + cosh z sinh w.I. sin (z + w) = cos z + w = 2 = cos z cos w sin z sin w = 2 2 = cos z sin w + sin z cos w = sin z cos w + cos z sin w: (c) Using the addition formula (a) and the formulas cos (iz) = cosh z and sin (iz) = i sinh z given on page 30 in CA we have. Solution (a) Using the de…nitions of sine and cosine functions given on page 29 in CA we have. and using that sin z = cos (z =2) and cos z = sin (z =2) we have.8. (c) cosh (z + w) = cosh z sinh w + sinh z sinh w.1 Establish the following addition formulae (a) cos (z + w) = cos z cos w sin z sin w. cos (z + w) = ei(z+w) + e i(z+w) 1 = = 2ei(z+w) + 2e i(z+w) = 2 4 1 i(z+w) i(z w) = e +e + e i(z w) + e i(z+w) + 4 + ei(z+w) ei(z w) e i(z w) + e i(z+w) = eiz + e iz eiw + e iw eiz e iz eiw e iw = = 2 2 2i 2i = cos z cos w sin z sin w: (b) Using the addition formula (a). 75 . cosh (z + w) = cos (i (z + w)) = cos (iz) cos (iw) sin (iz) sin (iw) = = cosh z cosh w i sinh z i sinh w = cosh z cosh w + sinh z sinh w: (d) Using the addition formula (b) and the formulas cos (iz) = cosh z and sin (iz) = i sinh z given on page 30 in CA we have. sinh (z + w) = i sin (i (z + w)) = i [sin (iz) cos (iw) + cos (iz) sin (iw)] = = i [i sinh z cosh w + cosh z i sinh w] = sinh z cosh w + cosh z sinh w: 76 . Use trigonometric formulas from page 29 and 30 in CA we have.I. Solution. cos z = 0 . sinh y = 0 . cos z = cos (x + iy) = cos x cos (iy) sin x sin (iy) = cos x cosh y i sin x sinh y: Now take the modulus squared. 77 . jcos zj2 = cos2 x cosh2 y + sin2 x sinh2 y = = cos2 x 1 + sinh2 y + sin2 x sinh2 y = = cos2 x + cos2 x + sin2 x sinh2 y = cos2 x + sinh2 y: The identity for jcos zj2 shows that the only zeros of cos z are the zeros of cos z on the real axis. the only periods of cos z are 2 n. 1. : : : . x = 2 + m . because cos x = 0 . Thus any period is an integral multiple of .8. 1 < n < 1. where z = x + iy. m = 0. 2. and use cosh2 y = 1 + sinh2 y. and since odd integral multiples are not periods.2 Show that jcos zj2 = cos2 x + sinh2 y. y = 0: Translation by any period of cos z sends zeros to zeros. Find all zeros and periods of cos z. 3 Find all zeros and periods of cosh z and sinh z. 1. 2. : : :. : : :. 2. Solution We have that cosh z = cos (iz). 2. m = 0. and the periods of cosh z are 2 mi. 78 . : : :.I. thus the zeros of sinh z are at z = m i. m = 0. 1. and the periods of cosh z are 2 mi. 1. 2. m = 0. 1.8. : : :. thus the zeros of cosh z are at z = i =2 +im . We have that sinh z = i sin (iz). m = 0. 4 Show that 1 1 1 + iz tan z= log . 2i 1 iz where both sides of the identity are to be interpreted as subsets of the complex plane. Solution Set z = tan w. we have that sin w eiw e iw e2iw 1 z = tan w = = = : cos w i (eiw + e iw ) i (e2iw + 1) Solve for e2iw 1 + iz e2iw = . show that tan w = z if and only if 2iw is one of the values of the logarithm featured on the right. In other words.I.8. 1 iz and take logarithm and solve for w 1 1 + iz w= log : 2i 1 iz 1 Because z = tan w then w 2 tan z and we have the identity 1 1 1 + iz tan z= log : 2i 1 iz 79 . the other from i to i1. Set 1 + iz = w1 1 iz and solve for z.5 z-plane I. 0] if and only if z 2 S. Show that the principal branch 1 1 1 + iz Tan z= Log 2i 1 iz maps the slit plane CnS one-to-one onto the vertical strip fjRe wj < =2g.5 w-plane −π/2 π/2 w = 2i1 w2 = 2i1 Log 11+iz iz We begin to show that 1+iz1 iz 2 ( 1.8. Solution I.8.5 I. i1).8.5 π i -1 -i −π 1+iz w1 = 1 iz w2 = Log w1 I. i] [ [i. thus 80 .I.8. one running from i to +i1.5 Let S denote the two slits along the imaginary axis in the complex plane.8. 0] if and only if z 2 ( i1. Show that (1 + iz) = (1 iz) lies on the negative real axis ( 1. Since we have that 1 + iz 1 w1 = w1 ) z = i 1 iz 1 + w1 the map is one to one. Remark that the map makes correspondence between the interval [ i: i1] in z plane and [ 1. se …gures. 1+w1 i goes from i1 to i along iR. 1 < n < 1. 81 . 1 w1 z= i 1 + w1 1 w1 As w1 goes from 1 to 1. The other branches are given by fn (z) = Tan 1 z + n . We have that w2 = Log w1 maps Cn( 1. 1] in w1 plane. i1] in z plane and [ 1. and between interval [i. 1+w 1 i goes from i to i1 along iR. 0] in w1 plane. thus w = w2i2 maps jIm w2 j < onto jRe wj < 2 . 0) onto fjIm wj < g. and when 1 w1 w1 goes from 1 to 0. where Tan 1 z is the principal branch for tan 1 z. We have that the function 1 1 1 + iz Tan z= Log 2i 1 iz maps CnS onto jRe wj < 2 . 6 w-plane ___________ +++++++++++ −π/2 π/2 1 1 1+iz w= w 2i 2 = 2i Log 1 iz We have that 1 1 1 + iz f0 (z) = Tan z= Log . so that fn (z) and fn+1 (z) have same value at the junction. and de…ne fn (z) = Tan 1 z + n on the nth sheet to the (n + 1) th sheet along one of the cuts. 2i 1 iz other branches of tan 1 z are fn (z) = f0 (z) + n . Use one copy of the double slit plane S for each integer n. Solution I.5 +++++++++++ ++++ π ____ i ++++++ _ _ _ _-1 __ -i ++++ ____ _ _ _ _ −π _______ 1+iz w1 = 1 iz w2 = Log w1 I.8.8.6 I.I.8.6 1 Describe the Riemann surface for tan z. 82 . where 1 < n < 1.8.8.6 z-plane I. Then fn on Sn continuous to fn+1 on Sn+1 . 1 < n < 1.Make in…nite many copies of S and call them Sn . and fn maps Sn onto vertical strip n 21 < Re z < n + 12 . and its image omits the sequence 12 + n. Note that composite function is not de…ned at i and 1 (endpoints of slits). 83 . De…ne fn (z) = Tan 1 z+n on Sn . Identify "+" side of cut on Sn to " " side of cut on Sn+1 . Show that h p i 1 cos w = i log w w2 1 .8. and set = eiz and take the logarithm and solve for z p z= i log w w2 1 : 1 Because w = cos z then z 2 cos w we have the identity 1 p cos w= i log w w2 1 : 84 .7 p Set w = cos z and = eiz . where both sides of the identity are to be interpreted as subsets of the complex plane. Solution Set w = cos z and = eiz . we have that eiz + e iz + 1= w = cos z = = : 2 2 Solve for p =w w2 1.I. Show that = w w2 1. the complex number whose sine is z. that is. 1 Let now z = sin w. because Im (sin w) = 0 ) y = 0 ) sin w = sin (x) and 1 < sin (x) < 1 for these x. +1) on the real axis. Show p that the inverse function is the branch of sin 1 z = i Log iz + 1 z 2 obtained by taking the principal value of the square root.I. by formulas on page 30 in CA we have that sin w = sin (x + iy) = sin (x) cos (iy) + cos (x) sin (iy) = = sin x cosh (y) + i cos (x) sinh y: If jRe wj = jxj < =2 then sin w is mapped on Cn ( 1. Note that eiw e iw z= : (1) 2i Now with = eiw and 1= = e iw in (1). so that the principal value of the square root is de…ned and continuous on the slit plane. 1] and [+1.8 Show that the vertical strip jRe (w)j < =2 is mapped by the function z (w) = sin w one-to-one onto the complex z plane with two slits ( 1. Solution Set w = x + iy. with argument in the open interval between =2 and =2. we …nd that 85 . we have 1= z= : 2i Mulitplying the above by 2i and doing some rearranging. The value of w is referred to as sin z. Hint. 1] [ [+1. First show that the function 1 z 2 on the slit plane omits the negative real axis. +1).8. 2 2 2iz = 1 or 2iz 1 = 0: With the quadratic formula. 1 p sin z= i log iz 1 z2 : 86 . we solve this equation for and …nd p p = iz + 1 z2 or eiw = iz + 1 z2 We now take the logarithm of both sides of the last equation and divide the result by i to obtain 1 p w= log iz + 1 z2 i 1 and. since w 2 sin z. II 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 X X X X X X X X X X X X X X X X 2 X X X X X X 3 X X X X X 4 X X X X X 5 X X X X X X X X 6 7 8 1 . 1. 2 .II. p > 1 n!1 n!1 nn +3n+1 (b) lim 2n = 0 (d) lim z = 0.1 Establish the following: n 2np +5n+1 (a) lim n+1 = 1 (c) lim p = 2. z 2 C: n!1 n +1 n!1 n! Solution (a) n 1 lim = lim = 1: n!1 n + 1 n!1 1 + 1=n (b) n 1=n lim = lim = 0: n!1 n2 + 1 n!1 1 + 1=n2 (c) 2np + 5n + 1 2 + 5n1 p +n p lim = lim = 2: n!1 np + 3n + 1 n!1 1 + 3n1 p+n p (d) Let m be a positive integer so that m > 2 jzj and let n > m. it follows by instängning that n! ! 0 as n ! 1 as was to be shown. then zn jzjm jzjn m 0 = = n! m! (m + 1) (m + 2) n jzjm jzj jzj jzj jzjm jzj jzj jzj = m! m + 1 m + 2 n m! m m m m m jzj 1 1 1 jzj 1 = !0 m! 2 2 2 m! 2 mn zn as n ! 1. 1.II. 3 .2 For which values of z is the sequence fz n g1 n=1 bounded? For which values of z does the sequence converge to 0? Solution The sequence is bounded for jzj 1. and the series converge to 0 for jzj < 1. 1.II. Solution If we choose 2 n> jzj then 2 < n jzj and jnn z n j = jn zjn > 2n ! 1 when n ! 1 for z 6= 0.3 Show that fnn z n g converges only for z = 0. 4 . II. N !1 N (N k)! Solution We have that N! N (N 1) : : : (N k + 1) 1 2 k 1 = = 1 1 ::: 1 !1 Nk (N k)! N N ::: N N N N as N ! 1 (k is …xed). k 0.4 N! Show that lim k = 1. 5 .1. 2 3 n is decreasing. n 1.II. n n t n thus an+1 > an . The limit of the sequence is called Euler’s constant. so the sequence an is increasing. We take the di¤erence bn+1 bn and by (1) and (2) we have that 6 . Remark. Show that the sequences both converge to the same limit .5 Show that the sequence 1 1 1 bn = 1 ++ + + log n.1. Solution We have that 1 1 1 1 bn = 1 + + + + + log n. while the sequence an = bn 1=n is increasing. It is not known whether Euler’s constant is a rational number or an irrational number. Show that 1 2 < < 35 . 2 3 n 1 n 1 1 1 an = 1+ + + + log n: 2 3 n 1 We trivially have that Z n 1 log n = dt ((1)) 1 t An elementary estimation gives Zk+1 1 1 1 < dt < ((2)) k+1 t k k We take the di¤erence an+1 an and by (1) and (2) we have that Z n+1 1 1 dt an+1 an = log (n + 1) + log n = > 0. if n 22. We have after some investigation on the decreasing sequence bn that 19 093 197 3 bn b22 = log 22 < . We have after some investigation on the increasing sequence an that 49 1 an a7 = log 7 > . n+1 n+1 t n thus bn+1 < bn . n 22 5173 168 5 thus bn 3=5. And this limit is in the interval 12 < < 53 . 7 . Z n+1 1 1 dt bn+1 bn = log (n + 1) + log n = < 0. if n 7. n 7. 20 2 thus 21 < an . so the sequence bn is decreasing. The both sequences converge to the same limit because bn an = 1=n have the limit 0 as n ! 1. then > for all n 0. n (b) ! 0 if and only if Re > 1. (Im )2 If Re = 1.6 For a complex number . we de…ne the binomial coe¢ cient " choose n" by ( 1) ( n + 1) = 1.1. we have that tk = k2 and n ! A then n ! 1. 2. = . : : :. n 1: 0 n n! Show the following (a) The sequence is bounded if and only if Re 1. n+1 n (e) If Re a > 1 and is not an integer. n (c) If 6= 0. then < n+1 n for n large Solution (a) We have that s Y n +1 Y 1 2 (1 + Re ) (1 + Re )2 (1 + Im )2 = 1 = = 1 + + : n k k k2 k2 k=1 k=1 Let (1 + Re )2 (1 + Im )2 2 (1 + Re ) tn = + k2 k2 k P If Re < 1. 8 . then ! 1. 6= 1.II. and +1 P P (Im )2 A 6= 0 becauxe for all k we have that 1 k > 1 and tk = k2 <1 so n is limited. 1. n+1 n (d) If Re 1. we have that tk > 0 and n ! 1 then n ! 1 because tk diverge is n unlimited. : : :. 1. Then we hav that M = lim n = n!1 n!1 n lim = M so M = M () M = 0 so ! 0. From (a) we see that the only possibility for this is Re > 1.n) e) Let be …xed and Re > 1. Then q j(n+1)j n)2 +(Im )2 j j2 1 = (Re (n+1)2 <1 for n > : j(n)j 2(1+Re ) Parts (a) and (b) seems to require some facts about products: Q n 1+ k ! 1 as n ! 1. d) j(n+1)j j nj distance( . 6= 1 j(n)j n+1 distance( 1. then n 6= 0 and we can divide. For Re > 1 we can see that n 0 is decreasing. Let M = lim n . >0 k=1 Qn 1 k2 ! A 6= 0 as n ! 1 k=1 9 . then n ! 1 if n!1 n n+1 n and only if Re > 1.If Re > 1 we have that it exists k 2 N so that 1 < tk < 0 for all k > K. >0 k=1 Qn 1 k ! 1 as n ! 1. 2. and so limited. b) We have that n ! 0 if and only if n ! 0. if Re 1.n) = = > 1. The conclution is that n is limited if and only if Re 1. c) If 6= 0. to get ( 1):::( (n+1)+1) n+1 (n+1)! n 1 = ( 1):::( n+1) = = n 1 ! 1 n n+1 1+ n n! as n ! 1. This gives that n is a decreasing series with respect to n. that is.7 De…ne x0 = 0.it is straightforward to show that x x2 + 4 2 . Since 1 0 xk 2 and xk+1 = x2k + 14 we have 2 1 1 1 1 0 xk xk+1 = x2k + + = 4 2 4 2 1 and hence 0 xk+1 2 . we have x0 = 0 and x1 = x20 + 4 = 4 and so 1 1 0 x0 and x0 = 0 = x1 : 2 4 Hence. and 1 xk+1 x2k+1 + = xk+2 : 4 Hence. It follows by Bounded Monotone Sequence Theorem that it converges. Now suppose that the assertion holds for n = k. Solution We …rst prove by induction that for every integer n 0 we have 1 0 xn and xn xn+1 : 2 1 1 For n = 0. Note that the real-valued function f given by f (t) = t2 + 41 for t 2 R is continuous and therefore we have. 10 .II. and that any limit satis…es x = x2 + 41 . xn ! 1 2 as required. 1 x = lim xn+1 = lim f (xn ) = f (x) = x2 + : n!1 n!1 4 1 It follows that x = x2 + 4 and so x = 12 . Hint. say xn ! x. Show that the sequence is bounded and monotone. Hence.1. 1 0 xk and xk xk+1 : 2 1 1 1 For 0 x 2 . Show that xn ! 12 . and de…ne by induction xn+1 = x2n + 14 for n 0. the assertion holds for n = 0. the inductive step holds and so the sequence fxn gn is bounded and increasing. then jsn sn 1 j ! 0. Solution We use standard "=2 proof and let sn ! s. then we have jsn+1 sn j = jsn+1 s+s sn j jsn+1 sj + jsn sj < "=2 + "=2 = ": Thus jsn+1 sn j ! 0 as n ! 1. 11 . choose N such that jsn sj < "=2 for n > N .II. Let " > 0.1.8 Show that if sn ! s. If n > N . : 0. : +1.9 Plot each sequence and determine its lim inf and lim sup.II. lim inf sn = 0. > > < 1. jxj = 1.1. x = 1. > > > > 1. x > 1: 12 . x = 1. jxj > 1. jxj < 1. jxj < 1: and 8 > > 1. lim sup sn = 1. (a) sn = 1 + n1 + ( 1)n (c) sn = sin ( n=4) n (b) sn = ( n) (d) sn = xn (x 2 R …xed) Solution (a) lim sup sn = 2 lim inf sn = 0 (b) lim sup sn = 1 lim inf sn = 1 (c) lim sup sn = 1 lim inf sn = 1 (d) 8 < +1. 1<x< 1. If we de…ne the function to be 0 at z = 0. (a) z (c) z 2 = jzj (b) z= jzj (d) z 2 = jzj3 Solution (a) continuous everywhere (b) Continuous except at z = 0.10 At what points are the following function continuous? Justify your answer.II. it is contionuous there. It is not de…ned at z = 0. where it is not de…ned (c) Continuous for jzj = 6 0. but it has a limit 0 as z ! 0.1. It has no limit as z ! 0 13 . (d) Contionuous for jzj = 6 0. ] as z ! 0. the function may take any value in intervall [ . 0]. 0]. 0]. It is continuous in Cn ( 1.11 At what points does the function Arg z have a limit? Where is Arg z continuous? Justify your answer. 0]. Values of function ! f g at points of ( 1.II. 14 . It is discontinuous at each point of ( 1.1. The value is depending in that direktion we approach z = 0. Solution Arg z have a limit at each point of Cn ( 1. II. At what points does h (z) have a limit? What is the limit? Solution No limit at 0. has limit at all other points of closed lower half-plane. 15 . 0).12 Let h (z) be the restriction of the function Arg z to the lower half- plane fIm z < 0g. limit = at points of ( 1.1. rRe ! 1 as r ! 0 thus jz j have no limit as z ! 0. then Re +i Im z = eLog r+i = eLog r Re +i Log r Im +i Re Im = = eLog r Re e Im ei(Log r Im + Re ) = rRe e Im ei(Log r Im + Re ) . and z ! 1 if = 0. thus we have limit if = 0. The conclution is z ! 0 if Re > 0.II. If Re > 0. 16 . and not z either. thus jz j = rRe e Im : Because e Im is bounded we look at rRe in three intervals If Re < 0. have no limit unless Im = 0.13 For which complex values of does the principal value of z have a limit as z tends to 0? Justify your answer.1. we get z = e Im . If Re = 0. otherwise we have no limit at 0. rRe ! 0 as r ! 0 thus jz j have limit 0 as z ! 0. and z ! 0 as z ! 0. Solution Set z = rei = eLog r ei = eLog r+i and = Re + i Im . 17 . R1 Then H(z) = h(t) t z dz is continuous for z 2 Cn [0. Then H (z) is also de…ned for z = z1 . 1].1. If h (z1 ) = 0 for some z1 2 [0. If zn ! z 2 Cn [0. Use the fact if jf (t) g (t)j < " for 0 t 1. then 0 jf (t) g (t)j dt < ". 1]. 0 then th(t) zn ! h(t) t z uniformly for 0 t 1. so the used proof shows that H (zn ) ! H (z). Solution We have that H (z) is de…ned for z 2 Cn [0.II. 1].14 Let h (t) be a continuous complex-valued function on the unit in- terval [0. and consider Z 1 h (t) H (z) = dt: 0 t z Where is H (z) de…ned? Where is H (z) continuous? Justify your answer. 1]. R1 Hint. 1]. 1) . 1] . (b) The exterior of the open unitdisk in the plane. Cn f0g . fjzj > 1g . fjzj < 1. (c) The exterior of the closed unitdisk in the plane. Im (z) 0g . 18 . C. Cn (0. (a) The punctured plane. (g) The complex plane.15 Which of the following sets are open subsets of C? Which are close? Sketch the sets. fjzj 1g . Cn [0. (f) The semidisk. Solution (a) open (c) open (e) open (g) both open and closed (b) closed (d) neither (f) neither Note: Sketches are di¢ cult to do reasonably. (e) The plane with the closed unitinterval removed.1. (d) The plane with the open unitinterval removed.II. II. 0] is not convex. To show that Cn ( 1. 1] is not star-shaped.16 Show that the slit plane Cn ( 1. For example take 1 + i and 1 i. (b) Cn [ 1. These two points are joined by a vertical line which contain 1. 0] is not convex we may choose any two points such that the stright line segment joining them contains a point in ( 1. We also have that if z belong to the domain then z belongs also to the domain. (c) Same argument as in (b) shows that Cn f0g is not star–shaped. Show that a punctured disk is not star-shaped. 1] is not star–shaped. 0].1. 19 . because we can see every point from z = 1. Show that the slit plane Cn [ 1. From this follows that Cn ( 1. Solution (a) Cn ( 1. 0] is star-shaped but not convex. 0] is star–shaped. Given any z we have that z can not bee seen from a stright line between z and z that must contain 0. Then since D is star-shaped with respect to z0 . We must show that D is star-shaped with respect to each of its points. z1 2 D be given. Let z 2 D be given. Suppose …rst that D is convex. D is star-shaped with respect to each of its points. Hence. We must show that for every point z 2 D the line segment connecting z0 to z is contained in D. To show that D is convex we must show that given two points in D. Solution (A. then since D is convex it contains the line segment joining z0 and z. 20 . Hence.II. suppose that D is star-shaped with respect to each of its points. D is star-shaped with respect to z0 . So …x z0 2 D. Kumjian) Let D C. So let z0 . since the empty set trivially satis…es both conditions we will assume that D 6= ?. D contains the line segment joining z0 and z1 . since z0 was chosen arbitrarily. Conversely. the line segment joining them is also contained in D.17 Show that a set is convex if and only if it is star-shaped with respect to each of its points. D is convex. to show that D is star-shaped with respect to z0 .1. then. V = fh 6= 0g. Solution Show that the following equivalent for an open subset U of the complex plane. (a) Any two points of U can be joined by a path consisting of straight linesegments parallel to the coordinate axis. (b) ) (c) Suppose rh = 0. some D 6= ?: 21 . then can connect to U . intervals parallel to coordinate axis. W [ V = U (c) ) (a) Suppose z0 2 U . U nD must be empty. In the context of topological spaces. so fh = 0g is open and W is open. ) h is constant in D. join by polygonal curve.1. Fix z0 2 D. Evidently V is open. If z1 2 D. (a) ) (b) Suppose rh = 0 in D. W \ V = ?. so U nD is open. Since rh = 0. this latter property is taken as de…nition of connectedness. (b) Any continuously di¤erentiable function h (x. Remark.II. By (c). h not constant. Also if can connected points near z0 2 U . (c) If V and W are disjoint open subsets of U such that U = V [ W .18 Show that the following are equivalent for an open subset U of the complex plane. y) on U such that rh = 0 is constant. Let D = points that can be joined to z0 by a polygonal curves. Let W = fh = 0g. rh = 0 ) h is constant on each segment of curve ) h (z0 ) = h (z1 ). say h (z0 ) = 0 for some z0 . then either U = V or U = W . h is constant in a neighborhood of each point. V 6= ?. Evidently D is open. W 6= ?. say a0 = 1. where m 1 and a 6= 0. and that p (z) = 1 + az m + . Thus jp (z)j attains its minimum in jzj R. Show that if p (z) is a nonconstant polynomial. a 6= 0. where we assume an 6= 0. We may assume it is attained at z0 = 0. 22 . Choose z = " a m (any m th root does the job).19 Give a proof of the fundamental theorem of algebra along the fol- lowing lines. Then p (z) = 1 "m + O "m+1 : Thus for " su¢ ciently small " we have jp (z)j < 1. Suppose a0 6= 0. which is a contra- diction. then jp (z)j attains its minimum at some point z0 2 C. Assume that the minimum is attained at z0 = 0. Solution (K. Seip) Set p (z) = an z n + an 1 z n 1 + + a0 . Since jp (z)j lim n = jan j .II. Then p (z) = 1 + az m + higher order 1 1 terms. Contradict the minimality by showing that P "ei 0 < 1 for an appropriate choice of 0 . jzj!1 jzj 9 R > 0 such that jp (z)j > ja0 j for all jzj > R.1. Conside zm = z0 + "e i 0 =m e i =m . Contradiction! We conclude that p (z0 ) = 0. Then jz0 j < R. Suppose p (z0 ) 6= 0. Suppose am = r0 ei 0 . " > 0: We have p (zm ) = 1 + r0 ei 0 em e i 0 e i + O "m+1 = 1 r0 "m + O "m+1 : Have p (zm ) 1 r0 "m < 1 for " > 0 small. Let z0 be a point in the disk jzj 6 R at which the continuous function for jp (z)j attains a minimum. aN 6= 0: Choose R so that jaN j RN > ja0 j + ja1 j R + jam+1 j Rm+1 + ::: + jaN 1j R N 1 : Then p (z) 6= 0 for jzj = R.Follow the suggestion. We can assume that p (z0 ) = 1. Write p (z) = 1 + am P (z z0 )m + O (z z0 )m+1 . where am 6= 0. Let p (z) = a0 + a1 z + am+1 z m+1 + ::: + aN z N . 23 . so that is a disk centred at z0 so that jp (z)j > jp (z0 )j on the disk. the term aN z N dominates and further jp (z)j > jp (0)j for jzj = R. 1 Find the derivatives of the following function.2. n (a) z 2 1 (c) (z 2 1) (e) 1= (z 2 + 3) (g) (az + b) = (cz + d) (b) z n 1 (d) 1= (1 z) (f) z= (z 3 5) (h) 1= (cz + d)2 Solution n 1 2 (a) 2z (c) n (z 2 1) 2z (e) 2z= (z 2 + 3) (g) (ad bc) = (cz + d)2 2 (b) nz n 1 (d) 1= (1 z)2 (f) ( 2z 3 5) = (z 3 5) (h) 2c= (cz + d)3 24 .II. 2.2 Show that 1 zn nz n 1 + 2z + 3z 2 + + nz n 1 = : (1 z)2 1 z Solution Use the geometric sum 1 z n+1 1 + z + z2 + z3 + + zn = . z 6= 1: 1 z Di¤erentiate both sides 1 + 2z + 3z 2 + + nz n 1 = (n + 1) ( 1) z n (1 z) + (1 z n+1 ) = = (1 z)2 nz n+1 nz n + z n+1 z n + 1 z n+1 = = (1 z)2 1 zn nz n+1 nz n 1 zn nz n = + = : (1 z)2 (1 z)2 (1 z)2 (1 z) 25 .II. if z= x = lim = : z!0 z i. 26 .II. if z=i y The functions x = Re z and y = Im z are not complex di¤erentiable at any point.2. Solution Di¤erentiation f (z) = x = Re z (from the de…nition) f (z + z) f (z) Re (z + z) Re (z) lim = lim = z!0 z z!0 z Re z 1. if z= x = lim = : z!0 z 0.3 Show from the de…nition that the functions x = Re z and y = Im z are not complex di¤erentiable at any point. if z=i y Di¤erentiation g (z) = y = Im z (from the de…nition) g (z + z) g (z) Im (z + z) Im (z) lim = lim = z!0 z z!0 z Im z 0. because the functions have di¤erent limit as z ! 0 through real and imaginary axis. otherwise is the function not analytic on any open set. Where is f (z) analytic? Solution d We will …nd dz f (z) using the limit de…nition f (z + z) f (z) = z 2 a (z + z)2 + b (z + z) (z + z) + c(z + z) (az 2 + bz z + cz 2 ) = = z 2 a z 2 +2z z + z2 + b z z + z z+z z + z z + c z 2 + 2z z + z (az 2 + bz z + cz 2 ) = z 2 2 2az z + a z + bz z + bz z + b z z + 2cz z + c z = = z 2 z z z = 2az + a z + bz + bz + b z + 2cz +c = z z z z z = 2az + a z + bz + b z + c z + (bz + 2cz) z z z We know z is not analytic at any open set of C and lim z does not ex- 4z!0 d ist.2.2) Suppose f (z) = az 2 + bz z + cz 2 . Therefore f (z) is analytic on C if b = c = 0. unless bz + 2cz = 0.4 (+ IV. 27 . and c are …xed complex numbers. By di¤erentiating f (z) by hand.8. show that f (z) is com- plex di¤erentiable at z if and only if bz + 2cz = 0.II. where a. the derivative dz f (z) would not exist. b. Therefore. 1 1 1 lim (g (z + h) g (z)) = lim f z+h f (z) = lim (f (z + k) f (z)) = f 0 (z) h!0 h h!0 h k!0 k where equation ( ) follows using the substitution k = h and basic algebraic properties of conjugation.5 Show that if f is analytic on D. We …rst show that g is di¤erentiable at z and that g 0 (z) = f 0 (z). Further. g 0 is the composite of continuous functions and therefore is itself continuous. Solution (A. since g 0 (z) = f 0 (z) for all z 2 D . 28 .2. Since f is analytic on the domain D. g is analytic on D . For a complex-valued function ' de…ned near a point z0 it is easy to show that lim ' (z) exists i¤ lim ' (z) exists. then g (z) = f (z) is analytic on the re‡ected domain D = fz : z 2 Dg. then the two limits are complex conjugates of each other. note that conjugation is a continuous function so h ! 0 i¤ h ! 0. and g 0 (z) = f 0 (z).II. its derivative f 0 is continuous on D. Hence. z!z0 z!z0 and if either exists. Let z 2 D be given. Kumjian) Let f be analytic on the domain D and de…ne g on D as above (note that D is also a domain). 1]. that is. 29 .II. use the de…ning identity (2. and de…ne Z 1 h (t) H (z) = dt. Hint. z 2 Cn [0.4) of com- plex derivative. 1] : 0 t z Show that H (z) is analytic and compute its derivative. Dif- ferentiate by hand.4) of derivative to …nd H 0 (z).6 Let h (t) be a continuous complex-valued function on the unit in- terval [0. Solution We use de…nition (2.2. also by uniform convergence of integrand H (z) is analytic for z 2 Cn [0. H 0 (z) = Z 1 Z 1 H (z + z) H (z) 1 h (t) h (t) = lim = lim dt dt = z!0 z z!0 z 0 t z z 0 t z Z 1 1 1 1 = lim h (t) = z!0 z 0 t z z t z Z 1 Z 1 h (t) h (t) = lim = dt = z!0 0 (t z z) (t z) 0 (t z) (t z) Z 1 h (t) = dt: 0 (t z)2 H 0 (z) is continuous in z. 1]. II.3.1 Find the derivatives of the following functions. sin z sinh z (a) tan z = cos z (b) tanh z = cosh z (c) sec z = 1= cos z Solution (a) d d sin z cos z cos z sin z ( sin z) cos2 z + sin2 z 1 tan z = = 2 = 2 = : dz dz cos z cos z cos z cos2 z (b) d d sinh z cosh z cosh z sinh z sinh z cosh2 z sinh2 z 1 tanh z = = 2 = 2 = : dz dz cosh z cosh z cosh z cosh2 z (c) d d 1 1 sin z sec z = = ( sin z) = = tan z sec z: dz dz cos z cos2 z cos2 z 30 . 2 Show that u = sin x sinh y and v = cos x cosh y satisfy the Cauchy- Riemann equations. f (z) = eix e ix ey e y eix + e ix ey + e y = u + iv = +i = 2i 2 2 2 eix+y eix y e ix+y + e ix iy eix+y + eix y + e ix+y + e ix y = i +i = 4 4 ix y ix+y i(x+iy) i(x+iy) e +e e +e e + e iz iz =i =i =i = 2 2 2 = i cos z: 31 .II. and now we calculate f (z). Do you recognize the analytic function f = u + iv? (Determine its complex form) Solution We have @u @ @ @v = sin x sinh y = cos x sinh y = cos x cosh y = . @x @x @y @y @u @ @ @v = sin x sinh y = sin x cosh y = cos x cosh y = : @y @y @x dx Hence.3. u and v satisfy Cauchy-Riemann equations. it follows by the theorem on page 50. then f is constant. Then both Re f = 21 f + f and Im f = 2i1 f f must also be analytic on D. Solution (A. that both are constant. Hence.3 1 2 3 P L K LLL Show that if f and f are both analytic on a domain D. Since Re f and Im f are also both real-valued. 32 . f = Re f + i Im f is constant. Kumjian) Let D be a domain and let f be a complex valued function such that both f and f are analytic on D.II.3. 33 . Solution (with use of hint) If f (z) = 0 for some z 2 D.II.4 Show that if f is analytic on a domain D. since jf j is constant. Then f 0 on D. Write f = jf j2 =f . then f is constant. Hint. Assume that f (z) 6= 0 for every z 2 D. and if jf j is constant. So f = Re f + i Im f is constant.3. From this follows that Re f and Im f are analytic and real-valued and therefore constant. Since f is analytic in D we have that 1=f and f = jf j2 =f (where jf j is constant) are analytic. and if jf j is constant. thus by (4) f is constant. 0 0 (2) uuy + vux = 0: In the simultaneusly systems of equations in (2) we …rst …rst multiply the …rst row with u and second row with v in (2) and add them together. Now suppose that jf j = k is constant. And if we in the same system multiply the …rst row 0 with v and the second row with u and add them we have (u2 + v 2 ) uy = 0. Solution Set f = u + iv. Hint. and we have that u = v = 0 and thus f is constant.4 Show that if f is analytic on a domain D. 34 . 0 0 (4) 0 = uy = v y . then f is constant. Write f = jf j2 =f . 0 0 Suppose that u2 + v 2 6= 0.3. where u and v are realvalued functions.II. We have that (3) gives that ux = uy = 0. and 0 have (u2 + v 2 ) ux = 0. Di¤erentiate both sides with respect to x respectively y 0 0 uux + vvx = 0. 0 0 (1) uuy + vvy = 0: 0 0 0 0 We use Cauchy-Riemanns equations ux = vy and uy = vx in (1) and have 0 0 uux vuy = 0. We have that u2 + v 2 = k 2 . 0 (3) (u2 + v 2 ) uy = 0: If in (3) u2 + v 2 = 0. Thus 0 (u2 + v 2 ) ux = 0. with says that u is constant. Cauchy-Riemanns equations gives 0 0 0 = ux = v y . @x @y @x @x @y @x from which it follows that jruj = jf 0 (z)j = jrvj : 35 . Then @u @u @u @v @v @v ru = .5 If f = u + iv is analytic. then jruj = jrvj = jf 0 j. = . Seip) Because f = u + iv is analytic. Solution (K.3. = .II. ru and rv are orthogonal. Kumjian) Let 2 R then rotation of a vector (x. y) 2 R2 by the angle (about the origin) is given by matrix multiplication: x x cos sin x x cos y sin 7! A = = : y y sin cos y x sin + y cos Hence. =A =2 ru: @x @y @y @x Thus. In Particular. Solution (A. x). It follows that ru and rv are orthogonal.6). then @u @v @u @v (0. then rv is obtained by rotating ru by 90 .6). = . vx 0 1 vy cos 2 sin 2 vy rv = = = =R ru vy 1 0 vx sin 2 cos 2 vx 2 36 . one can write the following. Keeping in mind (0.3.II. Solution (D. By the Cauchy-Riemann equations we have @v @v @u @u rv = . rotation by 90 is given by the transformation (x. y) 7! ( y. Jakobsson) Given since f is analytic. @y all we need to check is whether rv rv = 0.6 If f = u + iv is analytic on D.6) = and = @x @y @y @x Since ru = @u . @u and rv = @x @x @y @v @v . To prove that we use (0. rv is obtained by rotating ru by 90 . @v @u @v @u @v @v @v @v ru rv = + = =0 @x @x @y @y @x @y @y @x We conclude that ru and rv are orthogonal. Notice that the operation that sends ru ! rv is given by the counter-clock- wise rotation matrix with = 2 . We have that @v @v @u @u vx = . vy = . @x @y @x @y and R ( =2) = A =2 : 37 . uy = .Remark. ux = . (a) iz (b) z 2 (c) 1=z Solution II. 2x) (c) x cos 1 z x iy u = x2 +y 2 = r f (z) = = 2 = 2 ) y sin z jzj x + y 2 v = 2 x +y 2 = r 8 2 2 2 2 > > @u = (xy2 +yx2 )2 = sin r2cos = cos r2 2 > > @x < @u = 2xy 2 = 2 cos sin 2 = sin2 2 @y (x2 +y 2 ) r r @v 2xy 2 cos sin sin 2 > > = (x2 +y 2 )2 = r2 = 2 > > @x r : @v = y 2 x2 = sin2 cos2 = cos 2 @y (x2 +y 2 )2 r2 r2 1 ru = r2 ( cos 2 . 2y) f (z) = z 2 = x2 y 2 + 2ixy ) ) v = 2xy rv = (2y.3. 0) (b) u = x2 y 2 ru = (2x.3.7a II.7a II.II.7a y y 4 y 4 4 2 2 2 -4 -2 2 4 -4 -2 2 4 -4 -2 2 4 -2 x -2 x -2 x -4 -4 -4 (a) u= y ru = (0. cos 2 ) 38 .7 Sketch the vector …elds ru and rv for the following functions f = u + iv.3. sin 2 ) 1 rv = r2 ( sin 2 . 1) f (z) = iz = y + ix ) ) v=x rv = (1.3. @x @x @r = cos @ = r sin @y @y @r = sin @ = r cos Use Cauchy-Riemanns equations and the derivative to get the …rst equation @u @u @x @u @y @u @u @v @v = + = cos + sin = cos sin = @r @x @r @y @r @x @y @y @x 1 @v @v 1 @v @x @v @y 1 @v = r sin + r cos = + = .3.8 (see III. Solution Set x = r cos and y = r sin . @u = 1r @v @ . 39 .II. @u @ = r @v @r : Set u rei = rm cos (m ) and v rei = rm sin (m ) and use it in the Cauchy-Riemann’s equations. then take the partial derivative. the functions u rei = rm cos (m ) and v rei = rm sin (m ) satisfy the Cauchy-Riemann equations. r @x @y r @x @ @y @ r@ Use Cauchy-Riemanns equations and the derivative to get the second equa- tion @u @u @x @u @y @u @u @v @v = + = r sin + r cos = r sin r cos = @ @x @ @y @ @x @y @y @x @v @v @v @x @v @y @v = r cos + sin = r + = r : @x @y @x @r @y @r @r we get @u 1 @v @r = r@ .2) (36) Derive the polar form of the Cauchy-Riemann equations for u and v. @u @ = r @v @r : @r Check that for any integer m.4. @r @r r r@ r@ @u @ m @ @v = (r cos (m )) = mrm sin (m ) = rmrm 1 sin (m ) = r (rm sin (m )) = r : @ @ @r @r 40 .@u @ m 1 m 1 @ m 1 @v = (r cos (m )) = mrm 1 cos (m ) = r m cos (m ) = (r sin m ) = . ex sin y) = ex (cos y.1a NF FIGURE II.4. cos y) (in red).II.1b NF 41 . b) We have u + iv = Log z = log r + i and get 1! ru = ur . and rv = (ex sin y.4. ex cos y) = ex (sin y.1 Sketch the gradient vector …elds ru and rv for (a) u + iv = ez (b) u + iv = Log z Solution a) We have u + iv = ez = ex cos y + iex sin y and get ru = (ex cos y.4. sin y) (in black). r 1! rv = u : r FIGURE II. It follows by the chain rule that a af (z) f 0 (z) = cea Log z = : z z So. 42 . 0].II. and let f (z) be an analytic branch of z a on Cn ( 1. Show that f 0 (z) = af (z) =z. (Thus f 0 (z) = az a 1 . where we pick the branch of z a 1 that corresponds to the original branch of z a divided by z.) Solution We may write the branch as follows f (z) = z a = ea log z = ea(Log z+2 im) = ea Log z+2 iam = cea Log z .2 Let a be a complex number a 6= 0. where c = e2 iam for some m 2 Z. the desired result follows.4. same on f (z).4.3 p Consider the branch of f (z) = z (1 z) on Cn [0. 1]. 1] that has posi- tive imaginary part at z = 2. p Use branch of f 0 (z) = z (1 z) that it is i positive. p Jakobsson) Given f (z) = z (1 z).II. Take the principal branch with the positive imag- inary part. What is f 0 (z)? Be sure to specify the branch of the expression for f 0 (z).e. 43 . we can de…ne w = f (z) and get w2 = z (1 z). thus f 0 (z) is i positive for x > 0 large. i. 1 1 2z f 0 (z) = : 2 f (z) Solution (D. d d 2 (1 z) z = w dz dz dw 1 2z = 2w dz One gets dp 1 2z z (1 z) = p dz 2 z (1 z) and is de…ned on Cn [0.. Solution p Set f (z) = z (1 z) (principal branch) and di¤erentiate 1 1 d 1 1 2z f 0 (z) = p (z (1 z)) = p : 2 z (1 z) dz 2 z (1 z) Branch of f (z) is i increasing for x > 0 large. z 62 ( i1. Solution We have that d 1 1 1 1 1 1 iz + 1 + iz 1 Tan z= i ( i) = = : dz 2i 1 + iz 1 iz 2 (1 + iz) (1 iz) 1 + z2 1 Any two branches of tan z di¤er by a constant.II. Find the derivative of tan 1 z for any analytic branch of the function de…ned on a domain D. 44 . so derivatives are same. i1) : Find the derivative of Tan 1 z. i] [ [i.4 Recall that the principal branch of the inverse tangent function was de…ned on the complex plane with two slits on the imaginary axis by 1 1 1+iz Tan = 2i Log 1 iz .4. 45 . Suppose g (z) is an ana- lytic branch of cos (z). de…ned on a domain D. Find g 0 (z).II.4.5 p Recall that cos 1 (z) = i log z z 2 1 . Do 1 di¤erent branches of cos 1 (z) have the same derivative? Solution We have that d 1 cos z= dz i 1 2 1=2 = p 1 z 1 2z = z z 2 1 2 i z i = p 1 p =p = z z 2 1 z 2 1 z 2 1 1 =p : 1 z2 1 Derivatives of branches of cos z are not always the same. Then f 0 (z) = cos z.4. while another local inverse of f satis…es h2 (0) = . Kumjian) Regard h (z) as a local inverse of f (z) where f (z) = sin z. so using the formula given in the statement of theorem on page 51 we have 1 1 h0 (z) = 0 = f (h (z)) cos h (z) if cos h (z) 6= 0. Do di¤erent branches of sin 1 (z) have the same derivative? Solution (A. 46 . Find h0 (z). de…ned on a domain D.II. One local inverse of f satis…es h1 (0) = 0. di¤erent branches of sin (z) do not necessarily have the same deriv- ative.6 Suppose h (z) is an analytic branch of sin 1 (z). Using the formula above we see 0 1 0 1 h1 (0) = =1 and h2 (0) = = 1: cos 0 cos 1 Hence. Since f is assumed to be one-to-one.7 Let f (z) be a bounded analytic function. and since det Jf = jf 0 (z)j2 . jf (z)j M for some M < 1 when z 2 D. v (x. y)) ! (x. we may compute ZZ A (f (D)) = dudv f (D) by the change of variables (u (x.II. y) . y). Seip) We have f : D ! C. we get ZZ 2 Area (f (D)) = jf 0 (z)j dxdy: D 47 .4. de…ned on a bounded domain D in the complex plane. and suppose that f (z) is one-one- one. Show that the area of f (D) is given by ZZ 2 Area (f (D)) = jf 0 (z)j dxdy: D Solution (K. that gives f 0 (z) = 2z and Jf (z) = 4 jzj2 .4.4. get ZZ 4 t2 + 2t + 1 + y 2 dtdy = t2 +y 2 61 2 3 ZZ ZZ 6 7 = 44 t2 + y 2 dtdy + 0 + dtdy 5 = t2 +y 2 61 t2 +y 2 61 h i =4 + =6 : 2 Solution (D. Compute the area of the image. Solution II. Jakobsson) d Given w = f (z) = z 2 .4.8 Sketch the image of the circle fjz 1j 1g under the map w = z 2 . one can di¤erentiate f (z) and get dz f (z) = 2z = 2 (x + iy). we can use maple to integrate and get the following 48 .8 z-plane II.II.8 w-plane y 4 y 4 2 2 -4 -2 2 4 -4 -2 2 4 -2 x -2 x -4 -4 We have that f (z) = z 2 . The area of the image is ZZ ZZ Jf dxdy = 4 x2 + y 2 dxdy: (x 1)2 +y 2 61 Set t = x 1. Area (f (D)) = ZZ Z 2 Z p1 (x 1)2 2 = 0 jf (z)j dxdy = p 4 jx + iyj2 dx dy = 2 D 0 1 (x 1) Z p 2Z 1 (x 1) 2 Z 2 p 3=2 = p 4 x2 + y 2 dx dy = 8x2 2x x2 +8=3 2x x2 dx = 0 1 (x 1)2 0 =6 Change of variables x = 1 + sin t dx = cos tdt Z =2 p 8 3=2 8 (1 + sin t)2 1 sin2 t + 1 sin2 cos tdt = =2 3 Z ! =2 2 2 8 4 = 8 1 + 2sin|{z}t + sin t cos + 3 cos tdtdt = =2 odd Z =2 2 = 16 2 cos2 cos4 tdt = 0 3 Z =2 1 =8 2 (1 + cos 2t) 1 + 2 cos 2t + cos2 2t dt = 0 3 Z =2 5 4 1 =8 + cos 2t (1 + cos 4t) dt = 0 3 3 3 5 1 9 =8 =4 =6 : 2 3 6 6 49 . so by Exersice II.9 Compute ZZ 2 jf 0 (z)j dxdy. D 2 for f (z) = z and D the open unit disk fjzj < 1g.II.7 the integrals on top and bottom halves are each . Solution By the result above ZZ ZZ 0 2 2 jf (z)j dxdy = jf 0 (z)j dx dy = .4. 50 . Interpret your answer in terms of areas.4. The function maps the top and bottom halves of the unit disk one-to-one onto the unit disk. jzj<1 jzj<1 x>0 x<0 thus ZZ 2 jf 0 (z)j dx dy = 2 : jzj<1 Remark. 4.y).y)) @y @y = @g @u @g @v @h @u @g @u @g @v @u @x + @v @x @u @x + @h @v @v @x + @u @y + @v @y @h @u @u @y + @h @v @v @y = @g @h @u 2 @g @u @h @v @g @v @h @u @g @h @v 2 @u @u @x + @u @x @v @x + @v @x @u @x + @v @v @x + 2 2 @g @h @u @g @u @h @v @g @v @h @u @g @h @v @u @u @y + @u @y @v @y + @v @y @u @y + @v @v @y = 2 2 @g @h @u 2 @g @h @v 2 @g @h @u @g @h @v @u @u @x + @v @v @x + + = ! @u @u @y @v @v @y ! 2 2 2 2 @g @h @u @u @g @h @v @v @u @u + + @v @v + + @x @y @x @y | {z } | {z } jf 0 (z)j2 jf 0 (z)j2 @g @u @h @v @g @v @h @u @g @u @h @v @g @v @h @u + + + @u @x @v @x @v @x @u @x @u @y @v @y @v @y @u @y | {z } Use C-R equations RR RR rgrh jf 0 (z)j2 dxdy = rgrhdudv U U 51 .II. h) by ZZ @g @h @g @h DU (g. y) . y) . y)) dudv U @g(u(x. y)) h (u (x. we de…ne the Dirichlet form DU (g. h f ) : Remark.y).v(x. h) = r (g f ) r (h f )dudv = RR U g (u (x.y)) @h(u(x.y).10 1 2 3 P L K For smooth functions g and h de…ned on a bounded domain U .y)) @h(u(x. v (x. h) = + dx dy U @x @x @y @y Show that if z = f ( ) is a one-to-one analytic function from the bounded domain V onto U . g ???? U RR Du (g. h) = rgrhdxdy Assume h.y). then DU (g. This shows that the Dirichlet form is a "conformal invariant".y)) @x @x + @g(u(x.v(x. h) = DV (g f. v (x.v(x.v(x. Solution RR Du (g. 1 Show that the following functions are harmonic. y) is harmonic. uy = 2y uyy = 2 thus is the function u (x. y) = x2 y2 ) 0 ) 00 ) 4u = 0. uy = x + 3x2 3y 2 uyy = 6y thus is the function u (x. x > 0 2 2 (b) xy + 3x2 y y 3 (d) ex y cos (2xy) (f) x= (x2 + y 2 ) Solution (a) Set 0 00 ux = 2x uxx = 2 u (x. y) = 2 x3 + 3xy 2 + g (y) ) vy = 6xy + g 0 (y) 0 0 vy = ux = y + 6xy Thus we have that 52 .5. y) = xy+3x2 y y 3 ) 0 ) 0 ) 4u = 0. Cauchy Riemann give us 0 0 x2 0 vx = uy = x 3x2 + 3y 2 ) v (x. Cauchy Riemann give us 0 0 0 vx = uy = 2y ) v (x. y) = 2xy + C: (b) Set 0 00 ux = y + 6xy uxx = 6y u (x. y) is harmonic. thus v (x. and …nd its har- monic conjugates: (a) x2 y 2 (c) sinh x sin y (e) tan 1 (y=x) .II. y) = 2xy + g (y) ) vy = 2x + g 0 (y) 0 0 vy = ux = 2x Thus we have that g 0 (y) = 0 ) g (y) = C. y) is harmonic. y) = cosh x cos y + g (y) ) vy = cosh x sin y + g 0 (y) 0 0 vy = ux = cosh x sin y Thus we have that g 0 (y) = 0 ) g (y) = C. uy = sinh x cos y uyy = sinh x sin y thus the function u (x. thus v (x. y) = cosh x cos y + C In paranthesis we remark that the analytic function f (z) = u + iv is given by 53 . Cauchy Riemann give us 0 0 0 vx = uy = sinh x cos y ) v (x. 2 thus y2 x2 v (x. 0 y2 g (x) = y ) g (y) = + C. y) = + 3xy 2 x3 + C: 2 2 In paranthesis we remark that the analytic function f (z) = u + iv is given by y2 x2 f (z) = xy +3x2 y y 3 +i + 3xy 2 x3 + C = iz 2 (z + 1=2)+iC: 2 2 (c) Set 0 00 ux = cosh x sin y uxx = sinh x sin y u (x. y) = sinh x sin y ) 0 ) 0 ) 4u = 0. f (z) = sinh x sin y i (cosh x cos y + C) = ex e x eiy e iy ex + e x eiy + e iy = i + iC = 2 2i 2 2 i = ex e x eiy e iy + ex + e x eiy + e iy + iC = 4 ex+iy + e (x+iy) = i + iC = i cosh z + iC: 2 (d) Set 0 2 2 2 2 2 y2 ux = 2xex y cos (2xy) 2yex y sin (2xy) u (x. y) = ex sin (2xy) + g (y) ) 0 2 2 2 2 ) v = 2yex y sin (2xy) + 2xex y cos (2xy) + g 0 (y) : 0 y 0 2 2 2 2 vy = ux = 2xex y cos (2xy) 2yex y sin (2xy) Thus we have that g 0 (y) = 0 ) g (y) = C. thus 2 y2 v (x. y) = ex cos (2xy) ) 0 2 2 2 2 ) uy = 2yex y cos (2xy) 2xex y sin (2xy) 00 2 2 uxx = 2ex y (cos 2xy + 2x2 cos 2xy 2y 2 cos 2xy 4xy sin 2xy) ) 0 2 2 ) 4u = 0. y) = ex sin (2xy) + C In paranthesis we remark that the analytic function f (z) = u + iv is given by 2 y2 2 y2 2 f (z) = ex cos (2xy) + i ex sin (2xy) + C = ez + iC: (e) 54 . uyy = 2ex y (cos 2xy + 2x2 cos 2xy 2y 2 cos 2xy 4xy sin 2xy) thus the function u (x. y) is harmonic. Cauchy Riemann give us 8 0 0 2 2 2 2 2 y2 < vx = uy = 2yex y cos (2xy) + 2xex y sin (2xy) ) v (x. y) = ) ) 4u = 0. x > 0 ) 8 > 0 1 y y < ux = = =) 1 + (y=x)2 x2 x2 + y2 =) > x : u0y = 2 x + y2 8 > 00 2xy > < uxx = (x2 + y 2 )2 ) 2xy ) 4u = 0. x2 + y2 > > 0 2xy > > 0 3y 2 x2 : uy = : uyy = 2x 2 (x2 + y 2 )2 (x + y 2 )3 thus is the function u (x. thus 1 v (x. y) = log x2 + y 2 + C: 2 (f) Set 8 8 > x2 y 2 > 00 3y 2 x2 > 0 < ux = > u < xx = 2x x (x2 + y 2 )2 ) (x2 + y 2 )3 u (x. > > 0 : uyy = (x2 + y 2 )2 thus the function u (x. Cauchy Riemann give us 8 > 0 0 x 1 0 y < vx = uy = ) v (x. y) is harmonic. y) = a 2 log (x2 + y 2 ) + g (y) ) vy = + g 0 (y) x2 + y 2 2 x +y 2 > 0 0 y : v y = ux = 2 x + y2 Thus we have that g 0 (y) = 0 ) g (y) = C.Set 1 u (x. y) is harmonic. Cauchy Riemann give us 55 . y) = tan (y=x) . y) = +C (x2 + y2) In paranthesis we remark that the analytic funktion f (z) = u + iv is given by x y z z 1 f (z) = +i +C = 2 + iC = + iC = + iC: x2 + y 2 (x2 + y 2 ) jzj zz z 56 . thus y v (x.8 > 0 0 2xy y 0 x2 y 2 > < vx = uy = 2 ) v (x. y) = + g (y) ) vy = 0 2 + g (y) 2 2 2 2 (x + y ) 2 2 (x + y ) (x + y ) 2 2 > > 0 0 x y : v y = ux = (x2 + y 2 )2 Thus we have that g 0 (y) = 0 ) g (y) = C. y) iu (x. then u is a harmonic conjugate for v.II. y) + iv (x. Solution If f (z) = u (x.2 Show that if v is a harmonic conjugate for u. y) is analytic. y) is analytic then if (z) = v (x. Thus u is a harmonic conjugate of v.5. 57 . y) as (x. = lim = 1: n!1 n n n!1 2 58 . 2 2 (b) Show that @@xu2 + @@yu2 = 0.3 De…ne u (z) = Im (1=z 2 ) for z 6= 0. y). Solution (a) We …rst de…ne 8 < 1 z2 (x iy)2 2xy Im 2 = Im 4 = Im 2 = (x. y) 2 C1 (x). y) 6= (0. y) is not harmonic because the function itself. n1 goes to 1 as n ! 1 whereas we de…ned the function u to be 0 at the origin.II. @2u (d) Show that @x@y does not exist at (0.5. thus u (x. At (0. as do all partial derivative of u with respect to y. 0).y) 2 )n+2 with f (x. the n th derivative of u (z) will be (x2f+y (x. the function itself is 0. (c) Show that u is not harmonic on C. y) = (0. 0). (b) We have @2u @2u @u 3x2 y 2 @u 3y 2 x2 + 2 = 2y + 2x = @x 2 @y @x (x2 + y 2 )3 @y (x2 + y 2 )3 x2 y 2 x2 y 2 = 24xy + 24xy =0 (x2 + y 2 )4 (x2 + y 2 )4 (c) Notice that the function u (x. y) a polynomial of degree n + 2 in (x. The limit of u (x. 0). and set u (0) = 0. y) 6= (0. 0) does not exist. 0) u (z) = : z jzj 2 2 (x + y ) (x + y 2 )2 2 0 (x. 0) For all (x. y) ! (0. (a) Show that all partial derivatives of u with respect to x exist at all points of the plane C. y) 2 C1 (y). its …rst and second derivative are not continuous at the origin. By symmetry u (x. for instance u n1 . 1 1 n2 lim u . (d) 2 @u 2x (x2 + y 2 ) + 2xy 2 (x2 + y 2 ) 2y 2x3 + 6xy 2 = = @y (x2 + y 2 )4 (x2 + y 2 )3 thus 3 2 @2u ( 6x2 + 6y 2 ) (x2 + y 2 ) ( 2x3 + 6xy 2 ) 3 (x2 + y 2 ) 2x 6x4 36x2 y 2 + 6y 4 = = : @x@y (x2 + y 2 )6 (x2 + y 2 )4 We have ( 6x4 36x2 y 2 +6y 4 @2u (x2 +y 2 )4 (x. y) 6= (0. x 6= 0 (x. 59 . y) = (0. so @x@y does not exist at (0. x=0 @2u This partial derivative is not continuous at x = 0. . 0) = : @x@y 0. 0) = @x@y 0 (x. 0) @2u 6=x4 . 0). then both @ 2 (zh) =@x2 +@ 2 (zh) =@y 2 = 0 and @ 2 (h) =@x2 + @ 2 (h) =@y 2 = 0. Solution First partial derivative for zh (z) is. @ (zh) @z @h = h+z . @x2 @x2 @x @x @x @x @x2 @ 2 (zh) @2z @z @h @z @h @2h = h + + + z : @y 2 @y 2 @y @y @y @y @y 2 We have z = x+iy.5. thus @z=@x = 1. then h (z) is analytic. we have @h @h = i : @x @y 60 . therefore @ 2 (zh) @h @2h = 2 + z . @z=@y = i. @x @x @x @ (zh) @z @h = h+z : @y @y @y Second derivative for zh (z) is. @x2 @x @x2 @ 2 (zh) @h @2h = i2 + z : @y 2 @y @y 2 Add the equations @ 2 (zh) @ 2 (zh) @h @h @2h @2h 2 + 2 =2 + 2i +z + : @x @y @x @y @x2 @y 2 Because h (z) and zh (z) is harmonic. @ 2 (zh) @2z @z @h @z @h @2h = h + + + z .4 Show that if h (z) is a complex-valued harmonic function (solution of Laplace’s equation) such that zh (z) is also harmonic. and @ 2 z=@x2 = @ 2 z=@y 2 = 0.II. 61 .Write h = u + iv. then h (z) is analytic as was to be shown. get @u @v @u @v @v @u +i = i +i = i : @x @x @y @y @y @y Now we take real and imaginary parts and get @u @v @u @v @x = @y and @y = @x : Thus yields Cauchy-Riemann’s equations for u + v. = r cos : @r @ Taking the …rst derivative of u with respect to r @u @u @x @u @y @u @u @u @u = + = cos + sin = cos + sin : @r @x @r @y @r @x @y @x @y Taking the second derivative of u with respect to r @2u = @r2 @ 2 u @x @ 2 u @y @ 2 u @x @ 2 u @y = cos + cos + sin + sin = @x2 @r @y@x @r @x@y @r @y 2 @r @ 2 u @x @2u @2u @ 2 u @y = cos + sin cos + cos sin + sin = @x2 @r @y@x @x@y @y 2 @r @2u @2u 2 2 @ u = cos2 + 2 sin cos + sin : @x2 @x@y @y 2 Taking the …rst derivative of u with respect to @u @u @x @u @y @u @u = + = ( r sin ) + r cos : @ @x @ @y @ @x @y Taking the second derivative of u with respect to 62 .II.5. @r @ @y @y = sin .5 Show that Laplace’s equation in polar coordinates is @ 2 u 1 @u 1 @2u + + = 0: @r2 r @r r2 @ 2 Solution Let x = r cos and y = r sin . and get partial derivatives @x @x = cos . = r sin . one gets @ 2 u 1 @u 1 @2u + + = @r2 r @r r2 @ 2 @2u @2u @2u 1 @u @u = cos2 + 2 sin cos + sin2 + cos + sin + @x2 @x@y @y 2 r @x @y 1 @2u @2u @2u @u @u + 2 r2 sin2 2r2 sin cos + r2 cos2 r cos r sin = r @x2 @x@y @y 2 @x @y @2u 2 @ u 2 @ u 1 @u 1 @u = cos2 + 2 sin cos + sin2 + cos + sin + @x2 @x@y @y 2 r @x r @y @2u @2u @2u 1 @u 1 @u + sin2 2 sin cos + cos2 cos sin = @x2 @x@y @y 2 r @x r @y @2u @2u @2u @2u = sin2 + cos2 + sin 2 + cos 2 = + : @x2 @y 2 @x2 @y 2 Because u is a harmonic function we have @ 2 u=@x2 + @ 2 u=@y 2 = 0 and we have. @ 2 u 1 @u 1 @2u @2u @2u + + = + = 0: @r2 r @r r2 @ 2 @x2 @y 2 63 . @2u = @ 2 @ 2 u @x @ 2 u @y @u = 2 ( r sin ) + ( r sin ) + ( r cos ) + @x @ @y@x @ @x @ 2 u @x @ 2 u @y @u + (r cos ) + 2 (r cos ) + ( r sin ) = @x@y @ @y @ @y @2u @2u @u = 2 ( r sin ) ( r sin ) + (r cos ) ( r sin ) + ( r cos ) + @x @y@x @x @2u @2u @u + ( r sin ) (r cos ) + 2 (r cos ) (r cos ) + ( r sin ) = @x@y @y @y @2u @2u 2 2 @ u @u @u = r2 sin2 2 2r 2 sin cos + r 2 cos 2 r cos r sin : @x @x@y @y @x @y Combining these partial derivatives. 6 Show using Laplace’s equation in polar coordinates that log jzj is harmonic on the punctured plane Cn f0g.II. r 6= 0 because this case is taken out. thus we have @ 2 u 1 @u 1 @u2 1 11 log r = + + = + + 0 = 0: @r2 r @r r2 @ 2 r2 r r 64 .5. we have u (r.e. i. We can compute the Laplacian of log jzj without worrying about the origin. ) = log jzj = log r. Solution Set z = rei . Kumjamin) Set D := Cn ( 1. But such a function has no continuous extension to D. Seip) In Cn ( 1. ~ This contradicts the fact that v~ is harmonic on D ~ (since this implies v~ is ~ continuous on D). Therefore.5. Solution (K.7 Show that log jzj has no conjugate harmonic function on the punc- tured plane Cn f0g. though it does have a conjugate harmonic func- tion on the slit plane Cn ( 1. y) = Arg z + c: Such a function can not be made continuous in C since for x < 0. Moreover.II. Suppose that u~ does have a ~ Then its restriction to D would have to harmonic conjugate. say v~. 0] and let u : D ! R be given by u (z) := log jzj. it follows by Cauchy-Riemann equations that a harmonic conjugate v is given by v (z) := Im Log z = Arg z for z 2 D: Any other harmonic conjugate di¤ers from v by a constant. 0]. it is easily veri…ed that u (z) = Re Log z and this shows that u is harmonic on D by the main theorem in section II. Now if we set D~ := Cn f0g and de…ne u~ : D ! R by u~ (z) := log jzj. y) = + c: y!00 65 .5. log jzj has no conjugate harmonic function on the punctured plane Cn f0g. Solution (A. it can easily be shown by an argument similar to the one above that u~ is harmonic on D ~ even though it cannot be expressed as the real part of an analytic function on D~ (the property of being harmonic is local). we have lim v (x. y!00+ and lim v (x. be a harmonic conjugate of u and so it would be of the form z 7 ! Arg z + c for some c 2 R. y) = + c. 0). we know that any harmonic conjugate of log jzj must have the form v (x. on D. 8 Show using Laplace’s equation in polar coordinates that u rei = log r is harmonic.II.8) to …nd a harmonic conjugate v for u. Use the polar form of the Cauchy-Riemann equations. What is the analytic function u + iv? Solution Set ( ( 0 00 ur = urr = u rei = log r ) 0 r ) 00 r2 u = log r u =0 The function u rei = log r is harmonic because 1 1 u = urr + ur + 2 u = + = 0: r r r2 r2 Cauchy Riemann equations in polar form (page 50) gives us 8 < 0 1 0 log r (log r)2 0 vr = u = ) v (r.5. 2 thus (log r)2 2 v (r. ) = + + C: 2 2 We have that the analytic function f (z) = u + iv is given by ! 2 (log r)2 i i f (z) = log r+i +C = (log r + i )2 +C = (log z)2 +iC: 2 2 2 2 66 . ) = + g ( ) ) v = g0 ( ) : v 0 = rur0 = r 2 r Thus we have that 2 g0 ( ) = ) g( ) = + C. (Exercise 3. (Conformal everywhere if we view it as a mapping of the sphere).6. (x + y 2 ) = y. (x + y 2 ) = x. x + y2 = . x2 + y + = . Solution (a) We have that 8 x 1 z x iy < u= f (z) = = 2 = 2 ) x + y2 2 : v= y z jzj x + y2 x + y2 2 Conformal except at z = 0. We have 1 1 2 a 2 a2 v= . We have 1 1 2 c 2 c2 u= .II. a a 2 4 which is circles centred on y axis and tangent to x axis.1 Sketch the families of level curves of u and v for the following functions f = u + iv. c c 2 4 which is circles centred on x axis and tangent to y axis. (a) f (z) = 1=z (b) f (z) = 1=z 2 (c) f (z) = z 6 Determine where f (z) is conformal and where it is not conformal. (b) We have 1 z2 (x iy)2 x2 y 2 2ixy f (z) = 2 = 4 = = ) z jzj (x2 + y 2 )2 (x2 + y 2 )2 ( 2 2 2 2 cos(2 ) u = (xx2 +yy2 )2 = cos r2sin = r2 =) 2xy 2 cos sin sin(2 ) v = (x2 +y 2 )2 = r2 = r2 and 67 . where it is not de…ned. 1 1 u= .1b NF FIGURE II. Rotate of fu = 0g by =12 is fv = 0g. v= .6.1a NF FIGURE II. FIGURE II. (c) We have that u = r6 cos (6 ) f (z) = z 6 = r6 (cos (6 ) + i sin (6 )) ) v = r6 sin (6 ) Figure repeats itself. r2 = c cos (2 ) .6. Each grind repeats itself every =6. r2 = c sin (2 ) : c c Conformal everywhere except at z = 0.6. Conformal everywhere except at z = 0.1c NF 68 . 6.6.II.6. The sketch is of the inverse images of horizontal and vertical lines in that Figure. < < .6.2 Sketch the families of level curves of u and v for f (z) = Log z = u+iv. Refer to the Figure for the map w = Log z in section I. and if v = constanat and = constanat we have rays issuing from 0. 69 .2 π/2 3π/4 π/4 0 −3π/4 −π/4 −π/2 We have u = log jzj u = log r f (z) = Log z = log jzj+i Arg z = log r+i ) ) : v = Arg z v= If u = constant and r = constant we have circles centred at 0. Relate your sketch to a …gure in Section I. Solution II. The function f (z) is conformal everywhere.6. x = log (c sin y) = log c log (sin y) Level curves are invariant under ????? in x–direction. Determine where f (z) is conformal and where it is not conformal. Solution f (z) = ez = ex (cos y + i sin y) ) u = ex cos yv = ex sin y u = const. f (z) = e z = e(a+ib)(x+iy) = eax by (cos (ay + bx) + i sin (ay + bx)) ) u = eax by cos (ay + bx) v = eax by sin (ay + bx) Figure repeat themselves u = 0 at ay + bx = n + 2 .3b NF 70 . v = 0 at ay + bx = n .II. Level curves are in- variant under ????? in y–direction by .6. y = ab x + na + 2a . where is complex. y = ab x + na FIGURE II. (b) f (z) = e z .6. x = log (c cos y) = log c log (cos y) v = const.3 Sketch the families of level curves of u and v for the function f = u + iv given by (a) f (z) = ez .3a NF FIGURE II. 6:4aw It is clearly intended that A > 0 so we will assume this. or refer to the preceding problem. h is analytic with a nowhere vanishing derivative and so h is conformal.4 Find a conformal map of the horizontal strip f A < Im z < Ag onto the right half-plane fRe w > 0g. Hence. Hint. Note that the exponential functions maps the horizontal strip fz 2 C : =2 < Im z < =2g onto the right half-plane fw 2 C : Re w > 0g = fw 2 C : =2 < Arg w < =2g : Hence.II.6:4az FIGURE II. We de…ne the map z h : C ! C by h (z) = e 2A for all z 2 C. Solution (A. h maps the horizontal strip fz 2 C : A < Im z < Ag onto the right half-plane fw 2 C : Re w > 0g. Recall the discussion of the exponential function. 71 . Kumjian) FIGURE II. Then z h0 (z) = e 2A 2A for all z 2 C.6. 5az FIGURE II. so eiB ! ei =2 .w=z =(2B) does the trick FIGURE II. Solution Try w = z .II. Assume 0 < B < .6.5aw 72 .5 Find a conformal map of the wedge f B < arg z < Bg onto the right half-plane fRe w > 0g. multiply angels by 2B .6.6. f (z) maps the circle fjzj = rg onto an ellipse. which can be written u2 = ( + 1= )2 + v 2 = ( 1= )2 = 1. then w = u + iv = cos + i sin + cos = i sin = . and f (z) maps onto C. 2].6. Show that f (z) is one-to-one on the exterior domain D = fjzj > 1g. and sketch also the images of the parts of the rays farg z = g lying in D. f (z) is one–to–one on D. image is an ellipse. Show that for each w. 2].pNot de…ned 2 at z = 0.6 Determine where the function f (z) = z +1=z is conformal and where it is not conformal.II. Since f (z) maps @D onto the interval [ 2. Solution f (z) = z + 1=z. z = 1. Determine the image of D under f (z).e. f 0 (z) = 0 at z 2 = 1. i. the image of D is Cn [ 2. Sketch the images under f (z) of the circles fjzj = rg for r > 1. replace by 1= and get same ellipse. f 0 (z) = 1 1=z 2 . there are at most two values z for which f (z) = w. not conformal at z = 1. and that f (z) maps the circle fjzj = 1=rg onto the same ellipse. If f (z) = . If z = ei . then z = 4 =2. and f (z) is at most two–to–one. Since f (D) = f (D). Show that if r > 1. 73 . Determine the images under f (z) of the top half of the unit disk.6. the part of the upper half-plane outside the unit disk. Start by locating the images of the curves where u = 0. sketch the families of level curves of u and v. and the part of the lower half-plane outside the unit disk.7 For the function f (z) = z + 1=z = u + iv. where v = 0. the bottom half of the unit disk. Solution 74 . Note that the level curves are symmetric with respect to the real and imaginary axis.II. and where v = 1. Hint. and they are invariant under the inversion z 7! 1=z in the unit circle. Sketch roughly the images of the segments of rays outside the unit circle farg z = . Thus f (z) maps fjzj > 1g one–to–one onto the complement rotate of [ 2. and at what angles do they approach 1? Solution f (z) = z + ei =z . and a rotation by =2. +1) on the real axis. Sketch the images under f (z) of the unit circle fjzj = 1g and the intervals ( 1. The expression f (z) = ei =2 e i =2 z + 1= e i =2 z . 2] by =2. f 0 (z) = 0 at z 2 = ei . the function of Exercises 6 and 7.6. 75 . not conformal at z = ei =2 .II. jzj 1g under f (z).8 ( From Hints and Solutions) 1 2 3 P L K Consider f (z) = z + ei =z. 1] and [+1. Show that w = f (z) maps fjzj > 1g conformally onto the complement of a slit plane in the w plane. Determine where f (z) is conformal and where it is not conformal and where it is not conformal. Not de- …ned at z = 0. where 0 < < . At what angels do they meet the slit. shows that f (z) is a composition of the ro- tation by =2. f 0 (z) = 1 ei =z 2 . II. vy )i h(ux . 1) are mapped to the tangents in the directions (ux . with nonvanishing derivative. vy ) . same sign holds in ?????. vy )i = 0. can be rewritten h(ux . vx )i = 0: The systems of equations (1) ux u y + v x v y = 0 (2) u2y + vy2 (u2x + vx2 ) = 0 most hold to get orthogonality. 1) ! (uy . ts) . Solution Suppose f (0) = 0 and ux 6= 0 at 0. uy = vx ) anti conformal in neighborhood of 0.9 Let f = u + iv be a continuously di¤erentiable complex-valued func- tion on a domain D such that the Jacobian matrix of f does not vanish at any point of D. Either ux = vy . Put (1) into (2) and get u2x u2y + vy2 u2x vx2 = vx2 vy2 + u2x vy2 u4x u2x vx2 = = vx2 vy2 u2x + u2x vy2 u2x = vx2 + u2x vy2 u2x = 0: If ux 6= 0 we are lead to ux = vy and uy = vx . vx )i) t2 h(uy . or ux = vy . (uy . v (s. vx ) + t (uy . (ux . (0. vy )i+t (h(uy . t) and ( t. uy ) for some. vx ) . vx ) . ts) is tangent to image curve s + (u (s. The orthogonality of these directions for all t gives h(ux . vx ). A ????? gives C = 1. The tangents in the orthogonal directions (1. Show that if f maps orthogonal curves to orthogonal curves. vx ) + t (uy . (ux . t (ux . ts)) (1. vy ) = C ( vx . vx ) + (uy . orthogonal ) (uy .6. uy = vx ) conformal in neighborhood of 0. 0) ! (ux . ux = vy and uy = vx . (uy . So Cauchy Riemann is satis…ed for either f or f . vy ) . vy ) . vy ). vx ) + (uy . then either f or f is analytic. Tangent to curve s ! (s. 76 . vy ) and t (ux . vy ). 7. i) (d) ( 2. i. i) 7 ! (0. to obtain 77 . 1 + i. 1) Solution (w w0 ) (w1 w2 ) (z z0 ) (z1 z2 ) = (w w2 ) (w1 w0 ) (z z2 ) (z1 z0 ) (a) We have the points z0 = 1 + i w0 = 0 z1 = 2 w1 = 1 z2 = 0 w2 = i 1 The mapping is (w w0 ) (1 w2 =w1 ) (z z0 ) (z1 z2 ) = . 1) 7 ! (0. 1) 7 ! (1. 1. 1) 7 ! (0. 0) 7 ! (0. 0. 2) 7 ! (1 2i. 1.1 Compute explicitly the fractional linear transformations determined by the following correspondences of triples (a) (1 + i. 1. i. 1. i 1) (e) (1. (w w2 ) (1 w0 =w1 ) (z z2 ) (z1 z0 ) take limit as w1 ! 1. 1) (b) (0. 0. (w w2 ) (w1 w0 ) (z=z2 1) (z1 z0 ) take limit as z2 ! 1. 2. 1. 1. 1.II. 1 + 2i) (h) (1. 2) (f) (0. 1) (c) (1. 1. to obtain w 0 (z (1 + i)) (2 0) 2iz + (2 2i) = )w= : w (i 1) (z 0) (2 (1 + i)) z 2 (b) We have the points z0 = 0 w0 = 1 z1 = 1 w1 = 1 + i z2 = 1 w2 = 2 The mapping is (w w0 ) (w1 w2 ) (z z0 ) (z1 =z2 1) = . 2) 7 ! (0. 1) 7 ! (1. 1 + i. 2. 1) (g) (0. (w=w2 1) (w1 w0 ) (z z2 ) (z1 =z0 1) take limit as z0 . w2 ! 1. (w w2 ) (w1 w0 ) (z z2 ) (z1 z0 ) obtain (w (1 2i)) (0 (1 + 2i)) (z ( 2)) (i 2) = ) w = iz + 1: (w (1 + 2i)) (0 (1 2i)) (z 2) (i ( 2)) (e) We have the points z0 = 1 w0 = 0 z1 = 2 w1 = 1 z2 = 1 w2 = 1 78 . (w 1) ((1 + i) 2) z 0 2z (1 + i) = )w= : (w 2) ((1 + i) 1) 1 0 z (1 + i) (c) We have the points z0 = 1 w0 = 0 z1 = 1 + i w1 = 1 z2 = 2 w2 = 1 The mapping is (w w0 ) (w1 =w2 1) (z=z0 1) (z1 z2 ) = . to obtain w 0 (1 + i) 2 i 1 = )w= : 1 0 z 2 z 2 (d) We have the points z0 = 2 w0 = 1 2i z1 = i w1 = 0 z2 = 2 w2 = 1 + 2i The mapping is (w w0 ) (w1 w2 ) (z z0 ) (z1 z2 ) = . to obtain w 0 z 0 z = )w= 1 0 z i z i (g) We have the points z0 = 0 w0 = 0 z1 = 1 w1 = 1 z2 = 1 w2 = i The mapping is (w w0 ) (1 w2 =w1 ) (z z0 ) (z1 =z2 1) = .The mapping is (w w0 ) (w1 =w2 1) (z z0 ) (z1 =z2 1) = . to obtain w 0 z 0 iz = )w= : w i 1 0 z 1 (h) 79 . (w=w2 1) (w1 w0 ) (z=z2 1) (z1 z0 ) take limit as z2 . to obtain w 0 z 1 = )w=z 1: 1 0 2 1 (f) We have the points z0 = 0 w0 = 0 z1 = 1 w1 = 1 z2 = i w2 = 1 The mapping is (w w0 ) (w1 =w2 1) (z z0 ) (1 z2 =z1 ) = . w2 ! 1. w1 ! 1. (w=w2 1) (w1 w0 ) (z z2 ) (1 z0 =z1 ) take limit as z1 . w2 ! 1. (w w2 ) (1 w0 =w1 ) (z=z2 1) (z1 z0 ) take limit as z2 . (w w2 ) (w1 w0 ) (z z2 ) (z1 z0 ) obtain (w 1) (0 ( 1)) (z 1) (i ( 1)) iz + 1 = )w= : (w ( 1)) (0 1) (z ( 1)) (i 1) z+i 80 .We have the points z0 = 1 w0 = 1 z1 = i w1 = 0 z2 = 1 w2 = 1 The mapping is (w w0 ) (w1 w2 ) (z z0 ) (z1 z2 ) = . Solution It is obvious that the tripple (1 + i.1a we know that points in the w plane the tripple is mapped on.7. Without referring to an explicit formula. the image of the disk fjz 1j < 1g.2 z-plane II. i. By preservation of orientation the disk goes to the half-plane to lower left of the straight line through 0 and i 1. with slope 1. determine the image of the circle fjz 1j = 1g. z0 = 1 + i w0 = 0 z1 = 2 w1 = 1 z2 = 0 w2 = i 1 The circle jz 1j = 1 is mapped to the straight line through 0 and i 1. which maps 1 + i to 0. i. 0) lies on the circle jz 1j = 1 in the z plane.e.e.7. II.2 w-plane 81 .2 Consider the fractional linear transformation in Exercise 1a above.7. Image of the real axis is "circle" orthogonal to image of jz 1j = 1.e Im w = Re w since tree points determines the "circle". 2 to 1. and 0 to i 1. that must be the straight line through i 1.II. i. 2. and the image of the real axis. the line Im w = Re w + 2.7. From exercise II. we have that i on the unitcircle in the z plane is mapped to 1 in the w plane. 0 to 1 + i. that is orthogonal to image of real line. i. this is the straight line through 1+i and 0. it must be so that p the real axis in the z plane is mapped 1 1 2 onto the circle w 2 + 2 i = 2 in the w plane. 1) in the w plane. The tripple (1. 1) on the real axis is mapped to the points (i. 1 to 1.e Im w = Re w. the image of the open unit disk fjzj < 1g. 1 + i. 0. Determine the image of the unit circle fjzj = 1g.e Im w = Re w + 1 since tree points determines the "circle". Solution We have the points z0 = 1 w0 = i z1 = 0 w1 = 1 + i z2 = 1 w2 = 1 The mapping is (w w0 ) (w1 w2 ) (z z0 ) (z1 z2 ) = . By preservation of orientation the disk goes to the half-plane to upper left of the straight line through i and 1.7. and the image of the imaginary axis. The circle jzj = 1 is mapped to the straight line through i and 1. Imaginary axis must be mapped to a circle through 1+i. 82 .3 Consider the fractional linear transformation that maps 1 to i.II. (w w2 ) (w1 w0 ) (z z2 ) (z1 z0 ) obtain (w i) ((1 + i) 1) (z 1) (0 ( 1)) i 1 = )w= (w 1) ((1 + i) i) (z ( 1)) (0 1) z+i By our mapping w = (i 1) = (z + i). Illustrate with a sketch. i. II.3 w-plane 83 .3 z-plane II.7.7. We have that w (0) = 65 25 i must be an point on the arc. 1 to 2i. i.II. (w w2 ) (w1 w0 ) (z z2 ) (z1 z0 ) obtain (w ( i)) (2i 0) (z ( 1)) (1 i) ( 6 2i) z 2 + 6i = )w= (w 0) (2i ( i)) (z i) (1 ( 1)) 5z 3 + 4i It is obvious that the tripple ( 1. and the image of the interval [ 1. and i to 0. 84 . i) lies on the circle jzj = 1 in the z plane. And we can see that thiese three points are mapped to three points on the imaginary axis in the w plane. the image of the open unit disk fjzj < 1g. thus the arc is the circle w 21 i = 23 in the right half-plane from i to 2i. Determine the image of the unit circle fjzj = 1g. 1. this is an arc of the circle w 21 i = 32 . Illustrate with a sketch.4 Consider the fractional linear transformation that maps 1 to i. Solution We have the points z0 = 1 w0 = i z1 = 1 w1 = 2i z2 = i w2 = 0 The mapping is (w w0 ) (w1 w2 ) (z z0 ) (z1 z2 ) = . By preservation of orientation the disk goes to the right half-plane. +1] on the real axis.7. since tree points determines the "circle". The interval [ 1. 1] must be mapped to a circle through i and 2i that is orthogonal to image of the unit circle.e Re w = 0. 7.II.7.4 z-plane II.4 w-plane 85 . Im z = Re z + 1.II. The common point at in…nity for both real axis and the line through i must be mapped to the same point in the w plane. we have that the point is w (1) = i. The two lines in the z plane is parallell. i. Remark that the image can not be a circle because w (i) = 1. The real axis and the line through i have one common point at in…nity. 0. thus the images of the lines must go through i in the w plane and be parallell in that point. Solution We have the points z0 = 0 w0 = 1 z1 = 1 w1 = 0 z2 = 1 w2 = 1 + i The mapping is (w w0 ) (w1 w2 ) (z z0 ) (z1 z2 ) = . which is tangent line to circle at i. 1) on the real axis is mapped to the points (1.7. 1. 1 + i) in the w plane. 86 . so the image of the horizontal line must contain the point at in…nity.5 What is the image of the horizontal line through i under the frac- tional linear transformation that interchanges 0 and 1 and maps 1 to 1 + i? Illustrate with a sketch. (w w2 ) (w1 w0 ) (z z2 ) (z1 z0 ) obtain (w 1) (0 (1 + i)) (z 0) (1 ( 1)) iz i = )w= (w (1 + i)) (0 1) (z ( 1)) (1 0) z i The tripple (0. thus the mapping of the horizontal line through i must be the line through i.e. it must be so pthat the real axis in the z plane is mapped onto the circle w 1 2 + 12 i = 22 in the w plane. since tree points determines the "circle". II.5 z-plane II.5 w-plane 87 .7.7. 7. depending on whether the line passes through the origin. Seip) Set f (z) = z1 . If c = 0 we have that au + bv = 0. Then 1 2 f (S) . which proves the claim. Get ax + by c c au bv = 2 = 2 = 2 = c u2 + v 2 r r x + y2 The image is the solution of c (u2 + v 2 ) au + bv = 0.II. If c 6= 0 we have c u2 + v 2 au + bv = 0. which can be rewritten as 0s 12 2 2 a 2 b a 2 b u + v+ =@ + A . 0 2 S. 88 . this is a straight line through 0.6 Show that the image of a straight line under the inversion z 7! 1=z is a straight line or circle. Solution (K. Solution The image of the straight line ax+by = c is (kontrollera x iy i lösningshäftet) 1 1 z 1 w= = = 2 = (x iy) z zz jzj r2 where u = x=r2 and v = y=r2 . 2c 2c 2c 2c it is a circle through 0. az + b = cz 2 + dz .dc 2 R . and since one of a.7 Show that the fractional linear transformation f (z) = (az + b) = (cz + d) is the identity mapping z if and only if b = c = 0 and a = d 6= 0. At least one of theese rations is di¤erent from 0 and 1. Then 0 7! db 2 R . b. 89 . z = 1: From (2) we have b = 0 and if we add (1) and (3) we have c = 0. Solution Set R = R [ f1g. 0 . 1 7! ac 2 R . Thus the fractional linear transformation f (z) is the identity mapping if and only if b = c = 0. 1 .7. a = d 6= 0. Suppose f (z) = az+b cz+d maps R to R .II. c. z = 1: (2) b=0 z=0 (3) c + (d a) = 0. d can be chosen freely. Solution We have that az + b = z .ab 2 R . the result follows. cz 2 + (d a) z b = 0: cz + d Put in values on z for intance 0 and 1 the expession becomes (1) c (d a) = 0. 7. 90 .8 Show that any fractional linear transformation can be represented in the form f (z) = (az + b) (cz + d). where ad bc = 1.II. as can multiply all coe¢ cients by 1. Is this repre- sentation unique? Solution If w = ( z + )( z + ) divide each coe¢ cient by the square root of to obtain representation with ad bc = 1: Representation is not unique. and since one of a.ab 2 R .7. b. Seip) Set R = R[f1g. where a. 1 . 1 7! ac 2 R . b. the result follows. 0 . Then 0 7 ! db 2 R . Solution (K.II. d can be chosen freely.9 Show that the fractional linear transformations that are real on the real axis are precisely those that can be expressed in the form (az + b) = (cz + d).and d are real. c.dc 2 R . Suppose f (z) = az+b cz+d maps R to R . 91 . c. At least one of these is di¤erent from 0 and 1. x1 . Kumjian) Set R = R [ f1g. because we must have g = f 1 (again by the uniqueness part of the theorem). x1 6= 1 and the remaining three cases are x0 = 1. Hence. c and d as above. f is completely determined by the three extended real numbers x0 = f (0). The …rst case is when x0 . b. It remains to prove that given a fractional linear transformation f . c. where a. note that a fractional linear transformation f : C ! C is real on the real axis i¤ f (R ) R (since f is continuous). Suppose that f = fA where A is a 2 2 matrix with real entries a. f (1) = lim = z!1 cz + d 1 if c = 0: Hence. we must show that f = fA for some 2 2 matrix with real entries.II. There are four cases to consider. b. suppose that f (R ) R .7. g (x1 ) = 1 and g (x1 ) = 1 has the requisite properties. For A = ac db with det A 6= 0 we de…ne the fractional linear transformation fA : C ! C by the formula fA (z) = (az + b) = (cz + d). Let f : C ! C be a fractional linear transformation. Solution (A. f (R ) R . f = fA for some 2 2 matrix A with real entries i¤ its inverse f 1 has the same property. x1 = 1 and x1 = 1. By the unlikeness part of the theorem on page 64. Conversely. We have az + b a=c if c 6= 0.and d are real.9 Show that the fractional linear transformations that are real on the real axis are precisely those that can be expressed in the form (az + b) = (cz + d). Then for x 2 R we have ax + b f (x) = fA (x) = 2R : cx + d This is clear for x 2 R but requires a little work for x = 1. x1 = f (1) and x1 = f (1) (recall that we have assumed f (R ) R ). then fA 1 = (fA ) 1 . If A is a 2 2 matrix (with non-zero determinant). we have f (R ) R i¤ f = fA for some 2 2 matrix A with real entries. The formula for g (z) in the four cases is given as follows (in the …rst case k = xx11 xx10 2 R): 92 . So it su¢ ces to show that the fractional linear transformation g for which g (x0 ) = 0. x1 6= 1. x x0 x1 x1 k if x0 . if x1 = 1: x x1 x1 x0 In each case g (z) has the desired form. 93 . if x0 = 1. x x1 x x1 x x0 x x0 if x1 = 1. Hence. x1 . f = fA for some 2 2 matrix A with real entries as required. 7. and d are real or they are all pure imaginary. 94 . so a.10 Suppose the fractional linear transformation (az + b) = (cz + d) maps R to R. b.II. c. d are real. Solution Because R is mapped on R and the orientation is perserved. so a. f (z) = (az + b) = (cz + d). then f 0 (z) = 1= (cz + d)2 < 0 for all x ) c. then ax + b are pure imaginary for all x 2 R. Case 1 : Suppose f (x) is increasing. Show that a. then ax + b is real for all x 2 R. d are pure imaginary. f (x) will be either increasing or decreasing. Case 2 : Suppose f (x) is decreasing. then f 0 (x) = 1= (cx + d)2 > 0 for all x ) c. and ad bc = 1. b are also pure imaginary. b are also real. f (z) = (az + b) = (cz + d). then f f is conjugate to g g. We can think of f and g as the "same" map. (a) If f is conjugate to g.7. with appropriate domain and range.11 Two maps f and g are conjugate if there is h such that g = h f h 1 . Show the following. A point z0 is a …xed point of f if f (z0 ) = z0 . f and g have the same number of …xed points.II. the m fold composition f f (m times) is conjugate to g g (m times). then g is conjugate to f . and w0 = h (z0 ). then f1 is conjugate to f3 . after the change of variable w = h (z). Solution (a) If f is conjugate to g then 1 1 g=h f h )f =h g h thus g is conjugate to f (b) If f1 = h f2 h 1 and f2 = g f3 g 1 we get 1 1 1 f1 = h g f3 g h = (h g) f3 (h g) : (c) m times z }| { g=h f h 1 theng g : : : g = =h f h 1 h f h 1 ::: h f h 1 = m times z }| { 1 =h f f ::: f h (d) g=h f h 1 . (c) If f is conjugate to g. (b) If f1 is conjugate to f2 and f2 to f3 . In particular. Here the conjugating map h is assumed to be one-to-one. f (z0 ) = z0 . then 95 . and more generally. then the conjugating function h maps the …xed points of f to …xed points of g. (d) If f and g are conjugate. 1 g (w0 ) = h f h (w0 ) = h (f (z0 )) = h (z0 ) = w0 : We have that f and g have the same number of …xed points since every point of g is mapped to a …xed point of f by the function h 1 . 96 . we get w (z) = z. (b) If a fractional linear transformation f (z) has two …xed points. aw +b = w has no solutions and a = 1. Then h f h 1 is FLT with …xed points at 0 and 1. h (az) = Ah (z). there is a fractional linear transformation h (z) such that h (f (h 1 ( ))) = + 1. it has 2 solutions possibly one with multiplicity and 1 is not a …xed point. a 6= 1. it has one …nite solution z1 = b= (d a). so either h (z) = cz or h (z) = c=z. b 6= 0. (c) If a fractional linear transformation f (z) has exactly one …xed point. Thus (h f h 1 ) (w) = w +b. points satisfying f (z0 ) = z0 . If this is = 0 for every z. Is a unique? Hint. (h (ah 1 )) (w) = Aw. Hint. and z2 = 1 is a …xed point. i. such that h (f (z)) = h (z) + 1. there is a fractional linear transformation h (z) such that h (f (z)) = ah (z). It is not identically equall to zero. f is conjugate to a dilation Suppose aw is conjugate to Aw. a 6= 1. Consider a fractional linear transformation that maps the …xed points to 0 and 1. (h f h 1 ) (w) = aw for some a 6= 0. then h f h 1 has only one …xed point at 1. 97 . If c = 0. If c 6= 0. so they are two. h must map …xed points to …xed points. So either h (z) = cz ) h (az) = caz = Ah (z) = Acz ) A = a or h (z) = c=z ) h (az) = c= (az) = Ah (z) = Ac=z ) A = 1=a. In other words. 9h. then it is conjugate to the dilation z 7! az with a 6= 0. that is.II. b) Make a change of variables h (z) = ( z + ) = ( z + ) that maps …xed points to 0 and 1 . that is. az + b = cz 2 + dz . so it has either 1 or 2 …nite solutions. Thus a is not unique. Since this has no …nite …xed points. Solution a) az+b cz+d = z . or equivalently.12 Classify the conjugacy classes of fractional linear transformations by establishing the following (a) A fractional linear transformation that is not the identity has either 1 or 2 …xed points. so (h f h 1 ) (w) = aw+b. c) Suppose h maps the …xed points to 1.7.e. cz 2 + (d a) z b = 0. Consider a fractional linear transformation that maps the …xed point to 1. then it is conjugate to the translation 7! + 1. Now w + b is conjugate to w + 1. take A = 1=b. So 1 1 g h f h g (w) = w + 1 () () (g h) f (g h) 1 (w) = w + 1 () () ((g h) f ) (z) = (g h) (z) + 1: 98 . by a dilation g (w) = Aw. w ! w=A ! w=A + b ! w + Ab. III 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 2 3 4 5 6 7 8 1 . (b) the horizontal interval from (0. 4). (a) the arc of the parabola y = x2 . followed by the ver- tical interval from (2. 4). 4) to (2. Solution (a) Z Z 2 Z 2 2 2 x=t dx = dt. 0 t 2 2 2 72 y dx+x dy = = t dt+ t2 2t dt = : y = t2 dy = 2t dt 0 0 5 (b) Z Z 2 Z 2 2 2 x=t dx = dt 0 t 2 2 y dx + x dy = = 0 dt + t2 0 dt = 0 1 y=0 dy = 0 dt 0 0 Z Z 4 Z 4 2 2 x=2 dx = 0 dt 0 t 4 2 y dx + x dy = = t 0 dt + 22 dt = 16 y=t dy = dt 0 0 2 Z Z Z 2 2 2 2 y dx + x dy = y dx + x dy + y 2 dx + x2 dy = 16: 1 2 (c) Z Z 4 Z 4 2 2 x=0 dx = 0 dx 0 t 4 2 y dx + x dy = = y 0 dt + 02 dt = 0 1 y=t dy = dt 0 0 Z Z 2 Z 2 x=t dx = dt 0 t 2 y 2 dx + x2 dy = = 42 dt + t2 0 dt = 32 y = 4 dy = 0dt 0 0 2 Z Z Z y 2 dx + x2 dy = y 2 dx + x2 dy + y 2 dx + x2 dy = 32: 1 2 2 . 4). (c) the vertical interval from (0. 4).1. 0) to (0.III. 0) to (2.1 R 2 Evaluate y dx + x2 dy along the following paths from (0. followed by the horizontal interval from (0. 0). 0) to (2. 0) to (2. Note that the path integrals on 2 and 4 are zero because the edges are vertical.III. 0). (1.1.2 γ 1 γ γ 4 3 2 1 γ 1 Denote the square by D and note that xy dx = P dx + Q dy where P = xy and Q = 0. is the boundary of the square with vertices at (0.1. 0). hence by Green’s Theorem we have. 1). Z Z Z Z 1 Z 1 1 1 xy dx = xy dx + xy dx = x 0 dx x 1 dx = 0 = : 1 3 0 0 2 2 3 . Solution (A. Z Z ZZ Z 1 Z 1 @Q @P 1 xy dx = P dx+Q dy = dx dy = x dx dy = : @D D @x dy 0 0 2 Observe that = 1 + 2 + 3 + 4 where 1 and 3 are the bottom and top of the square while 2 and 4 are the last two sides taken in the order indicated by the order of the vertices in the statement of the problem (so the boundary is oriented counter clockwise). and (0.2 R Evaluate xy dx both directly and using Green’s theorem. Kumjian) VII. Then P and Q are continuously di¤erentiable on D and = @D. (1. where R 1). P (x. In this case. SolutionR Evaluate @D x2 dy directly. where D is the quarter-disk in the …rst quadrant bounded by the unit circle an the two coordinate axes.3 R Evaluate @D x2 dy both directly and using Green’s theorem.III. set @D = 1 + 2 + 3. Z Z 1 2 x=t dx = dt 0 t 1 x dy = = t2 0 dt = 0 1 y=0 dy = 0 dt 0 Z Z =2 2 x = cos t dx = sin t dt 0 t =2 2 x dy = = cos2 t cos t dt = 2 y = sin t dy = cos t dt 0 3 Z Z 1 x=0 dx = 0 dt 0 t 1 x2 dy = = 02 dt = 0 t y=1 dy = dt 0 3 Z Z Z Z 2 2 2 2 x dy = x dy + x dy + x2 dy = @D 1 2 3 3 R Now we evaluate @D x2 dy. y) = 0 and Q (x. y) = x2 .1. this time using Green’s theorem. Z ZZ 2 x dy = 2x dxdy = @D D ZZ Z 1 Z =2 x = r cos dx dy = r dr d 0 r 1 =2 x dxdy = =2 r cos rdrd = D y = r sin 0 =2 0 0 Z 1 Z =2 2 2 =2 r dr cos d = : 0 0 3 4 . this time using Green’s theorem.1. y) = y and Q (x. Z Z R x=t dx = dt R t R y dx = = 0 dt = 0 1 y=0 dy = 0 dt R Z Z x = R cos t dx = R sin t dt 0 t R2 y dx = = R sin t R sin t dt = 2 y = R sin t dy = R cos t dt 0 2 Z Z Z 2 2 2 R x dy = x dy + x2 dy = @D 1 2 2 R Now we evaluate y dx. Z ZZ y dx = dx dy = @D D ZZ Z R Z x = r cos dx dy = r dr d 0 r R = dx dy = = r dr d = D y = r sin 0 0 0 Z R Z 2 R = r dr d = : 0 0 2 5 . SolutionR Evaluate y dx directly. set = 1 + 2. In this case.4 R Evaluate y dx both directly and using Green’s theorem.III. y) = 0. P (x. where is the semicircle in the upper half-plane from R to R. 1. using Green’s theorem we have Z Z Z x dx = dx dy = Area D: @D D R For @D y dx.5 R R Show that @D x dy is the area of D. we have P = 0 and Q = x. we have P = y and Q = 0.III. Solution Using R in the two cases For @D x dy. while @D y dx is minus the area of D. using Green’s theorem we have Z Z Z y dx = dx dy = Area D: @D D 6 . (z = x + iy) z w z w is analytic for w 2 Cn . Express F 0 (w) as a line integral over . use uniform convergence of 1 1 1 1 1 = ! w z (w + w) z w (z w) (z (w + w)) (z w)2 as w ! 0. Get R P dx R Qdx F 0 (w) = 2 + .III.6 Show that if P and Q are continuous complex-valued functions on a curve . uniformly for z 2 . then R P dx R Qdy + . z = x + iy (z w) (z w)2 7 .1. Solution Di¤erentiate by hand. y) . (x. V U where F (x. y) = = @x @y @y @x ZZ @P @ @ @ @ = ( (x. y)) @F @ @ @x = @x + @x @F @ @ @y = @y + @y @ @ @ @ det JF (x. y) = ( (x. y)) dx dy @ @x @y @y @x U Using Green’s theorem 8 . and if F (x. then Green’s theorem holds for V . given by Z Z @ @ Pd = (P F) dx + dy . (x. y)) det (JF ) dx dy = @ @ V U F (x. in the sense that if the theorem holds for a bounded domain U with piecewise smooth boundary. The summand Q d is treated similarly.7 Show that the formula in Green’s theorem is invariant under coor- dinate changes. y) is a smooth function that maps U one-to-one onto another such domain V and that maps the boundary of U one-to-one smoothly onto the boundary of V . First note the change of variable formulae for line and area integrals.III. y) . @V @x @y ZZ Z@U Z Rd d = (R F ) det JF dx dy. and where JF is the Jacobian ma- trix R of F . y)). Use these formulae. y) = ( (x. (x. y) . (x. Solution ZZ ZZ @P @P d d = ( (x. y) .1. with R = @P=@ . Hint. Z Pd = @V Z Z @ @ @ @ = (P F ) dx + dy = P ( (x. y)) +P dx dy = @y @x @y@x @x @y @x@y U ZZ @P @ @P @ @ @P @ @P @ @ = + + dx dy = @ @y @ @y @x @ @x @ @x @y U ZZ @P @ @ @P @ @ = + dx dy = @ @y @x @ @x @y U ZZ @P = JF dx dy: @ U R Similar argument as above hold for Qd Here we use that Z Z @ @ Qd = (Q F ) dx + dy @V @U @x @y and replace R = @P=@ by R=@Q=@ to conclude that Z ZZ @Q Qd = d d : @V V @ 9 . (x. y) . y)) P + P ( (x. (x. y) . y) . y)) dx + dy = @U @x @y @U @x @y ZZ @ @ @ @ = P + P dx dy = U @y @x @x @y ZZ @ @ @2 @ @ @2 = P ( (x. (x. 1. Solution VII.8 Prove Green’s theorem for the rectangle de…ned by x0 < x < x1 and y0 < y < y1 (a) directly.8a VII. We have Z Z Z Z Q dy = P dx = Q dy = P dx = 0: 1 2 3 4 Integrate around the rectangle 10 .1. y0 ) to 1 and proceed in this way in counterclockwise direction.8b III III IV R II IV S II T I I (a) Make …gure and set the linesegment between (x0 . and (b) using the result for triangles.1. y0 ) and (x1 .III. y) Q (x0 . y0 )] dx = y0 Z y1 Z y1 Zx0x1 Z x1 = Q (x1 . The integrals on the diagonal are in di¤erent directions and will cancel Z (P dx + Q dy) = @R Z Z = (P dx + Q dy) + (P dx + Q dy) = @S @T ZZ ZZ @Q @P @Q @P = dx dy + dx dy = S @x @y T @x @y ZZ @Q @P = dx dy: R @x @y 11 . y0 ) dx = y0 y0 x0 x0 Z x1 Z y1 Z x1 Z y1 = P (x. y) dy P (x. y) dy P (x. y) dy Q (x0 . y)] dy [P (x. y1 ) dx + P (x. y1 ) P (x. y0 ) dx+ Q (x1 . ZZ @Q @P dx dy = R @x @y Z y1 Z x1 Z x1 Z y1 @Q @P = dx dy dy dx = y0 x0 @x @y Z y1 Zx0x1 y0 = [Q (x1 . y1 ) dx Q (x0 . y) dy = x0 y0 x0 y0 Z Z Z Z = P dx+ Q dy + P dx+ Q dy = 1 2 3 4 Z = (P dx + Q dy) : @R (b) Write R = S [ T . (d) y dx x dy. If it is not. If it is. Solution We compute the partial derivatve because if h exist we have that dh = P dx + Qdy where @P=@y = @Q=@x. (a) P =x Q=y @P @Q @y =0 =0 R x2 R@x y2 P dx = 2 + g (y) Qdy = 2 + f (x) x2 +y 2 )h= 2 (b) P = x2 Q = y5 @P @Q =0 =0 R@y 3 @x R y6 P dx = x3 + g (y) Qdy = 6 + f (x) 2x3 +y 6 )h= 6 (c) P =y Q=x @P @Q =1 =1 R@y R@x P dx = xy + g (y) Qdy = xy + f (x) ) h = xy (d) P =y Q= x @P @y = 1 @Q @x = 1 The line integral is not independent of path.III. (a) x dx + y dy.2. we use Greens’theorem Z ZZ (y dx x dy) = 2 dx dy = 2 : jzj=1 jzj<1 12 . (b) x2 dx + y 5 dy. …nd a closed path around which the integral is not zero. …nd a function h such that dh = P dx + Q dy.1 Determine whether each of the following line integrals is indepen- dent of path. (c) y dx + x dy. 2 Show that the di¤erential ydx + xdy . x2 + y 2 is closed. 0) . y) 6= (0.2. Solution y x P = x2 +y 2 Q= x2 +y 2 @P (x2 +y2 )+y(2y) y 2 x2 @Q (x2 +y2 ) x(2x) y 2 x2 @y = (x2 +y 2 )2 = (x2 +y 2 )2 @x = (x2 +y 2 )2 = (x2 +y 2 )2 Because @P=@y = @Q=@x. (x. 13 . Show that it is not independent of path on any annulus centered at 0.III.the di¤erential is closed and we can use Green’s Theorem I P dx + Q dy = jzj=r I I y x 1 = dx+ 2 dy = 2 y dx + x dy = x2 + y 2 x + y2 r jzj=r jzj=r ZZ 1 1 = 2 (1 + 1) dx dy = 2 2 r2 = 2 6= 0 r r jzj<r Thus the line integral is not independent of path. @D consists of the two circles with opposite ori- entation. as noted above. both P and Q are continuously di¤erentiable on D = D [ @D. Solution (A. Green’s Theorem applies and we obtain Z ZZ @Q @P P dx + Qdy = dxdy = 0. r2 be given so that a < r1 < r2 < b. 14 .3 γ1 γ2 D Let r1 . Observe that D is a bounded domain with piecewise smooth boundary @D = 1 [ 2 where 1 denotes the circle fz 2 C : jzj = r1 g with clockwise orientation while 2 de- notes the circle fz 2 C : jzj = r2 g with counter clockwise orientation.III. equation ( ) holds and jzj=r P dx + Qdy is independent of the radius r.2. Hence. More- over. Show directly using Green’s theorem that jzj=r P dx + Qdy is independent of the radius r. for a < r < b. Kumjian) III.2. Z I I P dx + Qdy = P dx + Qdy P dx + Qdy: @D jzj=r2 jzj=r1 H Hence.3 Suppose P and Q are smooth functions on the annulus fa < jzj < bg that Hsatisfy @P=@y = @Q=dx. We must prove that I I P dx + Qdy = P dx + Qdy: jzj=r1 jzj=r2 Let D denote the annulus fz 2 C : r1 < jzj < r2 g. for a < r < b. @D D @x @y and since. so the …rst part applies. 15 . Obtain the result for the annulus above by parameterizing the circles jzj = r0 and jzj = r1 by 0 (t) = r0 e2 it . Let 0 and 1 be two closed paths in D such that the straight line segment Rfrom 0 (t) to 1 R(t) lies in D for every parameter value t. Use the straight lines to deform 0 to 1 . The straight line segments joining 0 (t) to 1 (t) are radical are in the annulus. a1 6 t 6 b1 . De…ne s (t) = s 1 (t) + (1 s) 0 (t). Then P dx + Q dy = P dx + Q dy. 1 (t) = r1 e2 it .4 Let P and Q be smooth functions on D satisfying @P=@y = @Q=@x. Solution Use the theorem on page 81. The theorem on page 81 applies.2.III. 0 6 t 6 1. Use this to give another 0 1 solution to the preceding exercise. 0 6 s 6 1. 5 Let 0 (t) and 1 (t). when < 0 (t) < . it deform 0 to 1 continuously in D. 1 (t)) s 1 (t) + (1 s) 0 (t) max ( 0 (t) .III. r1 (t)) sr1 (t) + (1 s) r0 (t) max (r0 (t) .2. be paths in the slit annulus fa < jzj < bg n ( b. a) from A to B. Solution Write 0 (t) = r0 ei 0 (t) . Suggestion. 0 6 t 6 1 . and each s is a path in D from A to B. r1 (t)) < b and < min ( 0 (t) . 0 s 1: This does the trick. r0 (t). (Use the fact that (t) = Arg 0 (t) is continuous on the slit annulus) Also 1 (t) = r1 ei 1 (t) . Deform separately the modulus and the principal value of the argument. 0 (t) continuous. 1 (t)) < 16 . We see that s (t) belongs to the slit annulus for 0 s 1 and 0 t 1 since for every 0 5 1 we have a < min (r0 (t) . Write down explicitly a family of paths s (t) from A to B in the slit annulus that deforms continuously to 1 . Then consider s (t) = [sr1 (t) + (1 s) r0 (t)] ei[s 1 (t)+(1 s) 0 (t)] . 0 t 1. III.2. Therefrom the curve (t) may be deformed to any point of this region. Suppose j (t)j = r for 0 6 t 6 1 and (0) = (1) = 1 where a < r < b. 0 t 1. The second case is when (t) does not enclose the origin. write (t) = rei j (t) . Add multiples of 2 to the j ‘s. 17 . Follow hint. Then start by …nding a subdivision 0 = t0 < t1 < < tn = 1 such that arg (t) has a continuous deter- mination on each interval tj 1 t tj . 0 t 1. Deform by s (t) = rei[(1 s) (t)+2 sm] . Note (1) = 2 m for some integer m. Hint.6 Show that any closed path (t). Reduce to the case where j (t)j j (0)j is constant. in the annulus fa < jzj < bg can be deformed continuously to the circular path (t) = (0) e2 imt . since (t) is continuous. for some integer m. For this case (t) the counter of a region that lies entirely inside the annulus. Note j (tj ) j+1 (tj ) is a integral multiple of 2 . In particular it can be continuously deformed to the point (0) = (t) = (0) e2 imt when m = 0. For the closed path (t) enclosing the origin we can deform (t) to a curve with constant modulus. obtain (t) continuous for 0 6 t 6 1 such that (0) = 0 and (t) = rei (t) . tj 1 6 t 6 tj . Solution We …rst note that there are two cases. we must eventually end where we started. S1 . j R we will have d = 2 on the j ´s we get by moving from 0 to @F . Three paths must be disjoint. Show that @F is a …nite union of a closed paths in D. Either a vertex in @F have one edge coming in and one out. while @S d = 0 for any other square in PR the grid. and every point of E has distance at least 5 from every point of CnD not in E. that is the edges that form @F . : : : . the corresponding integrals along the edge cancel. or it have two coming in and two out. Lay down a grid of squares in the plane with side length . then there is a closed path in D such that d 6= 0.7 Show that if 0 and 1 lie in di¤erent connected components of the complement C nD of D in the extendedR complex plane. Since d = 2 . P R except for vertices with Rfour edges. We have thus @F d = 2 . The hypothesis means that there are > 0and a bounded subset E of CnD such that 0 2 E. Solution Proceed as in the hint.) j 18 . where we sum over squares. and that @F d = 2 . @F D.and let F be the union of the closed squares in the grid that meet E or that border on a square meeting R E. Sn in F . m . (In fact.III. R Assume that 0 Ris the centre of one of the squares S0 in the grid.2. Then this is the sum over the integrals over the edges of the Sj ´s that have F on oneRside and CnF on the other. Now note how edges @F can meet. Then @S0 d = 2 . with a closed path. By starting on following edges. and making a left turn at vertices where 4 edges meet. S0 . we must have d 6= 0 for one of the j ‘s. Note that by construction. If two squares in F are ?????. Call the paths 1 . : : : . Thus @Sj d = 2 . Hint. y) = 3x2 y y 3 (c) Use Cauchy-Riemanns equations page 83.1 For each of the following harmonic functions u. u (x. the conjugate harmonic functions of u. y) = x y (c) u (x. u (x. y) = x3 3xy 2 du = (3x2 3y 2 ) dx 6xy dy dv = 6xy dx + (3x2 3y 2 ) dy v (x. y) = sinh x cos y du = cosh x cos y dx sinh x sin y dy dv = sinh x sin y dx + cosh x cos y dy v (x. and …nd v. y) = x y du = dx dy dv = dx + dy v (x. y) = x2 +y 2 Solution (a) Use Cauchy-Riemanns equations page 83. y u (x.3. y) = x3 3xy 2 (d) u (x. y) = x2 +y 2 19 . (a) u (x. y) = x2 +y 2 2xy x2 +y 2 y(2y) 2xy x2 y 2 du = (x2 +y 2 )2 dx + (x2 +y 2 )2 dy = (x2 +y 2 )2 dx + (x2 +y 2 )2 dy 2 2 dv = (xy2 +yx2 )2 dx 2xy (x2 +y 2 )2 dy x v (x.III. …nd du. y) = sinh x cos y y (b) u (x. y) = x + y (b) Use Cauchy-Riemanns equations page 83. …nd dv. u (x. y) = cosh x sin y (d) Use Cauchy-Riemanns equations page 83. analytic such that u = Re '. Solution Write h = u + iw where u.3.III. w = Re . Assume that h (z) = f (z) + g (z) where f (z) and g (z) are analytic on D. To show the oposite direction. so thus there are '. By the Theorem on page 83 both u and w have harmonic conjugate in D. then h (z) = f (z) + g (z). g = 12 [' i ]. (since D is star-shaped).2 Show that a complex-valued function h (z) on a star-shaped domain D is harmonic if and only if h (z) = f (z) + g (z). u = (' + ') =2. w= + =2. gives 1 h= 2 '+'+i +i = 12 [' + i ] + 1 2 '+i : Take f = 12 [' + i ] . w are harmonic. where f (z) and g (z) are analytic on D. 20 . Then h = Re f + Re g + i (Im f + Im g) and from Cauchy-Riemanns equations follows that @ 2 (Re f ) @ 2 (Re g) @ 2 (Im f ) @ 2 (Im g) + +i + @x2 @x2 dx2 @x2 @ 2 (Re f ) @ 2 (Re g) @ 2 (Im f ) @ 2 (Im g) + + + i = @y 2 @y 2 @y 2 @y 2 @ 2 Im f @ 2 Im g @ 2 Re f @ 2 Re g @ 2 Im f @ 2 Im g = + +i + + @x@y @x@y @x@y @x@y @y@x @y@x @ 2 Re f @ 2 Re g +i =0 @y@x @y@x so h (z) is harmonic. Show that any harmonic function on D has a harmonic conjugate on D. then from z1 to little disk. a). Integral is well-de…ned. an annulus slit along the negative real axis. continuous. 21 .3.III. Solution Integrate the C–R equations we have Z z Z Z 1 @u @u @u @u v (z) = dr + r d = dx + dy z0 r@ @r @y @x integrate along r interval. Or map the slit annulus to a rectangle by w = Log z. and de…ne v (z) explicitly as a line integral along the path consisting of the straight line from c to jzj followed by the circular arc from jzj to z. Perception is same as getting from z0 to z1 . Fix c between a and b. harmonic and is harmonic conjugate for u. Suggestion.3 Let D = fa < jzj < bg n ( b. then interval. By appropriate choice of C. use the polar form of the Cauchy-Riemann equations to convert the r–derivative of u to a –derivative of v. a). a).4 (From Hints and Solutions) Let u (z) be harmonic on the annulus fa < jzj < bg. Solution u has harmonic conjugate v1 on the annulus slit along ( b. For the identity. 22 . Arg z also jumps by a constant across the slit. and u C log jzj has a harmonic conjugate v1 C log jzj on the annulus. and also a harmonic conjugate v2 on the annulus slit (a.3. and also constant below the slit. 2 0 @r where r is any …xed radius. a < r < b. Since v1 v2 is constant above the slit ( b. v1 jumps by a constant across the slit. b). v1 C Arg z is continuous across the slit ( b. Show that there is a constant C such that u (z) C log jzj has a harmonic conjugate on the annulus. a).III. Show that C is given by Z 2 r @u C= rei d . Show that if a harmonic function u on a domain D has a conjugate har- monic function v on D. Solution ! t = dx . Further. this is v (Q) v (P ).3. If goes from P to Q. dy . which shows that the integral is independent of path in D.III. 23 . dx ds . then the integral giving the ‡ux is inde- pendent of path in D.5 The ‡ux of a function u across a curve is de…ned to be Z Z @u ds = ru n ds. @n where n is the unit normal vector to and ds is arc length. ! ds ds n = dy ds . the ‡ux across a path in D from A to B is v (B) v (A). R R R @u ru! n ds = @u dy @x ds @u dx @y ds ds = @y dx + @u @x dy |{z} = R @v @v R @x dx + @y dy = dv ( ) C–R equations. Show that if f (z) has the mean value property with respect to circles. 0 < r < 1. Then h (z) is the average of h (z) over any disk fjz z0 j = r0 g. as de…ned above.4.1 Let f (z) be a continuous function on a domain D. What is then is that if h (z0 ) is the average of h (z) for any circle fjz z0 j = rg. Solution Suppose that f satis…es the mean value property. Express dx dy in polar coordinates at z0 and integrate …rst with respect to . ZZ ZZ 1 1 x = x0 + r cos dx dy = r dr d 0 r R f (z) dx dy = f (z) dx dy = A D0 R 2 D0 y = y0 + r sin 0 2 Z 2 Z R Z R Z 2 Z R 1 1 1 = 2 f z0 + rei r dr d = 2 f z0 + rei d r dr = 2 f (z0 ) r d R 0 0 R 0 0 R2 0 Z R 2f (z0 ) R 1 r2 = r dr = 2 f (z0 ) = f (z0 ) : R2 0 R 2 0 Another note: It’s better to use something other than 1=A as A was already used as the circle average. then f (z) has the mean value property with respect to disks. 0 < r < r0 . 24 . A = R2 and parametrizing in polar coordinates.III. and use the mean value property with respect to circles. Note: This MVP for area is not exactly what is de…ned as the MVP for disks. then f (z0 ) = A D0 f (z) dx dy. We have. Z 2 1 f (z0 ) = f z0 + rei d 2 0 for all z0 2 D and r > 0 such that Br (z0 ) D. that is if z0 2 D and D0 is a disk RR centered at z0 with area A and 1 contained in D. Let D0 = fz 2 C : jz z0 j Rg D. 4.III. Solution The polar form of Cauchy-Riemann’s equations (from Exercise II.3. = r : @r r@ @ @r Now we have Z 2 Z 2 @u @v r z0 + rei d = z0 + rei d =0 0 @r 0 @ 25 .3.2 Derive (4.8).2) from the polar form of the Cauchy-Riemann equations (Exercise II.8) @u 1 @v @u @v = . Next suppose that f is continuous and that it has the mean value property. Suppose …rst that f is an a¢ ne function. t 2 I be given. b) has the mean value property if s+t f (s) + f (t) f = . Show that any continuous function on I with the mean value property is a¢ ne. equation ( ) holds for t = k=2m for all k = 0. m are nonnegative integers with k 2m . We do this by induction on m. d 2 I be given with c < d. Now suppose that the assertion holds for some integer m 0 (that is. We must prove that f is a¢ ne. By the inductive hypothesis we may assume k = 2j + 1 for some integer j = 0. f has the mean value property. We claim that for any c. For m = 0. 1 and the assertion is obvious. We will …rst prove that equation ( ) holds for t = k=2m where k. : : : . that is. and t 2 [0.3 A function f (t) on an interval I = (a. This inductive step entails showing that equation ( ) holds for t = k=2m+1 for any nonnegative integer k with k 2m+1 . : : : . Now let s. b) and let f : I ! R be given. s. d 2 I. we have t = (t0 + t1 ) =2 and 1 t = ((1 t0 ) + (1 t1 )) =2. t 2 I: 2 2 Show that any a¢ ne function f (t) = At + B has the mean value property. Solution (A. then s+t s+t 1 1 f (s) + f (t) f =A +B = (As + B) + (At + B) = : 2 2 2 2 2 Hence. Then setting t0 = j=2m and t1 = (j + 1) =2m . with c < d.III. thus. Kumjian) Let I := (a. B 2 R such that f (t) = At + B for all t 2 I. 2m ). there are A. 2m 1 (for if k were even we could rewrite t as j=2m for some integer j with 0 j 2m ). 26 . 1] we have f (tc + (1 t) d) = tf (c) + (1 t) f (d) : So let c.4. we have t = k = 0. there are unique constants A and B such that f (x) = Ax + B for all x 2 [c. It is now straightforward to verify that for any c. 27 . f (tc + (1 t) d) = 1 =f ((t0 c + (1 t0 ) d + (t1 c + (1 t1 )d) = 2 1 = (f (t0 c + (1 t0 ) d) + f (t1 c + (1 t1 ) d)) = by the mean value property 2 1 = (t0 f (c) + (1 t0 ) f (d) + t1 f (c) + (1 t1 ) f (d)) = by inductive hypothesis 2 = tf (c) + (1 t) f (d) : So we have proved that equation ( ) holds for t = k=2m where k. m are nonnegative integers with k 2m . Since such numbers are dense in [0. d. d 2 I. d] (take A = (f (c) f (d)) = (c d) and B = f (c) Ac ). with c < d. 1 tc + (1 t) d = ((t0 c + (1 t0 ) d + (t1 c + (1 t1 ) d)) 2 hence. 1] the claim follows by the continuity of f . The desired result now follows by observing that A and B do not depend on c. with area element d and unit outward normal vector n. get R R! R R ! ::: V ! n d surface = : : : r V volume. R R @u R R @u ::: @ d ::: @ d ! ! j x x0 j=R 1 R j x x0 j= 0 @u @ : : : ud_ is constant. Get directional derivative of Strokes formula. Solution Prove MV theorem for harmonic functions in Rn . ! j x x 0j= 28 . Better: Apply stroke’s theorem to shell f 0 < < 1 g. R R R VolumeR get : : : ru! n d surface = : : : r2 ! n d volume = 0. Apply the Gauss divergence theorem ZZ ZZZ F nd = r Fd @Br Br to F = ru. For A 2 R3 and r > 0. with volume element d . x = unit vector.4.III. let Br be the ball of radius r centered at A. Apply to ru = ! v. get R @u ! R @u ! R R R @2u ! @r (R x ) d @r (r x ) d = 0 0 (r x ) drd (! S @r2 x ). Let d = area measure. Hint. as follows u (R! x) ! R R @u ! ! u 0 = 0 @r (r x ) dr.4 Formulate the mean value property for a function on a domain in R3 . and show that any harmonic function has the mean value property. and let @Br be its boundary sphere. Assume that u (z0 ) = c > b fore some z0 2 D. This show that a u b on D. then a u b on D. Then it follows by the maximum principle that u c on D. Show that if a u b.5. 29 . but this is a contradiction to the asumption that u is continuously extended to @D. Solution Since D [ @ is a compact set we have that u attains both maximum and minimum values. Then by the maximum principle u c0 on D. Assume that u (z0 ) = c0 > a for seme z0 2 D. For u we have that b u a on @D.1 Let D be a bounded domain. but this contradicts that u is continuously extended to @D with u a on @D. and let u be a real-valued harmonic function on D that extends continuously to the boundary @D. Thereby u b on D.III. So u a on D. equivalently a u b on D. For z 2 C with jzj = r we have by the triangle inequality n jz n + j jz n j + j j = rn + = (rn + ) ei' = j + j: so the claim is proven. Kumjian) We claim that the maximum modulus of z n + over the disk fz 2 C : jzj rg is rn + and this is attained at = rei where = (2k + ')=n for k = 0. n 1. 1.5. What is the maximum modulus of z n + over the disk fjzj rg? Where does z n + attain its maximum modulus over the disk? Solution (A. r > 0 and = ei' .III. By the maximum modulus principle the maximum modulus occurs on the boundary fz 2 C : jzj = rg. 30 . : : : .2 Fix n 1. Seip) If p (z) has no zeros.5. Note that if p (z) = an z n + an 1 z n 1 + + a0 . p(z) is a bounded entire function. so p(z) is bounded for jzj R by continuity. Then p(z) is bounded when jzj R. unless p (z) is a constant. Solution It is su¢ cient to show that any p (z) has one root. that any polynomial p (z) of degree n 1 has a zero. if we denote by m (R) the maximum of 1=p (z) on the circle jzj = R. p (z) is constant.III. Also. Thus. a contradiction which implies p (z) must have a zero. 1 1 Also. 31 . which means 1=p (z) = 0. for by division we can then write p (z) = (z z0 ) g (z). jp (z)j ! 1. By the maximum principle j1=p (z)j m (R) when jzj R. with g (z) of lower degree. p (z) 6= 0. then 1=p (z) is an entire function. and so 1=p (z) is not an entire function. by applying the maximum principle to 1=p (z) on a disk of large radius. Solution (K. then m (R) ! 0 when R ! 1. which must be constant. Thus.3 Use the maximum principle to prove that the fundamental theorem of algebra. then as jzj ! 1. This is impossible. This follows as an 1 a0 p (z) = z n an + + + n : z z 1 Assume p (z) is non-zero everywhere. which can only happens if f (z) is a constant.4 Let f (z) be an analytic function on a domain D that has no zeros on D. Solution (a) If jf (z)j attains its minimum on D. by the maximum principle. (b) Show that if D is bounded. (a) Show that if jf (z)j attains its minimum on D. 32 . Assume that jf (z)j does not attain its minimum on @D. and if f (z) extends con- tinuously to the boundary @D of D. then f (z) attains its minimum on @D. Then jf (z)j attains its minimum on D and it follows by (a) that f is constant so jf (z)j attains its minimum on @D.III.5. then j1=f (z)j attains its maximum on D. (b) Since D [ @D is compact and f (z) is continuous it follows that jf (z)j attains both maximum and minimum value on D [ @D. then f (z) is constant. Now let " ! 0. By the maximum principle.III. Solution (K. we have " C (z + 1) f (z) M (R 1)" for all jzj R. Seip) We assume that jf (z)j C for Re z > 0 and jf (iy)j M . when Re z 0. Hint. Solution Now we consider the function (z + 1) " f (z) in the right half-plane. For " > 0 set " F" (z) = (1 + z) f (z) : 33 . we have " (z + 1) f (z) M for jzj R. we assume that jf (z)j C in this domain. Suppose that f (z) extends continuously to the imaginary axis and satis…es jf (iy)j M for all points iy on the imaginary axis. On a semidisk of radius r in the right half-plane we have " C (z + 1) f (z) jr 1j" For r = R su¢ cient large. Show that jf (z)j M for all z in the right half-plane. because f (z) is bounded here.5 Let f (z) be a bounded analytic function on the right half-plane. On the imaginary axis we have that " (z + 1) f (z) M because jf (iy)j M for all points iy on the imaginary axis. when Re z 0.5. consider (z + 1) " f (z) on a large semidisk. For " > 0 small. we have jf (z)j M for all z in the right half-plane. 34 . Re z > 0 we have jF" (z)j M. thus jf (z)j M. Thus for arbitary z.Then jF" (iy)j M and F" Rei C R " M for su¢ ciently large R. or jf (z)j M (1 + jzj)" : This holds for every " > 0. Remark. to ob- tain jg (z)j 6 M + ". jzj < R. 35 . Solution Set " g (z) = (z + 1) f (z) : Then jg (z)j 6 jf (z)j becauce jz + 1j > 1 for all z in the right half-plane. Suppose that lim sup jf (z)j M as z approaches any point of the imaginary axis. Then g (z) ! f (z) as " ! 0.1).6 Let f (z) be a bounded analytic function on the right half-plane.III. 0 < Re z 6 : Apply the maximum principles to the domain fzj jzj < R. Show that jf (z)j M for all z in the half-plane. Take R large so that jg (z)j 6 M for jzj > R. This is a technical improvement on the preceding exercise for students who can deal with lim sup (see Section V.5. The lim sup condition implies that there is > 0 such that jg (z)j 6 M + ". Re z > g. jf" (z)j 6 M on @D. We have that M + =2 M + =3 j(z z1 )" : : : (z zn )" j and j(z z1 )" : : : (z zn )" j M+ M Note that g (z) extends continuously to @D and that g (z) is analytic in D. Then [ (z aj )" ] f (z) = f" (z) satis…es f" (z) is continuous in D [ @D. In the case that there is only one exceptional point z = 1. Then (z aj )" . Let " ! 0. Let g (z) = (z z1 )" : : : (z zn )" f (z) where " is chosen so that 8z 2 D [ @D.5. Show that jf (z)j M on D. an . consider the function (1 z)" f (z). : : : . Let z1 . Now suppose that jf (z)j M for all z 2 @D except fz1 . Suppose there are a …nite number of points on the boundary such that f (z) extends continuously to the arcs of @D separating the points and satis…es f ei M there. and since M + =2 jg (z0 )j = j(z0 z1 )" : : : (z0 zn )" j f (z0 ) (M + ) M+ we have that 36 . We have that jz aj j ! 0 as z ! aj for 1 j n. obtain jf (z)j 6 M on D. : : : . By de…nition jf (z)j C. Hint.7 Let f (z) be a bounded analytic function on the open unit disk D.III. where 1 j n is analytic " (take analytic Q branch). zn 2 @D be all the points for which f (z) does not extend continuously. By the maximum modulus principle. : : : . Solution Let C = supz2D jf (z)j. jf" (z)j 6 M on D. then there is a z0 2 D such that jf (z0 )j = M + with 0< C M. zn g and to the contrary M < C. Solution Let the points be a1 . M + =2 M + =3 (M + ) jg (z0 )j sup j(z z1 )" : : : (z zn )" j jf (z)j M M+ z2@D M which is a contradiction. 37 . z = eiw . Modify proof idea of Exercise III.8 Let f (z) be a bounded analytic function on a horizontal strip in the complex plane.III. so it is mapped to right half-plane by w = i log z. Solution Assume strip is 2 < Im z < 2 . Suppose that f (z) extends continuously to the boundary lines of the strip and satis…es jf (z)j M there.5.5.5 to get jgj 6 M on the half–plane. Then g (w) = f (eiw ) hold on right half–plane. Find a conformal map of the strip onto D and apply Exercise 7. Hint. 38 . 6 M at all point of iR except at w = 0. Show that jf (z)j M for all z in the strip. Sup- pose that u (z) is bounded below on D. Hint. there is also R0 > 0 such that for R > R0 u (z) + log jzj 0. 39 . Since u is bounded below. hence for all z 2 D such that jzj > . Take R > 0 so large that u (z) + log jzj > 0 for jzj > R. 0 < jzj < . D 6= C. to obtain u (z) > 0 on D. Solution (K. > 0) u (z) + log jzj ". and let u (z) be a harmonic function on D that extends continuously to the boundary @D. we can …nd and so that ( . and that u (z) 0 on @D. Take > 0 such that u (z) > " for z 2 D. Since " and can be chosen arbitarily small.5. Seip) We may assume 0 2 @D. Show that u (z) 0 on D. z 2 D \ fjzj = Rg . and consider functions of the form u (z) + log jzj on D \ fjzj > "g. Let > 0. then let " ! 0. < jzj < R.III. Suppose 0 2 @D. Solution (From Hints and Solutions) Let " > 0. Let ! 0.9 Let D be an unbounded domain. By the maximum principle. Applying the maximum principle to u (z) log jzj in D \ f < jzj < Rg. we get u (z) " log jzj. z 2 D \ fjzj = g for arbitary " > 0. the result follows. u (z) + log jzj > " + log for z 2 D. By continuity of u at 0. w (z) ! +1 as z ! z0 . z 6= z0 . assume R > 1. By the maximum principle. Show that u (z) 0 on D. Let " ! 0. u > 0 (@D) n fz0 g. Let u (z) be a har- monic function on D that extends continuously to each boundary point of D except possibly z0 .5. z0 2 @D. Then w (z) > " log (1=R) at all points of (@D) n fz0 g. then take " > 0 small. Or: consider w (z) = u (z) + log (1= jz z0 j). The map ' (z) = 1= (z z0 ) maps D onto an unbounded domain. Solution D hold. get w (z) > 0 on D. and let z0 2 @D. 40 . to which the result of Ex- ercise 9 applies. Suppose u (z) is bounded below on D. u > M on D. w (z) > " log (1=R) on D. and that u (z) 0 for all z 2 @D. Assume jz z0 j 6 R on D.10 Let D be a bounded domain.III. call it M . Consider fm + ni : u (m + ni) = M g. Solution Suppose u (z) attains its maximum at m + ni. follow same approach as in book. It have a point with largest m. This is a …nite set. u attains value M at that point. 41 . A point m + ni of E is an interior point of E if its four immediate neighbors m 1 + ni. Show that a harmonic function on a bounded set of lattice points attains its maximum modulus on the boundary of the set. Then u (m + ni) is the average of its values at its four neighbors. For complex–valued case. so all their values must coincide with M.5. A function E is harmonic if its value at any interior point of E is the average of its values at the four immediate neighbors. u = c at some point of boundary of E. m + ni i belong to E. Remark: Should just consider real-valued u. Otherwise m + ni is a boundary point of E.III. and show it attains its maximum and minimum at boundary points of E.11 Let E be a bounded set of integer lattice points in the complex plane. This point have a neighbor in the boundary of E. 1). ‡ux across [0. @@y = @@x ) = 2y x f (z) = + i = 2z iz = (z i) z Streamlines are straight line with slope 1=2. Solution V = (2. i] = (i) (0) = 2. Find the velocity potential (z). = 2x + y. 1] = (1) (0) = 1. the stream function (z). and the complex velocity potential f (z) of ‡ow. Sketch the streamlines of the ‡ow. ‡ux across [0.6. i] on the imaginary axis. 42 . @@x = @@y .1 Consider the ‡uid ‡ow with constant velocity V = (2.III. V = r . 1] on the real axis and across the interval [0. Determine the ‡ux of the ‡ow across the interval [0. 1). 6. r r where ur and u are the unit vectors in r and directions. 43 . and consider the vector …eld given in polar coordinates by V (r. respec- tively. sin u : > 0. is the velocity vector …eld of a ‡ow b) = Re ( log z + arg z) stream function = = Im ( log z + arg z) = arg z log jzj complex velocity of a point f (z) = ( i ) log z c) Flux = of around a circle centred at 2 .III. = 1. (b) Find the stream function (z) and the complex velocity potential f (z) of the ‡ow. h ( ) = )V =r = log r + is harmonic locally. ) is the velocity vector …eld of a ‡uid ‡ow. ) = r ur + r u r = @@r ur + 1r @@ u = r ur + r u Get @@r = r .2 Fix real numbers and . Solution a) V (r. = log r + h ( ) 1@ r@ = h0 ( ) = r . (c) Determine the ‡ux of the ‡ow emanating from the origin. When is 0 a source and when is 0 a sink? (d) Sketch the streamlines of the ‡ow in the case = 1 and = 1. (a) Show that V (r. Note: and are only locally de…ned. h0 ( ) = . sin u : > 0. ) = ur + u . and …nd the velocity potential (z) of the ‡ow. d) = 1. 6. where (r. Determine the ‡ux of the ‡ow emanating from the origin. and real axis. Solution = cosr = r cos r2 = Re 1 z = Re z 2 = Im 1 z = sin r R jzj Since is single-valued. and there is no source or sink at 0. ) = (cos ) =r.3 Consider the ‡uid ‡ow with velocity V = r . which correspond to the curves Im w = constant get modi…ed to tangent to real line at 0 = z (1). z = 1=w. Show that the streamlines of the ‡ow are circles and sketch them. Set w = 1=z. ‡ux = jzj=" d = 0. r = @@r ur + 1r @@ u = cos r2 ur sin r2 u. 44 . these streamlines are the curves Im (1=z) = constant.III. and this correspond to arcs of circles from 1 to 1 in the z–plane. z = 1 is a sink. (since = ew ).6. The stream function is = ImH(log (z 1) = H(z + 1)) = arg ((z 1) = (z + 1)) . then is not single-valued. Solution is the composition of an analytic function w = (z 1) = (z + 1) and the har- monic function log jwj . Hence streamlines are arcs of circle from 1 to 1. and = Re log ((z 1) = (z + 1)). w = log .4 Consider the ‡uid ‡ow with velocity V = r . where z 1 (z) = log : z+1 Show that the streamlines of the ‡ow are arcs of circles and sketch them. jz 1j=" d = jz 1j=" d arg (z 1) = 2 = ‡ux H H emanating from +1. Set = (z 1) = (z + 1). so is harmonic. The line Im w = constant in the w plane correspond to rays issuing from 0 in the plane.III. 45 . jz 1j=" d = jz 1j=" d arg (z + 1) = 2 = ‡ux ema- nating from 1. Determine the ‡ux of the ‡ow emanating from each of the singularities at 1. since FLT´s maps circles to circles. including the circle through 1. z = 1 is a source. C1 = C2 require A + B B = z [ A B ]. A (y ) + B [arg (ez 1) ]. Map the strip to a half-plane by = ez and solve a Dirichlet problem with constant boundary values on the three intervals in the boundary separating sinks and sources. Map to half-plane = ez . Sketch the streamlines of the ‡ow.6.III. A [ (y ) + 2 [arg (e 1) ]] = A [ ( + y) + 2 arg (ez 1)] 46 . but no source at 1 . Hint. Hint : Map the half-plane by. Find the stream function (z).5 Consider the ‡uid ‡ow in the horizontal strip f0 < Im z < g with a sink at 0 and equal sources at 1. c3 = 0. Find the complex vector …eld V (z). Find the stream function (z) and the velocity vector …eld V (z) of the ‡ow. Sketch the ‡ow lines. A arg + B arg ( 1) + C. Solution Consider ‡uid ‡ow in a horizontal strip. requires A+ B+C = 0 ) C = (A + B). f0 < Im z < g with a sink 0. arg = arg ez = y. Ay+C+B arg (ez 1). 6.6 For a ‡uid ‡ow with velocity potential (z).III. What is the stream function of the conjugate ‡ow? What is the complex velocity potential of the conjugate ‡ow? Solution Stream function of conjugate ‡ow in direction r is . we de…ne the conju- gate ‡ow to be the ‡ow whose velocity potential is the conjugate harmonic function (z) of (z). Complex velocity potential is g (z) = i = if (z) = i ( + i ) 47 . 7 Find the stream function and the complex velocity potential of the conjugate ‡ow associated with the ‡uid ‡ow with velocity vector ur =r. which are circles centered in origin. Speed = jV (z)j = jf 0 (z)j = 1= jzj. = log jzj.III. Do the particles near the origin travel faster or slower than particles on the unit circle? Solution Flow ur =r. Sketch the streamlines of the conjugate ‡ow. Particles travel faster near z = 0. Streamlines of the conjugate ‡ow are = log jzj = constant . Complex velocity potential of conjugate ‡ow is i log z. r = jzj = constant.6. = arg z. 48 . The velocity potential is = log r + .III.) Solution This is the vector …eld from 2d. (The steam function of the conjugate is always the negative of the velocity potential of the ‡ow.8 Find the stream function of the conjugate ‡ow of 1 V (r. The conjugate ‡ow has velocity potential = log r = Re ((i 1) log z). (See Exercise 2d. 49 .6) Streamline of conjugate ‡ow are orthogonal to the stream of the ‡ow (Exercise 2d) and velocity vector …eld. The stream function of the conjugate ‡ow is then Im ((i 1) log z) = + log r = .6. Hence V = r ( log r + ). ) = ( ur + u ) : r Sketch the streamlines of both the ‡ow and the conjugate ‡ow on the same axis. Conjugate ‡ow is the rotation by 90 of the velocity vector …eld of the ‡ow. so the origin is a sink of both ‡ows in this case. See Ex. ) T2 = B A = T/ 1 T2 . 1 on bottom T1 = A + B half and is 0 on the interval ( 1. Find and sketch the isothermal curves for the heat distribution. where T2 = (T1 + T3 ) =2. 1) is held at temperature T2 .7.1 Find the steady-state heat distribution u (x. these are the curves 1+z Arg = konsant 1 z then (1 + z) (1 z) = 1 + 2i Im z jzj2 and Im z = konstant. Solution (z) = 2 (Arg (1 + z) Arg (1 z)) has temp +1 on top. y) in a laminar plate corresponding to the half disk fx2 + y 2 < 1. get u = (T1 T2 ) (z) + T2 B = T2 u = (T1 T2 ) 2 [Arg (1 + z) Arg (1 z)] + T2 The isothermal curves are the curves where u = konstant. Consider the steady-state heat distribution for the full unit disk with the top held at temper- ature T1 and the bottom temperature T3 . 1). Hint. y > 0g.III. where the semi- circular top edge is held at temperature T1 and the lower edge ( 1. Take u = A + B. 1 jzj2 50 . where the two edges on the coordinate axis are grounded (that is. x > 0. y > 0g. Solution (z) = 2 (Arg (1 + w (z)) Arg (1 w (z))).III. y) for the electric …eld for a con- ducting laminar plate corresponding to the quater-disk fx2 + y 2 < 1. Hint. = 0 on the edges). Find and sketch the equipotential lines and the lines of force for the electric …eld.7.2 Find the potential function (x. Use the conformal map = z 2 and the solution to the preceding exercise. and the semicircular edge is held at constant potential V1 . 1 = V1 2 (Arg (1 + z 2 ) Arg (1 z 2 )) 51 . y) for the electric …eld for a con- ducting laminar plate corresponding to the unit disk where the boundary quater-circles in each quadrant are held at a constant voltages V1 . u0 (w) = 1 Arg w. Example: V2 = 1. w ( 1) = 1.7. w ( i) = 1.III. Solution Let w = i (1 z) = (1 + z). we get a system we can back solve. w (1) = 0. V1 = V3 = V4 = 0. Su¤ers to …nd a har- monic function u (w) in upper half-plane with boundary values constant on each interval. u1 (w) = 1 Arg (w 1). Any such function is a linear combination of u 1 (w) = 1 Arg (w + 1). V2 . this solution is 2 (z) = 1 1 Arg i 11+zz . and if = A 1 u 1 + A0 u0 + A1 u1 + C. V3 and V4 . w (i) = 1. and then (z) = vj j (z). 4 (z). Map the disk to the upper half- plane by w = w (z) and consider potential functions of the form Arg (w a). 1 (z). Hint. 3 (z). 52 . z ! w (z) maps unit disk to upper half–plane. 8 8 > > A 1 + A 0 + A 1 + C = V3 > > C = V2 < < A0 + A1 + C = V3 A1 = V1 V2 ) > > A1 + C = V1 > > A0 = V4 V1 : : C = V2 A 1 = V3 V4 That does it-just plug in.3 Find the potential function (x. and we P can get other elementary solutions. as above. want a b + c = 100 ) : 8 a b+c=0 < a = 50 b = 50 : c=0 (z) = 50 Arg (sin z + 1) 50 Arg (sin z 1). Hint.4 Find the steady-state heat distribution in a laminar plate corre- sponding to the vertical half-strip fjxj < =2.III. Use w = sin z to map the strip to the upper half-plane. 53 . where the ver- tical sides at x = =2 are held at temperature T0 = 0 and the bottom edge ( =2.7. =2) on the real axis is held at temperature T1 = 100. Make a rough sketch of the isothermal curves and the lined of heat ‡ow. y > 0g. and make use of harmonic functions of the form Arg (w a). Solution 8 < a+b+c=0 Try = a Arg (w + 1) + b Arg (w 1) + c. y) = ax + b is harmonic. no heat passes through the bottom edge. Solution u (x. y > 0g.7.5 Find the steady-state heat distribution in a laminar plate corre- sponding to the vertical half-strip fjxj < =2. @u@y =0 T1 T0 2 a + b = T0 a= ) 2 a + b = T1 b = T0 +T 2 1 u = 1 (T1 T0 ) x + T0 +T 2 1 . y) is parallel there to the x axis. Hint. and the bottom edge ( =2. =2) on the real axis is insulated.III. so the gradient ru of the solution u (x. where the side x = =2 is held at constant temperature T0 . 54 . Try linear functions plus constants. that is. the side x = =2 is held at constant temperature T1 . III. 1) is held at temperature T1 . Take z = sin w. 1) is held at temperature T0 .6 Find the steady-state heat distribution in a laminar plate corre- sponding to the upper half-plane fy > 0g. w = sin 1 (z). and the interval (1. Hint. this has desired mapping property. w = w (z) : upper half–plane ! vertical semi strip.7. where we take branch sin z with values in the vertical semi-strip. Make a rough sketch of the isothermal curves and the lines of heat ‡ow. Solution Let z = z (w) map the vertical half-strip to the upper half-plane with z ( =2) = 1 and z ( =2) = 1. Note : This conformal map was used in Exercise 4. where the interval ( 1. 1) is insulated. Use the solution to Exercise 5 and the conformal map from Exercise 4. 55 . Then solution is composition of sine function with 2 (T1 T0 ) Re w + T0 +T 2 1 . 2 1 T0 +T1 1 Get u = (T1 T0 ) Re sin (z) + 2 . the interval ( 1. III. y. 0) . . 0. Show that F is conservative.7 The gravitational …eld near the surface of the earth is approxi- mately constant. z) space and the surface of the earth is represented by the plane where z = 0 (the ‡at earth theory). of the form F = ck. @x @y @z we have that F is conservative.7. and …nd a potential function for F . (0. c) = (0. y. r = c k Since @ @ @ r F = . z) = cz. where k is the unit vector in the z direction in (x. 0. Solution ! (x. 56 . III. y. z) = 1 r 57 . Solution ! ! r = ur2r = r .8 Show that the inverse square force …eld F = ur =r2 on R3 is con- servative. Find the potential function for F . (x. and show that is harmonic.7. show that the function 1=rn 2 is harmonic on Rn n f0g. Solution n 1 1 2 n 1 rn 2 = n 2 = u = r = (x21 + : : : + x2n ) 2 (1 x n) 2 +:::+x2 2 n @u n @xj = 1 2 (x21 + : : : + x2n ) 2 2xj = (2 n) xj r n n @2u n n 1 @x2j = (2 n) r + (2 n) xj 2 (x21 + : : : + x2n ) 2 2xj = 2 n (2 n)( n)x2j rn + r n+2 2 n (2 n)n P 2 u = n rn r n+2 xj = 0 this shows that u is harmonic on Rn n f0g P @u P x F = u= e = (2rnn) xj ej = r2n n1 r j ej = (2rnn)~1ur @xj j ! F = (2r) rrn = rcnu 1 R P F d = rnc 1 Area (s) = (n 2) Area(S) rn 1 58 .III.7.9 For n 3. Find the vector …eld F that has this function as its potential. IV 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 2 3 4 5 6 7 8 1 . i]. And integrate along each side in the triangle = 1 + 2 + 3 We parametrizize 1 = [0. then Re (1 t + it) = 1 t and dz = ( 1 + i) dt.1. thus 2 (t) = 1 t + it. 0]. for 0 t 1. then Z Z 1 Re z dz = (0) ( i) dt = 0: 3 0 2 . thus 1 (t) = t. then Re (t) = t and dz = dt Z Z Z 1 1 Re z dz = Re (t) dz = t dt = : 1 1 0 2 We parametrizize 2 = [1. Re (i (1 t)) = 0 and dz = i dt. 1]. for 0 t 1. 0 < x < 1g. then Z Z 1 1 1 Re z dz = (1 t) ( 1 + i) dt = + i: 2 0 2 2 We parametrizize 3 = [i.1 1 2 3 P L K F KKK FFF Let be the boundary of the triangle f0 < y < 1 x. then. with the usual counterclockwise R orientation. for 0 t 1. R EvaluateR the following integrals (a) Re z dz (b) Im z dz (c) z dz Solution a) We begin do split the contur into three parts so that = 1 + 2 + 3.IV. thus 3 (t) = i (1 t). 1 0 Z Z 1 1 1 Im z dz = t ( 1 + i) dt = + i. 0 2 2 Z2 Z 1 1 Im z dz = (1 t) ( i) dt = i: 3 0 2 thus Z 1 Im z dz = : 2 c) For details see a) Z Z 1 1 z dz = t dt = . 0 2 Z Z 11 z dz = (1 t + it) ( 1 + i) dt = 1 2 0 Z Z 1 1 z dz = (i (1 t)) ( i) dt = : 3 0 2 thus Z z dz = 0: 3 .And now we take the parts in the integral together Z Z Z Z i Re z dz = Re z dz + Re z dz + Re z dz = 1 2 3 2 b) For details see a) Z Z 1 Im z dz = 0 dt = 0. (a) z dz (b) z m dz (c) z m jdzj Solution Set z = ei .1. and thus dz = iei d . (a) Z Z 2 Z 2 2 m im i i(m+1) ei(m+1) z dz = e ie d = i e d = = 0 0 m+1 0 1 0. 2. m 6= 1 = ei2 (m+1) 1 = m+1 2 i m= 1: (b) Z Z 2 Z 2 m im i z dz = e ie d = i ei(1 m) d = 0 0 2 ei(1 m) 1 0. with the usual counterclockwise orientation. m 6= 0 z jdzj = e d = = ei2 m 1 = 0 im 0 im 2 m = 1: 4 .2 1 2 3 P L K 111 LLL Absolutbeloppet i c) nagot a •r f el Let be the unit circle fjzj = 1g. for m = 0. 1. m 6= 1 = = ei2 (m 1) 1 = 1 m 0 1 m 2 i m = 1: (c) Z Z 2 2 m im eim 1 0.IV. : : :. R m EvaluateR the following Rintegrals. where 0 2 . 1. R mEvaluate theR following integrals. = e 1 = m+1 = 2 1 m 2 iR . m = 1: 5 .IV.3 1 2 3 P L K 111 Let be the circle fjzj = Rg. 2 iR . R m for m = 0. 1. m 6= 1. (a) jz j dz (b) jz m j jdzj (c) z dz Solution Set z = Rei . m 6= 1. m = 1. (a) Z Z 2 Z 2 m m im i m+1 2 jz j dz = R e iRe d = iR ei d = Rm+1 ei 0 = Rm+1 ei2 1 = 0: 0 0 (b) Z Z 2 Z 2 m jz j jdzj = m im R e iRe i d =R m+1 d = Rm+1 [ ]20 = Rm+1 (2 0) = 2 Rm+1 : 0 0 (c) Z Z 2 Z 2 2 m m im i m+1 i(1 m) m+1 ei(1 m) z dz = R e iRe d = iR e d =R = 0 0 1 m 0 Rm+1 i2 (1 m) 0. : : :. where 0 2 . 0. with the usual counterclockwise ori- entation. and thus dz = iRei d and jzj = R. 2. and apply Green‘s theorem. dz = dx + idy.IV.1. 6 . then Z z dz = 2i Area (D) : @D Solution Substitute z = x iy.4 1 2 3 P L K LLL Show that if D is a bounded domain with smooth boundary. IV. as z = 1+eit = 1+cos t+ i sin t where 0 t 2 .5 1 2 3 P L K 138 111 KKK Show that I ez dz 2 e2 : jz 1j=1 z+1 Solution We begin with parametrizize the circle jz 1j = 1. For the nominator in the integrand ez = (z + 1) we have jez j = eRe z+i Im z = eRe z ei Im z = eRe z e2 . M L estimate gives I ez dz 2 e2 : jz 1j=1 z + 1 7 .1. And for the denominator in the integrad we have j1 + zj = j2 + cos t + i sin tj Re (2 + cos t + i sin t) = 2 + cos t 1. Thus ez jez j e2 M= = = e2 : 1+z jz + 1j 1 Becauce the integrationcontour is a circle with radius 1 we have that L = 2 . since Re z 2 on the circle. M L estimate gives p H Log z 2 log R p log R jzj=R dz 6 2 R = 2 2 . R>e : jzj=R z2 R Solution For the integrand Log z=z 2 . Becauce the integrationcontour is a circle with radius R we have that L = 2 R.IV. with the nominator Log z = log R + i with the principalargument we have p p p p Log R log2 R + 2 log2 R + 2 log2 R + log2 R 2 log R = 6 6 = : R2 R2 R2 R2 R2 there the last inequalitys follows by given fact R > e that gives 2 < log2 R. R>e : z2 R2 R 8 .6 1 2 3 P L K 111 KKK Show that there is a strict inequality I p log R Log z dz 2 2 .1. } .1. 9 . m > 1. th root of w R L = 2 R. R > 1. then f (z) dz = R R f (z) dz + f (z) dz 6 m0 length ( 0 )+mlength ( 1 ) = (m0 m) length ( 0 )+ 0 1 mlength ( ) < mlength ( ) R Thus if f (z) dz = ML. jz m j = Rn . )M= ik=m with strict inequality unless z = R e|2 {z Rn Rm 1 . then jf (z)j = M on . In this case. = 0 [ 1. R > 1. get zn jzj=R z m 1 dz < Rn Rm 1 2 R= 2 Rn+1 Rm 1 . 1 zm 1 6 1 Rm 1 .7 1 2 3 P L K Show that there is a strict inequality I zn 2 Rn+1 m dz .IV. m 1. n 1: jzj=R z 1 Rm 1 Solution R If jf (z)j 6 m0 < m on an arc 0 of the circle. n > 0. 1. Take real parts. Show that f (z) = cz for some constant c with modulus jcj = M . R2 then 0 < 2 M . If have strict inequality somewhere.IV. Since f ei iei 6 M . have 0 Re f ei iei d = 2 M . Solution H Multiply f by a unimodular constant. ) f e ie i i M .8 1 2 3 P L K LLL Suppose the continuous function f ei on the unit circle satis…es R f ei M and jzj=1 f (z) dz = 2 M . f (z) iM z. we have Im f e ie di i 0. Re f e ie d 6 f e i i i 6 M . 10 . ) f (z) = cz for a constant. jcj = M . assume jf (z)j 6 M . jzj=1 f (z) dz = R2 i R2 0 f e d = 2 M . f ei ie i M . We conclude that Re f ei iei d M . di¤erentiate h by hand. Solution For the analyticity. and …nd H 0 (w). z w is analytic on the complement of .6). Show that Z h (z) H (w) = dz. (See i Exercise III. The Deriv- R h(z) h(z) ative is H 0 (w) = lim 1w z (w+ w) z w dz = w!1 R h(z) w R lim 1w (z (w+ w))(z w) dz = (zh(z)dz w)2 w!1 11 .1.IV.9 1 2 3 P L K Suppose h (z) is a continuous function on a curve .1. w 2 Cn . R R R (a) z 4 dz (b) ez dz (c) cos z dz (d) sinh z dz Solution The integrals are all independent of path and can be evaluated by …nding a primitive. for a path that travels from i to i in the right half-plane.1 1 2 3 P L K K 111 LLL KKK 246 Evaluate the following integrals.2.IV. (a) Z i z5 ( i)5 ( i)5 2 5i z 4 dz = = = : 5 i 5 5 5 (b) Z i ez dz = [ez ] i =e i e i = 1 ( 1) = 0: (c) Z i cos z dz = [sin z] i = ei( i) e i( i) ei( i) e i( i) = sin ( i) sin ( i) = = 2i 2i 2 (e e ) = =i e e : 2i (d) Z i sinh z dz = [cosh z] i = cosh ( i) cosh ( i) = e i+e i e i +e ( i) = = 0: 2 2 12 . and also for a path from i to i in theR left half-plane. Solution In right half–plane use f (z) = Log z = log jzj + i . and also for a path from i to i in the left half-plane. where < < as a primitive.IV. evaluate 1=z dz for a path that travels from i to i in the right half-plane. get Z 1 dz = [Log z] i i = log j ij + i log j i j i = i: z 2 2 In left half–plane use f (z) = log0 z = log jzj + i . get Z 1 3 dz = [log0 z] i i = log j ij + i log j i j + i = i: z 2 2 13 .2.2 1 2 3 P L K K 111 LLL 246 R Using an appropriate primitive. For each path give a precise de…nition of the primitive used to evaluate the integral. where 0 < < 2 as a primitive. then z m has a primitive on Cn f0g. Solution z m+1 The primitive is m+1 .2. which is analytic for z 6= 0 and m 2 Z.3 1 2 3 P L K 111 LLL KKK Show that if m 6= 1.IV. m 6= 1. 14 . 0]. (c) Evaluate Z dz p . 1) or using dz z = z c) 15 . z2 1 where is the path from 2i to +2i in D counterclockwise around the circle jzj = 2. 1]. 1]. z 2 1 = t2 2z + z 2 .IV. b) p p Log z + z 2 1 is the composition of z+ z 2 1 and Log w on Cn ( 1. But values of the branch are > 0 on (1.) Solution p D = Cn ( 1. (a) p Show that z + z 2 1 omits the negative real axis. d p 1=2 so it’s analytic. where we use everywhere the constant branch z+ z 2 1 z2 1 z2 1 p d a az a of z 2 1. z = (t2 + 1) =2.2. By the chain role.4 1 2 3 P L K p Let D = Cn ( 1. z 2 1 a) p p z + z 2 1 = t (t > 0) ) z 2 1 = t z. 0] on the real axis. (b) p p Show that Log z + z 2 1 is a primitive for 1= z 2 1 on D. by ????? it on (1. oriented counterclockwise. that is. the range of the function on D does not include any values in the interval ( 1. (d) Evaluate the integral above in the case is the entire circle jzj = 2. dz Log z + z 2 1 = z+p1z2 1 1 + 21 (z 2 1) 2z = p1 1+ p z = p 1 . ) It assumes no negative values. 2z = t2 + 1. 1) = D \ R. 1). and consider the branch of z 2 1 on D that is positive on the interval (1. (Note that the primitive is discontinuous at z = 2. ) z + z 2 1 !p 2+( 1) 3 = 2 3. jz 1j ! p p p p 3. haveLog z + z 2 1 ! p p log 2 + 3 + i Arg z + z 2 1 at z = 2. have arg z 1! . arg z + z 2 1 ! . jz 2 1j ! 3. have p p p p arg z 2 p1 ! + . and also Log 2i + i 5 Log 2i i 5 ?????. d) R dz p 2+i0 p 2 z 1 = Log z + z 2 1 2+i0 p At 2 + 0t . As z ! 2 + i0. As z ! 2+i0.R p 2i p p p dz = Log z + z 2 1 2i = Log 2i + i 5 Log 2i i 5 = i z2 1 p Use fact that p z 2 1 is positive imaginary p on positive imaginary axis. Values R dz coincide ) p 2+i0 p z2 1 = Log z + z 2 1 2+i0 = log jj log jj + i ( i) = 2 i 16 . 2 2 2 arg z + z 1 ! . ) z + z 2 1 ! 2 + ( 1) 3 = 2 3. 2. Solution (A. Conversely. Kumjian) Let f be an analytic function de…ned on a domain D.5 1 2 3 P L K Show that R an analytic function f (z) has a primitive in D if and only if f (z) dx = 0 for every closed path in D. by the Lemma on page 77. there is a continuously di¤erentiable function F on D @ @ such that dF = f dx + if dy. It follows that the real and imaginary parts of F satisfy the Cauchy-Riemann equations: @ @F @F @ Re F = Re = Re f = Im if = Im = Im F. It follows that the integral on the left is path independent for paths which are not necessarily closed and hence. suppose f (z) dx = 0 for every closed path 2 D. 102. Then by the Fundamental R Theorem of Calculus for Analytic Functions (Part I) we have f (z) dx = 0 for every closed path 2 D (since for anyRsuch path A = B). that is. we have Z Z f (z) dx + if (z) dy = f (z) dx = 0 for every close path 2 D. 17 .1) on p. namely F. @x @x @y @y @ @F @F @ Re F = Re = Re if = Im f = Im = Im F: @y @y @x @x Moreover. By formula (1. Hence.IV. @x F = f and @y F = if . Suppose …rst that f has a primitive in D. F 0 (z) = f (z) for all z 2 D. f has a primitive in D. show that Z 1 1 2 2 p e x =2 e itx dx = e t =2 .1 1 2 3 P L K LLL 2 By integrating e z =2 around a rectangle with vertices R.IV. uniformly for 0 6 y 6 1 as R ! 1. For more see the next exercise. 0. 2 = 1 e x =2 dx = 2 R1 R1 e (x +2ixt t )=2 dx = 2 2 (x+it)2 =2 1R e dx = 1 2 1 2 et =2 1 e x =2 e ixt dx R1 ) p12 2 2 1 e x =2 dx = e t =2 18 . it R.and sending R to 1. Solution R RR R t ( R+iy)2 =2 R t (R+iy)2 =2 RR 2 z 2 =2 x2 =2 @D e dz = R e dx 0 e idy+ 0 e idy+ R e (x+it) =2 dx = 0 p R1 2 e ( R+iy) =2 . Remark.3. This 2 shows that e x =2 is an eigenfunction of the Fourier transform with eigenvalue 1. 1 < t < 1: 2 1 Use the known value of the integral for t = 0. This shows that the n ’s form an orthogonal system of eigenfuctions for the (normalized) Fourier transform operator F with eigenvalues 1 and i. Further. F 4 is the identity operator. show that Z 1 n (x) m (x) dx = 0. and the inverse Fourier transform is given by (F 1 f ) (x) = (Ff ) ( x). n 6= m: 1 Remark. (b) By integrating the function it)2 =2 dn 2 e(z n e z dz around a rectangle with vertices R.3. (d) R R 00 00 Using n m dx = n n dx and (c). Use the identity from Exercise 1. Solution 19 . show that Z 1 1 p n (x) e itx dx = ( i)n n (t) .2 1 2 3 P L K We de…ne the Hermite polynomials Hn (x) and Hermite orthogonal functions n (x) for n 0 by 2 dn x2 x2 =2 Hn (x) = ( 1)n ex e . it R and sending R to 1. 1 < t < 1: 2 1 Hint.IV. and also justify and use the identity dn (x+it)2 1 dn (x+it)2 e = e : dxn in dtn (c) 00 Show that n x n + (2n + 1) n = 0. n (x) = e Hn (x) : dxn (a) Show that Hn (x) = 2n xn + is a polynomial of degree n such that is even when n is even and odd when n is odd. Thus F extends to a unitary operator on square-integrable functions. p12 ^ n (t) = ( i)n n (t) 2 2 PartRB 2 dn z2 0 = @DR ez =2 dz ne dz R 1 x2 =2 d z2 R 1 x2 =2 dn (x+it)2 nR1 2 n (x+it)2 1 e dx e dx = 1 e dx n e dx = ( i) 1 ez =2 dtd n e dx R 1 (x it)2 =2 d x2 R1 2 2 1 e dz n e dx = 1 ex =2 dzdn e (x+it) dx = R1 2 d 2 R1 2 2 e t =2 1 e itx ex =2 n e x dx + ( i)n 1 ex =2 dxdn e (x+it) dx 2 | dx {z } ( 1)n n (x) dn R1 e t2 =2 ^ (t) = in n e + ( i)n t2 d et 2 e x2 =2 dxe 2ixt n dt dxn 1 p p ^ n (t) = ( i)n (t) + ( i)n 2 d et e 2 2t2 = ( i)n 2 d t2 e dxn dxn dn x2 =2 Hn (x).pwe get ( i)n e t =2 ^ n (t) = 2 e t =2 in n (t).PartRA 2 dn z2 0 = @DR e(z it) =2 dz ne dz R 1 (x it)2 =2 dn x2 R1 2 2 dn x2 ! 1e dxne dx = 1 e t =2 e itx ex =2 dx ne dx = R t2 =2 1 itx n t2 =2 n^ e e ( 1) (x) dx = e ( 1) n (t) R 1 R 1 x2 =2n dn (x+it)2 R 1 x2 =2 dn 2 (x+it)\@DR ! 1 e dx n e dx = 1 e d(it)n e (x+it) dx = n R1 2 2 n R1 2 2 ( i) dtd n 1 ex =2 e (x+it) dx = ( i) dtd n 1 ex =2 e 2itx et dx = Z 1 n 2 p n p e x =2 e 2itx dx = 2 ( i)n dtd n e t = 2 e t =2 in n (t) 2 2 2 ( i) dtd n e t | 1 {z } p 2 e (2t)2 =2 If we set these equal. 1 n (x) e itx dx = 1 e x =2 ( 1)n ex dx 2 x2 ixt n e e d R 1 R 1 ( 1)n 1 Dxn ex ex =2 ixt dx = 1 e x =2 Dxn ex =2 ixt dx = 2 2 2 2 R1 x2 n (x it)2 =2 t2 =2 R1 2 e dx = (i)n 1 e x Dtn e(x it) =2 et =2 dx = 2 2 e Dx e 1 2 R1 2 2 2 R1 2 2 et =2 in Dtn 1 e x e(x it) =2 dx = et =2 in Dtn 1 e x =2 e ixt e t =2 dx = p 2 2 2 2 et =2 in Dtn e t =2 e t =2 20 . Hn (x) = ( 1)n ex 2 x2 n (x) = e dxn e = x2 d x2 x2 x2 H0 (x) = 1. H1 (x) = e dx e = e ( 2x) e = 2x x2 d2 x2 x2 x2 x2 H2 (x) = e dx2 e =e 4x2 e 2e = 4x2 2 R1 R1 2 dn Hn (x) = 2n xn + lower order polynomial. Solution (a) R1 R R 2 f (x)2 dx = f (z)2 dz 6 f ei d 1 0 R1 R 1 R2 2 1 f (x)2 dx = f (z)2 dz 6 f ei d R1 1 R 2 2 Add.IV. that X n cj ck X n jck j2 . : : : . show that Z 1 Z 2 X n 2 i d f (x) dx f e = c2k : 1 0 2 k=0 Hint.3. Hint. by from f e = c k e cm e im k=0 k=0 k=0 P P 2 P 2 P 2 (b) f (x)2 = ak x k + i b k x k = ak x k + bk xk R1 2 R1 P k 2 R1 P k 2 P k P f (x) dx = a k x dx+ b k x dx = a k x + bk x k = P 1 k 1 1 ck x 21 .k=0 j+k+1 k=0 with strict inequalityRunless the complex numbers c0 . 1 Start by evaluating 0 f (x)2 dx. show that Z 1 Z 2 X n 2 d jf (x)j dx f e i = jck j2 : 1 0 2 k=0 (c) Establish the following variant of Hilbert’s inequality.3 1 2 3 P L K Let f (z) = c0 + c1 z + + cn z n be a polynomial. cn are all zero. j. For the …rst inequality. get 2 1 f (x)2 dx =6 f ei d R2 i 2 P 2 n i 2 P n ik Pn 0 f e d = 2 c k . (a) If the ck ’s are real. apply Cauchy’s theorem th the function f (z)2 separately on the top half an the bottom half of the unit disk. (b) If the ck ’s are complex. hR P R1 P i 1 d d 1 ak eik 2 + 1 bk eik 2 " #1 R1 P 2 R1 P n P n xj+k+1 P n cj ck (c) 0 ( cj xj ) dx = 0 cj ck xj+k dx = cj ck j+k+1 = j+k+1 j.k=0 j.k=0 R1 R1 P 0 jj 6 0 jf (x)j dx 6 0 jf (x)j dx 6 2 2 jck j . so latter R2 2 R1 R2 integral is i 0 P ei ei ( ) d .k=0 0 k=0 @D R1 R1 R2 2 gives 0 P (z) log x dx 0 P (z)2 (log x + 2 i) dx+ 0 P ei 2 iei d = 0. must have2 R1 R1 0 jf (x)j = 0 jf (x)j . By integrating P (z)2 log z.k=0 j. Note that P ei d . R1 R 2 R Get 2 i 0 P (x)2 dx = i 0 P ei ei d . P (z)2 log z dz = 0. for 1 6 x 6 0. Thus 0 P (x)2 dx = 0 P ei ei ( ) 2d R1 R2 2 d Pn ) 0 P (x)2 dx 6 0 P ei 2 = jak j2 k=0 22 . j. around a keyhole contour. so f (x) = 0. for equality. for an appropriate branch of log z. and then f (x) 0 2 2 Suppose P (z) = a0 + a1 z + ::: + an z n is a polynomial. Pn R1 Pn R 6 aj ak 2 show that j+k+1 = P (x) dx jak j2 . write P (z) = P (0) + zQ (z).3. L R. divide by zP (z). M = max zP1(z) 1 Rn+1 . along the following lines. Solution P (z) is a polynomial with no zeros. Assume P (0) = 1 . when n = det P z=R ML 1 Rn ! 0. H 1 H P (z) H 1 Q(z) H 1 jzj=R z dz = zP (z) dz = zP (z) + P (z) dz = zP (z) dz2 = i jzj=R jzj=R jzj=R ! 0 by ML-estimate if deg P > 1. Write p (z) = 1 + zQ (z). if n > 1. and integrate around a large circle. this will lead to a contradiction if P (z) has no zeros. 23 .4 1 2 3 P L K Prove that a polynomial in z without zeros is constant (the fundamental theorem of algebra) using Cauchy’s theorem. if R ! 1. If P (z) is a polynomial that is not a constant.IV. Hint. Show that Area (D) sup jz f (z)j 2 : z2@D Length (D) Show that this estimate is sharp.IV.5 1 2 3 P L K Suppose that D is a bounded domain with piecewise smooth boundary. and use Exercise 4 in section 1. take D = D = unit disk.3. Solution R R [z f (z)] dz = zdz = 2iArea (D) R @D @D @D [z f (z)] dz 6 sup jz f (z)j Length (@D) z2@D ) sup jz f (z)j > Area(D) 2 Length(@D) z2@D To see its sharp. Consider @D jz f (z)j dz. get sup jz f (z)j = 1 z2@D Area(D) 2 Length(@D) =2 2 =1 24 . and that f (z) is analytic on D [ @D. and that R in fact there exist D and f (z) for which equality holds. Since f is continuous in the closed disk fjzj Rg. it must be uniformly continuous there (since the close disk is both closed and bounded). 105. So let " > 0 be given. Solution (A.IV. Hence.6 1 2 3 P L K H fjzj Rg and analytic Suppose f (z) is continuous in the closed disk on the open disk fjzj < Rg. Kumjian) H H To prove jzj=R f (z) dz = 0 it su¢ ces to show that jzj=R f (z) dz ". for every " > 0. we have " jf (z1 ) f (z2 )j < if jz1 z2 j < : 2 R For r 2 (0. there is a > 0 such that for all z1 . 25 . Then since H fr is analytic on a domain containing the closed disk fjzj Rg. we have jzj=R f (z) dz = 0 by Cauchy’s theorem. Now let r be chosen such that 1 =R < r < 1. Show that jzj=R f (z) dz = 0.3. 1). Then for all z with jzj = R we have jz zrj = R (1 r) < and so by the choice of we have " jf (z) fr (z)j = jf (z) f (rz)j < : 2 R Hence we have I I I " f (z) dz = (f (z) fr (z)) dz j(f (z) fr (z))j jdzj 2 R = ": jzj=R jzj=R jzj=R 2 R by the ML-estimate theorem on p. Hint. Approximate f (z) uniformly by fr (z) = f (rz). z2 in the closed disk. de…ne fr on the disk fz : jzj < R=rg by fr (z) = f (rz) for all z such that jzj < R=r. 4. H zn jzj=1 dz = 0: z 2 (c) The nominator has one zero at z = 0 inside jzj = 1.IV. thus by by Cauchy´s theorem. Then for any z0 2 D and positive integer m we have. Husingezthe Cauchy integral formula: (a) jzj=2 zz 1 dz. Let D be bounded domain with piecewise smooth boundary @D and let f be an analytic function de…ned in a domain which contains D. n 0 (f) jz 1 ij=5=4 (z 1)2 dz H sin z H (c) jzj=1 z dz (g) jzj=1 z2 (z2dz 4)ez H H (d) jzj=1 cosh z 3 z dz (h) dz jz 1j=2 z 2 (z 2 4)ez Solution We rearragnde the Cauchy integral formula. 26 . thus by by Cauchy´s theorem. thus by by Cauchy´s theorem. n 0 (e) jzj=1 zm dz. thus by by Cauchy´s theorem. H zn 2 i n jzj=2 dz = [z ]z=1 = 2 i: z 1 0! (b) The nominator has no zeros inside jzj = 1. Z f (z) 2 i dm dz = f (z) : @D (z z0 )m (m 1)! dz m z=z0 (a) The nominator has one zero at z = 1 inside jzj = 2. H sin z 2 i jzj=1 dz = [sin z]z=0 = 0: z 0! (d) The nominator has one tripel zero at z = 0 inside jzj = 1.1 1 2 3 P L K 111 LLL Evaluate H then following integrals. 1<m<1 H zn H Log z (b) jzj=1 z 2 dz. thus by by Cauchy´s theorem. and one zero at z = 2 inside jz 1j = 2. f (z) = Log z for all z 6= 0 with Arg z 6= . H Log z 2 i d jz 1 ij=5=4 2 dz = (Log z) = 2 i: (z 1) 1! dz z=1 (g) The nominator has one double zero at z = 0 inside jzj = 1. The nominator has one double zero at z = 1 inside jz 1 ij < 5=4. If m 1. Then f is analytic on a domain jz 1 ij 5=4. I ez 2 i dm 1 z 2 i m dz = m 1 e = : jzj=1 z (m 1)! dz z=0 (m 1)! Thus I ez 0 if m 0. thus by by Cauchy´s theorem. the nominator has a zero of order m at z = 0 inside jzj = 1. then the integrand z m ez de…nes an analytic function on C. If m 0. by Cauchy’s Theorem the integral must be zero. I dz 2 i 1 2 i 1 i i = + = + 2: jz 1j=2 z (z 2 4) ez 0! (z 2 4) ez z=0 0! z (z + 2) ez z=2 2 4e 27 . that is. H dz 2 i d e z i jzj=1 2 = = : z (z 2 4) ez 1! dz z2 4 z=0 2 (h) The nominator has one zero at z = 0. hence. hence by by Cauchy´s theorem. thus by by Cauchy´s theorem. thus by by Cauchy´s theorem. dz = 2 i jzj=1 zm (m 1)! if m 1: (f) Let f be the principal branch of the logarithm de…ned on the slit plane. H cosh z 2 i d2 jzj=1 dz = cosh z = i: z3 2! dz 2 z=0 (e) The nominator has one tripel zero at z = 0 inside jzj = 1. 2 1 2 3 P L K LLL Show that a harmonic function is C 1 . 0 28 . and note that @x u = Re f 0 . @y u = @u @y @v = @x = Re (f 0 ). that is. @x = Re ( if ). Now take successive derivatives.4. @v v = Im f = Re ( if ). a harmonic function has partial derivatives of all orders.IV. get m @ u @xp @y q a multiple of Re f is Re (if 0 ). Solution @ @ Write u = Re f . 29 .IV. which is the MVP. 0 2 whenever u (z) is harmonic in a domain D and the closed disk jz z0 j is contained in D.3 1 2 3 P L K LLL Use the Cauchy integral formula to derive the mean value property of harmonic functions.4. Solution H H2 f (z) d Write f (z0 ) = jz z0 j= z z0 dz = 0 f z0 + ei 2 . that Z 2 d u (z0 ) = u z0 + ei . z0 2 D. Let n ! 1. and letting n ! 1.@D) = d1 . L = length @D. By applying this estimate to f (z)n . and let z0 2 D. z2@D By ML–estimate.IV. 30 . Solution R Write f (z0 ) = 1 2 i f (z) @D z z0 dz . Use kf n k = kf kn . get jf (z0 )j 6 L 1=n kf k 2 d . get n jf (z0 )jn 6 2 1 d kf n k L. taking nth roots. Using the Cauchy integral formula.4. show that there is a constant C such that jf (z0 )j D sup fjf (z)j : z 2 @Dg for any function f (z) analytic on D [ @D. Letkf k = max jf (z)j. This provides an alternative proof of the maximum principle for analytic function. get jf (z0 )j 6 kf k. show that the estimate holds with C = 1. jf (z0 )j 6 2 1 d kf k L. Remark.4 1 2 3 P L K Let D be a bounded domain with smooth boundary @D. 1 jz z0 j 6 1 dist(z0 . Apply to f . take nth roots. Then set g = eu+iv where g is an entire function. for some constant C. Now we have that jgj = eu+iv = eu eiv = jeu j eiv = eu : We can se that becauce u is bounded above. then u is constant.IV.1 1 2 3 P L K 111 LLL Denna beh• over N og f ixas till Lite Show that if u is a harmonic function on C that is bounded above. by Liouville’s theorem g is constant. so is g bounded above. 31 . Express u as the real part of an analytic function. Hint. Solution Let u be the real part of the analytic function f = u + iv.5. where we know that u C. Now we now that g is both analytic and bounded. Thus f is constant. and expo- nentiate. then f (z) is constant. Solution If f (z) does not attain values in the disk jw cj < ".2 1 2 3 P L K 111 LLL Show that if f (z) is an entire function. and there is a nonempty disk such that f (z) does not attain any values in the disk. Then 1= (f (z) w0 ) 32 .5. and f (z) is constant. hence constant by Liouville’s theorem.IV. then 1= (f (z) c) is a bounded entire function. Let w0 be the center of the disk that f omits. We de…ne an entire function g by the formula 1 g (z) = for all z 2 C: f (z) z0 It follows that jg (z)j 1=r for all z 2 C. 33 . Solution (A.5. Kumjian) Let f be such an entire function. The desired result now follows immediately. then f (z) is constant. That is.IV. jf (z) z0 j r for all z 2 C.2 1 2 3 P L K LLL Show that if f (z) is an entire function. Then there is a disc D = Dr (z0 ) for some z0 2 C and r > 0 such that f (z) 62 D for all z 2 C. and there is a nonempty disk such that f (z) does not attain any values in the disk. Hence. g is a bounded entire function an so by Liouville’s Theorem g must be a constant function. If f does not attain values in the disk jw cj < ", then 1= (f c) is bounded, hence constant by Liouville‘s theorem, and f is constant. 34 IV.5.3 1 2 3 P L K LLL A function f (z) on the complex plane is double periodic if there are two periods w0 and w1 of f (z) that do not lie on the same line thought the origin (that is, w0 and w1 are linearly independent over the reals, and f (z + w0 ) = f (z + w1 ) = f (z) for all complex numbers z). Prove that the only entire functions that are doubly periodic are the constants. Solution The "lattice" of points mw0 +nw1 , m; n 2 Z are the corners of parallelograms that cover C, each parallelogram being a translate F + mw0 + nw1 of "the fundamental region" F as shown in the …gure: Figure IV.5.3 Thus each 2 C can be written = z + mw0 + nw1 with z 2 F . Since we assume f ( ) = f (z), it follows that f i bounded, thus a constant. 35 If jf (z)j 2 M for jzj 6 R. Suppose jw0 j, jw0 j 6 R. Let P = parallelogram with vertices 0; w0 ; w1 ; w0 + w1 , and suppose for z 2 P . Any z 2 C have form + mw0 + nw1 where 2 P . Thus jf (z)j = jf ( )j 6 M , so f (z) is bounded and entire. 36 IV.5.4 1 2 3 P L K Suppose that f (z) is an entire function such that f (z) =z n is bounded for jzj R. Show that f (z) is a polynomial of degree at most n. What can be said if f (z) =z n is bounded on the entire complex plane? Solution Apply the Cauchy estimates for f (m+1) (z) to disk jz z0 j < R, and let R ! 1, to obtain f (m+1) (z0 ) = 0. If f (z) is a polynomial of degree 6 m such that f (z) =z m is bounded near 0, then f (z) = cz m . 37 IV.5.5 1 2 3 P L K Show that if V (z) is the velocity vector …eld for a ‡uid ‡ow in the entire complex plane, and if the speed jV (z)j is bounded, then V (z) is a constant ‡ow. Solution Use V (z) = f 0 (z) (section III.6 p92), when f is entire. Get f 0 constant, by Liouville‘s theorem, i.e. V (z) is constant. 38 IV.6.1 1 2 3 P L K LLL Let L be a line in the complex plane. Suppose f (z) is a continuous complex-valued function on a domain D that is analytic on DnL. Show that f (z) is analytic on D. Solution Rotate and apply result in text. 39 IV.6.2 1 2 3 P L K Let h (t) be a continuous function on the interval [a; b]. Show that the Fourier transform Z b H (z) = h (t) e itz dt a is an entire function that satis…es jH (z)j CeAjyj ; z = x + iy 2 C; for some constants A; C > 0. Remark. An entire function satisfying such a growth restriction is called an entire function of …nite type. Solution The analyticity of H (z) follows from the theorem in the text. If jhj 6 M , take C = M (b a) and A = max (jaj ; jbj). We have je itz j = ety . jH (z)j 6 M (b a) max ety 6 M (b a) eAjyj a t b 40 IV.6.3 1 2 3 P L K Let h (t) be a continuous function on a subinterval [a; b] of [0; 1). Show that the Fourier transform H (z), de…ned as above, is bounded in the lower half-plane. Solution 41 IV.6.4 1 2 3 P L K Let be a smooth curve in the plane R2 , let D be a domain in the complex plane, and let P (x; y; ) and Q (x; y; ) be continuous complex-valued func- tions de…ned for (x; y) on and 2 D. Suppose that the functions depend analytically on for each …xed (x; y) on . Show that Z G ( ) = P (x; y; ) dz + Q (x; y; ) dy is analytic on D. Solution Reduce to Riemann integral and apply theorem in the text. 42 IV.7.1 1 2 3 P L K Find an application for Goursat’s theorem in which it is not patently clear by other means that the function in question is analytic. Solution Answer to this is not known (by me). 43 IV.8.1 1 2 3 P L K LLL Show from the de…nition that @ @ @ @ @z z = 1; @z z = 0; @z z = 0; @z z = 1: Solution We use the …rst-order di¤erential operators (page 124) @ 1 @ @ @ 1 @ @ = i and = +i : @z 2 @x @y @z 2 @x @y We have that @ 1 @ @ 1 z = i (x + iy) = (1 + 1) = 1 @z 2 @x @y 2 @ 1 @ @ 1 z = +i (x + iy) = (1 1) = 0 @z 2 @x @y 2 @ 1 @ @ 1 z = i (x iy) = (1 1) = 0 @z 2 @x @y 2 @ 1 @ @ 1 z = +i (x iy) = (1 + 1) = 1 @z 2 @x @y 2 44 IV.8.2 1 2 3 P L K LLL Compute @@z (az 2 + bz z + cz 2 ). Use the result to determine where az 2 + bz z + cz 2 is complex-di¤erentiable and where it is analytic. (See Problem II.2.3) Solution By the Leibnitz rule and the results of Exercise 1, the z –derivative is bz+2cz. The function is complex di¤erentiable at z if and only if bz + 2cz = 0. If b = c = 0, the function is entire. Otherwise this locus is either a point f0g or a straight line through the origin, and there is no open set on which the function is analytic. 45 IV.8.3 1 2 3 P L K LLL Show that the Jacobian of a smooth function f is given by 2 2 @f @f det Jf = : @z @z Solution @u @v @x @x @u @v @u @v Jf = det @u @v = @x @y @y @x @y @y 2 2 @f 2 @f 2 @(u+iv) @z @z = 1 4 @x i @(u+iv) @y 1 4 @(u+iv) @x + i @(u+iv) @y 2 2 2 2 1 @u @v @v @u 1 @u @v @v @u = 4 @x + @y + @x + @y 4 @x @y + @x + @y h i 1 = 4 2 @u @v @x @y @v @u 2 @x @y + 2 @u @v @x @y @v @u 2 @x @y = @u @v @x @y @v @u @x @y = det Jf 46 IV.8.4 1 2 3 P L K LLL Show that @2 @2 @2 + = 4 : @x2 @y 2 @z@ z Deduce the following, for a smooth complex-valued function h. (a) h is harmonic if and only if @ 2 h=@z@ z = 0. (b) h is harmonic if and only if @h=@z is conjugate-analytic. (c) h is harmonic if and only if @h=@ z = 0. (d) If h is harmonic, then any mth order partial derivative of h is a linear combination of @ m h=@z m and @ m h=@ z m . Solution @2 @ @ @ @ @2 2 @ 2 @ @2 @2 @2 4 @z@ z = @x i @y @x + i @y = @x2 i @x@y i @y@x + @y 2 = @x2 + @y 2 (a) clear @2h @h (b) h harmonic , @ z@z =0, @z is analytic @2h @ @h (c) h harmonic , @z@ z =0, @z @z = 0. Since @h @z = @@hz , this ????? @h @z is @h analytic , @ z is conjugate analytic. @ k+l h (d) If h is analytic, then @z k @ z l = 0 unless l = 0. Only non-vanishing deriva- @mh k+l tive is @zm = h (z). If h is conjugate analytic, ????? @@zk @zhl = 0 unless l = 0. m If h is harmonic, then h = analytic + conjugate analytic, so all derivatives k l of h vanish except for @@zhk = @@zhk . 47 R (f) jz 1 ij=5=4 (zLog1)z2 dz = 21!i dz d Log zjz=1 = 2 i z1 z=1 = 2 i R d e z ( e z )(z2 4) e z (2z) (g) jzj=1 z2 (z2dz 4)ez = 2 i dz z2 4 =2 i (z 2 4)2 z=0 z=0 2 i4 i = 16 = 2 R dz 1 1 (h) jz 1j=2 z(z 2 4)ez =2 i (z2 4)ez + 2 i (z+2)ez = z=0 z=2 1 2 i 4 + 8e2 = i 12 1 + 1 4e2 48 . 5 1 2 3 P L K LLL With dz = dx idy.IV.8. show that for a smooth function f (z) that @f @f df = dz + : @z @z Solution df = @f @x dx + @f @y dy @f @f @z dz + @z dz = 12 @fi @f @x @y (dx + idy) + 21 @f @x + i @f @y (dx idy) = h i 1 @f 1 @f 1 @f 1 @f i @f @f @f @f 2 @x dx + 2 @y dy + 2 @x dx + 2 @y dy + 2 @x dy @y dx @x dy + @y dx = df 49 . get @D z dz = 2i D dxdt = 2iArea 50 .8.IV. take f = 1. @D f (z) g (z) dz = 2i D @E (f g) dz = 2i D f gdz @g @f @g Using the Leibnitz rule.6 1 2 3 P L K Show that if D is a domain with smooth boundary. and if f (z) and g (z) are analytic on D [ @D.4.4. g (z) = z.4). obtain @@z (f g) = RRf @z = f g R f @z + g @z = 0 To get Ex1. Solution R RR @ RR By (8. then Z ZZ f (z) g (z) dz = 2i f (z) g 0 (z) dx dy: @D D Compare this formula with Exercise 1. 3 3 The term O jzj has Taylor series 0 + O jzj . @@ zf2 (0) = 0.8. @f @z (0) = 0. That ????? it. @z@ z (0) = 1. z 2 . f (0) = 0. 51 . z. is given by @f @f 1 @2f @2f f (z) = f (0)+ (0) z+ (0) z+ 2 2 (0) z + 2 (0) jzj2 +o jzj3 : @z @z 2 @z @z@ z Solution Since a smooth function f is a linear combination of the functions 1. z 2 . through the quadratic terms.IV.7 1 2 3 P L K LLL Show that the Taylor series expansion at z0 = 0 of a smooth function f (z). for jzj2 = z z 2 @2f are harmonic. For instance. it su¢ ces to check the formula for these six functions. @f@z (0) = 0. z. jzj2 and a reminder term O jzj3 . The formula holds fro each of these. = @z (0). @z @w @ z @w @z @ @h @w @h @ w (h w) = + : @z @w @z @ w @z Solution Suppose w (0) = 0. b = @ z (0) then h (w (z)) = a z + z + O jzj2 + b z + z + O jzj2 + O jwj2 = a z + a z + b z + b z + O jzj2 = a + b z + (a + b ) z + O jzj2 This gives. @h 2 @w @w b = @ w (0) and w (z) = w + w + O jzj . a = @w@h (0). @ @h @w @h @ w (h w) = + .8.IV. h (0) = 0. write h (w) = aw + bw + O jwj2 .8 1 2 3 P L K LLL Establish the following version of the chain rule for smooth complex- valued functions w = w (z) and h = h (w). @h w @h @w @z (0) = a + b = @w @z + @@hw @w @z @ w=@ z @h @w = @w @z + @@hw @@wz @h w @h @w @h @w @h @w @h @ w @z (0) = a + b = @w @z + @w = @w @z + @ w @z @z |{z} @ w=@z 52 . then @h@ zw = 0 + 0 = 0 so h w is analytic. (h w)0 (z) = @z @ @h @w h w = @w @z + 0 = h0 (w (z)) w0 (z) 53 .IV.8. then (h w) (z) and (h w)0 (z) = h0 (w (z)) w0 (z). Solution If the functions h and w are analytic.9 1 2 3 P L K LLL Show with the aid of the preceding exercise that if both h (w) and w (z) are analytic. in the sense that it is equal to 0 away from w.10 1 2 3 P L K Let g (z) be a continuously di¤erentiable function on the complex plane that is zero outside of som compact set. Show that ZZ 1 @g 1 g (w) = dx dy. Solution Apply Pompeeiu‘s formula to a large disk fjzj < Rg. we obtain ZZ 1 @ 1 g (w) = g (z) dx dy: C @z z w Thus the "distribution derivative" of 1= ( (z w)) with respect to z is the point mass w ("Dirac delta-function"). w 2 C: C @z z w Remark. If we integrate this formally by parts.8. 54 .IV. and it is in…nite at w in such a way that its integral (total mass) is equal to 1. V 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 2 3 4 5 6 7 8 1 . 1 1 2 3 P L K (Harmonic Series) Show that X n 1 log n: k=1 k P1 1 Deduce that the series k=1 k does not converge. we have for all x 2 [k. for n 1 X n n Z X k+1 Z n+1 1 1 1 dx = dx = log (n + 1) log n: k=1 k k=1 k x 1 x Where the last inequality follows from the fact that x 7 P ! log x is an increas- ing function on the positive reals. Use the estimate Z k+1 1 1 dx: k k x Solution Let k be a nonnegative integer. Hint. 2 . k + 1] that 1=x 1=k . k + 1] Z k+1 1 1 1 ((k + 1) k) = dx: k k k x Therefore. Then since the function x 7! 1=x is decreasing on the positive reals.1. And thus is the sum nk=1 k1 have not limit as n ! 1 and does not converge.V. Hence locking at the oversum in the interval [k. 3 .2 1 2 3 P L K P1 Show that if p < 1.V. Use Exercise 1 and the comparison test. Hint. then 1 1 p > : k k Thus we have X1 1 X1 1 p > : k=1 k k=1 k P By the comparison test. we have that Pthe since the sum nk=1 1=k diverge as n ! 1 by Exercise 1.1. Solution If k > 1 then. also the sum nk=1 1=k p diverge as n ! 1. k p < k for 0 < p < 1. which was to be shown. then the series k=1 1=k p diverges. Now we have that XN Z N N 1 dx x p+1 n p+1 N p+1 n p+1 p 6 =1+ = ! k=n+1 k n xp p+1 n p 1 p 1 as N ! 1. k] that 1=k 1=x. where 1 p S 1+ = . p 1 p 1 since its terms are > 0. Use the estimate k1p < k 1 dx xp .1. Then since the function x 7! 1=x is decreasing on the positive reals. we have for all x 2 [k 1. Thus we have 4 . Hence locking att the undersum in the interval [k 1.3 1 2 3 P L K P1 Show that if p > 1. then the series k=1 1=k p converges to S. Becauce p > 1. k] Z k 1 1 1 p (k (k 1)) = p p dx k k k 1 x Therefore Xn n Z X k Z n n 1 dx dx x p+1 1 n p+1 1 = 1+ = 1+ = 1+ 1+ : k=1 kp k=1 k 1 xp 1 x p p+1 1 p 1 p 1 p 1 Hence the serie converge to S. it follows that 1=k p 1=xp .V. where Xn 1 1 S p < : k=1 k (p 1) np 1 Rk Hint. Solution Let k be a nonnegative integer. XN 1 Xn 1 n p+1 1 =S 6 = k=n+1 kp k=1 k p p 1 (p 1) np 1 thus XN 1 1 S : k=n+1 kp (p 1) np 1 5 . so series converge.V. Hint. 6 . and terms ! 0. Show that the partial sums of the series satisfy S2 < S 4 < S 6 < < S5 < S3 < S1 . jtermsj # .1.4 1 2 3 P L K Show that the series X1 ( 1)k+1 1 1 1 =1 + + k=1 k 2 3 4 converges. which is standard signs alternate. Solution We de…ne the sum Sn by Xn ( 1)k+1 1 1 1 Sn = =1 + + :::: k=1 k 2 3 4 This is the “alternating series test”. . Organize the terms in the series in Exercise 4 in groups of four. 7 .) Hint. where S is the sum of the series in Exercise 4. and relate it to the groups of three in the above series. thus also 3S=2 converge by . and so on 3 1 1 1 1 1 1 1 1 S = 1+ 2 + + 2 + + 2 + :::+ = 2 3 4 5 7 8 9 11 12 1 1 1 1 1 1 1 1 = 1+ + + + + + ::: 3 2 5 7 4 9 11 6 Since series S in Exercise 4 converge. and divide the terms by 2 S 1 1 1 1 1 1 = + + + ::: 2 2 4 6 8 10 12 Now add one group of four from the sum S with a group of two from the sum S=2. Solution We have the series S from exercise V...5 and arrange the terms in groups of four 1 1 1 1 1 1 1 1 1 1 1 S=1 + + + + + + ::: 2 3 4 5 6 7 8 9 10 11 12 And now we arrange the terms in groups of four 1 1 1 1 1 1 1 1 1 1 1 S= 1 + + + + + + ::: 2 3 4 5 6 7 8 9 10 11 12 Now we rearrange the terms in groups of two.1.. (It turns out that S = log 2.V..1.5 1 2 3 P L K Show that the series 1 1 1 1 1 1 1 1 1+ + + + + + 3 2 5 7 4 9 11 6 converges to 3S=2. so these 2n terms have sum less than 2(n+1) log 2 . Solution 2 (Integral Test) Note that x log x and x (log x)2 are increasing for x > 1. Thus series converges. Thus series diverges. each of the terms greater 1 1 than 2n+1 log 2n+1 .6 1 2 3 P L K Show that Z R 1 = [log (log k)]R e = log (log R) log e (log e) ! 1 e k log k P diverges while 1 1 k=1 k(log k)2 converges. each of the terms less 1 1 than 2n+1 log 2n+1 . using an upper estimate.1. Solution 1 (Comparison Test) For the …rst sum n+1 2X 1 k=2n +1 k log k there are 2n terms between k = 2n +1 and k = 2n+1 . Vi can use Cauchy’s integralkriterium and look att them corresponding integrals.V. The other series is treated similarly. so these 2n terms have sum greater than 2(n+1) log 2 . so the terms of both series are positive and decreasing. For the other sum n+1 2X 1 k=2n +1 k (log k)2 there are 2n terms between k = 2n + 1 and k = 2n+1 . by comparison with the harmonic series. by comparison with the harmonic series. e x log x 8 . Becauce of di¢ cult with the integrand we split the integral into two parts Z R 1 = [log (log x)]R e = log (log R) log (log e) ! 1. while the series X 1 1 k=1 k (log k)2 konverges. e x (log x)2 log x e log e log R as R ! 1. 9 . Z R R 1 1 1 1 = = ! 1.as R ! 1. We have that the series X 1 1 k=1 k log k diverges. V. ) p Function " for 0 6 x 6 k 1=2k . k = xk . # for x > k 1=2k . when Sn = ak is the nth partial-sum of k=m k=1 the series.1 dx k+x2k = 2 = 0 at k 2 xk 1 + kx3k 1 = 2kx3k 1 . This follows imme- k=1 k=m P n P n diately from ak = S n Sm 1 .1. then Sn Sm 1 = ak . ) xk k 1 p 2 k Converge to 0 uniformly. ! 0 as x ! 1. d xk kxk(k+x2k ) xk (2k)x2k 1 1 V. k = x2k . Solution P n P n If Sn is the partial-sum ak . k+x 2k 6 k+k = = worst-case estimation. (1+x p 2k ) k 2 xk 1 = k 3k 1 .7 1 2 3 P L K P Pk=n 1 Show that the series ak converges if and only if k=m k(log k)2 converges.2. 10 . Hint. We have x 0 k 1=(2k) 0 fk (x) 0 + 0 fk (x) % & and the function is increasing for 0 x k 1=(2k) . We have that fk (x) attains its maximum for x = k 1(2k) . (k + x2k )2 0 there fk (x) = 0 for x1 = 0 and x2 = k 1=(2k) . We have xk fk (x) = . Determine the worst-case estimator "k by calculus. k + x2k thus 0 kxk 1 k + x2k xk (2k) x2k 1 fk (x) = . p xk k 1 f (x) = lim lim = lim p = 0: k!1 k + x2k k!1 k + k k!1 2 k The series fk (x) = xk = k + xk converge uniformly to 0 on [0.V. 1).2. We determine f (x) applying the works case estimator "k to the function fk (x). 11 . Solution Suppose fk (x) converge pointwise to a function f (x) as k ! 1. and decreasing for x for x k 1=2k .1 1 2 3 P L K Show that fk (x) = xk = k + xk converges uniformly to 0 on [0. and fk (x) ! 0 as x ! 1. 1). in this interval we have that xk ! 0 as k ! 1. 12 . now we take the limit in three intervals. where 8 < 0 for 0 x < 1.2 1 2 3 P L K Show that gk = xk = 1 + xk converges pointwise on [0. in this interval we have that x ! 0 as k ! 1. we start to show that is is an increasing function 0 kxk 1 xk kxk 1 kxk 1 gk (x) = = >0 (1 + xk )2 (1 + xk )2 also is the function gk (x) increasing. If 0 x < 1. and we have xk 0 g (x) = lim k = = 0: k!1 1 + x 1+0 If x = 1. +1] for any " > 0. What is the limit function? On which subsets of [0.V. 1) but not uniformly. : 2 1 for x > 1 Convergence is uniform on [0. 1 g (x) = for x = 1. 1]. Convergence is uniform on [1 + ". 1 "] for any " > 0. because limit function is not continuous at 1. and we have xk 1 1 g (x) = lim = lim = = 1: k!1 1 + xk k!1 1=xk + 1 0+1 We also have that gk (x) ! g (x). 1) does the sequence converge uniformly? Solution Suppose gk (x) converge pointwise to a function g (x) as k ! 1.2. then 1k 1 1 g (1) = lim k = = : k!1 1 + 1 1+1 2 k If x > 1. Not uniform on [0. f (z) = lim z k 1 = k!1 1 for jzj = 1: 0 If jzj = 1. What can be said 0 about uniform convergence of fk (z)? Solution Suppose fk (z) converge pointwise to a function f (z) as k ! 1 in the interval jzj 1.2. 13 .3 1 2 3 P L K Show that fk (z) = z k =k converges uniformly for jzj < 1. then fk (z) = 1 so the series can not converge uniformly for 0 jzj < 1. Show that 0 fk (z) does not converge uniformly for jzj < 1. But for any " > 0 the series fk (z) = z k 1 converges uniformly for jzj 1 ". 0 0 Now suppose fk (z) converge pointwise to a function f (z) as k ! 1 in the interval jzj 1. in this domain we have that z k =k 1=k thus zk 1 f (z) = lim lim = 0: k!1 k k!1 k Thus fk (z) converges uniformly for jzj < 1. now we must split the interval when we take the limit 0 0 for jzj < 1.V. 14 . Hence. We have xk 1 1 + x2k = 1 + x2k for jxj 1 xk x2k < 1 + x2k = 1 + x2k for jxj > 1 This establishes the fact that xk 1 1 + x2k for all 1 < x < +1.V.2. Solution We = k12 . where the sum P1apply the Weierstrass M test (p. First we show that x 1 + x for all 1 < x < +1. 1 xk 1 = Mk : k 2 1 + x2k k2 Thus convergence is uniform on 1 < x < +1 by the Weierstrass M test. 135. for each k 1 and 1 < x < +1.4 1 2 3 P L K Show that X1 1 xk k=1 k 2 1 + x2k converges uniformly for 1 < x < +1.)k with Mk 2k k=0 Mk converges. then xk = x2k = 1 then X 1 1 xk X 1 1 = k=1 k 1 + x2k k=1 2k which is a divergent series.2. and we have X 1 1 xk X 1 1 X 1 1X 1 < < = xk k=1 k 1 + x2k k=1 1=x k + xk k=1 1=x k k=1 which is a convergent series. 15 . If x > 1. and we have X 1 1 xk X11 1 X 1 1 = < k=1 k 1 + x2k k=1 k 1=xk + xk k=1 xk which is a convergent series. and take limit in three intervals If x < 1.V.5 1 2 3 P L K P1 xk For which real numbers x does k 1+x2k converge? Solution Suppose gk (x) converge pointwise to a function g (x) as k ! 1. in this interval we have that xk ! 1 as k ! 1. in this interval we have that xk ! 0 as k ! 1. If x = 1. If x = 1.V. and decreasing for x 1.6 1 2 3 P L K P1 xk Show that for each " > 0.2. If x > 1. We have x 0 1 0 fk (x) 0 + 0 fk (x) % & and the function is increasing for 0 x 1. and we have X 1 1 xk X 1 1 (1 + ")k X 1 1 1 = 2k = k=1 k 1 + x2k k=1 k 1 + (1 + ") k=1 k 1= (1 + ") + (1 + ")k k X1 1 1 k=1 k (1 + ")k 16 . 1 + x2k thus 0 kxk 1 1 + x2k xk (2k) x2k 1 xk 1 x3k 1 fk (x) = = (1 + x2k )2 (1 + x2k )2 0 there fk (x) = 0 for x1 = 0 and x2 = 1. Solution We have xk fk (x) = . then xk = x2k = 1 then X 1 1 xk X 1 1 = k=1 k 1 + x2k k=1 2k which is a divergent series. and fk (x) ! 0 as x ! 1. in this interval we have that xk ! 1 as k ! 1. the series k 1+x2k converges uniformly for x 1 + ". We have that fk (x) attains its maximum for x = 1. the series 1 n=1 an n z converges uniformly for Re z 1 + ". jan j 6 C. Solution P1 an n z . Show that for each " > 0.2. Re z > 1 + ". jan n z j 6 n=1 P P Cn Re z 6 n1+" C = Mn .V. Apply Weierstrass M-test.7 1 2 3 P L K LLL Let an be aPbounded sequence of complex numbers. 17 . Mn < 1 ) an n z converges uniformly for Re z > 1 + ". Here we choose the principal branch of n z . k=1 k2 k=1 k2 k=1 k2 which is a convergent series by Weierstass M test.2.8 1 2 3 P L K LLL P zk Show that k2 converges uniformly for jzj < 1.V. 18 . Thus the series con- verges uniformly for jzj < 1. Solution Apply the Weierstrass M test with M = 1=k 2 in jzj 1. we …nd that X1 zk X1 jzjk X 1 1 = < . we …nd that X1 zk X1 jzjk X 1 1 = 2 < . k=1 k k=1 k k=1 k which is a divergent series by Weierstass M test. then the series must converge for jzj 1 Apply the Weierstrass M test with M = 1=k in jzj 1. 19 .2. Solution Suppose that we have uniform convergence for jzj < 1.V.9 1 2 3 P L K LLL P zk Show that k does not converges uniformly for jzj < 1. Thus the series do not converge uniformly for jzj < 1. Solution Use Cauchy criterion. Then jfm (x) fk (x)j < " for m.2. Let N = max fN1 . then the sequence converges uniformly on the union E = E1 [ E2 [ [ En . Let " > 0. 20 . ) ffm g converges uniformly. Nn g.10 1 2 3 P L K Shoe that if a sequence of functions ffk (x)g converges uniformly on Ej for 1 j n.V. : : : . k > N and x 2 E1 [ : : : [ En . k > Nj and x 2 Ej . Choose Nj such that jfn (x) fk (x)j < k for m. Solution Follow the hint. because the collection of all disks centred at (m + ni) =10 of radius =10 covers the complex plan.11 1 2 3 P L K Suppose that E is a bounded subset of a domain D C at a positive distance from the boundary of D. that is > 0 such that jz wj for all z 2 E and w 2 CnD. Hint. @D) > .V. and they cover E.2. Consider all closed disks with centers at points (m + ni) =10 and radius =10 that meet E. 21 . There are only …nitely many of the disks in ????? (because E is bounded). Show that E can be covered by a …nite number of closed disks contained in D. and they are contained in D. because dist (". Show that for each > 0 and m 1.12 1 2 3 P L K Let f (z) be analytic on a domain D.V. Use this to show that if ffk (z)g is a sequence of analytic functions on D that converges to f (z) on D.2. 22 . @D). and suppose jf (z)j M for all z 2 D. Solution Use the Cauchy estimates for f (m) (w) on a disk centred at w of radius < d (w. then for (m) each m the derivatives fk (z) converge uniformly to f (m) (z) on each subset of D at a positive distance from D. f (m) (z) m!M= m for all z 2 D whose distance from @D is at least . get f (m) (w) 6 m!m m . thus ak k2 k2 1 R = lim = lim 2 = lim 2 = lim = 1: k!1 ak+1 k!1 (k + 1) k!1 k + 2k + 1 k!1 1 + 2=k + 1=k 2 (d) We apply the ratio test.V. thus ak 2k 1 1 R = lim = lim k+1 = lim = : k!1 ak+1 k!1 2 k!1 2 2 (b) We apply the ratio test.3. thus ak k 6k+1 6k 6 R = lim = lim k = lim = lim = 6: k!1 ak+1 k!1 6 k + 1 k!1 k + 1 k!1 1 + 1=k (c) We apply the ratio test.1 1 2 3 P L K LLL Find the radius of convergence of the following power series: P 1 P1 3k z k P 1 kk (a) 2k z k (d) k 4 +5 k (g) 1+2k kk zk k=0 k=0 k=1 P1 P1 2k z 2k P1 (b) k k 6k z (e) k2 +k (h) (log k)k=2 z k k=0 k=1 k=3 P1 2 k P1 z 2k P1 k!z k (c) k z (f) 4k k k (i) kk k=1 k=1 k=1 Solution (a) We apply the ratio test. thus ak 3k 4k+1 + 5k+1 R = lim = lim k = k!1 ak+1 k!1 4 + 5k 3k+1 4 k+1 4 k+1 3k 5k+1 5 +1 5 5 +1 5 = lim = lim = : k!1 3k+1 5k 4 k +1 k!1 3 4 k +1 3 5 5 23 . 2 jzj2 < 1 = lim = : k!1 1 2 jzj2 1. thus ak 2k z 2k (k + 1)2 + k + 1 k+2 1 R = lim = lim 2 = lim = k!1 ak+1 k!1 k + k 2k+1 z 2(k+1) k!1 k 2 jzj2 1 + 2=k 1 0. (f) We apply the ratio test. 2 jzj2 > 1 p The convergence radius R = 1= 2. thus ak kk 1 + 2k+1 (k + 1)k+1 k k 1 + 2k+1 (k + 1)k+1 R = lim = lim = lim = k!1 ak+1 k!1 1 + 2k k k (k + 1)k+1 k!1 (k + 1)k (1 + 2k k k ) (k + 1) ! k 1=k 1 2k+1 (k + 1)k = lim + = k!1 1 + 1=k (1 + 2k k k ) (k + 1) 1 + 2k k k ! k k 1=k 1 k+1 1 = lim +2 = 2: k!1 1 + 1=k (1 + 2k k k ) (k + 1) k 1= (2k k k ) + 1 (h) We apply the root test. thus 24 . thus ak z 2k 4k+1 (k + 1)k+1 4 (k + 1) (k + 1)k R = lim = lim k k = lim = k!1 ak+1 k!1 4 k z 2(k+1) k!1 z2 kk k 4 (k + 1) 1 + 1=k = lim = 1: k!1 jzj2 1 (g) We apply the ratio test.(e) We apply the ratio test. thus k k ak k! (k + 1)k+1 k+1 1 R = lim = lim k = lim = lim 1+ = e: k!1 ak+1 k!1 k (k + 1)! k!1 k k!1 k 25 . 1 1 1 R= p = q = = 0: lim k jak j lim log k (log k)k=2 k k!1 lim k!1 k!1 (i) We apply the ratio test. Now consider z such that jz 1j = 1 then jz 1j = 1: P1 Since k=1 1 diverges. the given series converges for all z satisfying jz 1j < 1. (b) We apply the ratio test for ak = 1=k!. the series converges for all z satisfying jz 2j < 1=2 and diverges for all z satisfying jz 2j > 1=2. P 1 P1 P 1 (a) (z 1)k (c) 2m (z 2)m (e) nn (z 3)n k=1 m=0 n=1 P1 (z i)k P1 (z+1)m P1 2n (b) k! (d) m2 (f) n2 (z 2 i)n k=0 m=0 n=3 Solution (a) We apply the ratio test for ak = 1. Therefore. thus ak 1 R = lim = lim = 1: k!1 ak+1 k!1 1 hence the radius of convergence is R = 1. Hence. thus am 2m 1 1 R = lim = lim m+1 = lim = : m!1 am+1 m!1 2 m!1 2 2 hence the radius of convergence is R = 1=2. Now consider z such that jz 2j = 1=2 then 26 .V.3. thus ak 1 (k + 1)! R = lim = lim = lim (k + 1) = 1: k!1 ak+1 k!1 k! 1 k!1 Therefore. (c) We apply the ratio test for am = 2m .2 1 2 3 P L K LLL Determine for which z the following series converge. Therefore. the series converges for all z satisfying jz 1j < 1 and diverges for all z satisfying jz 1j > 1. the series converges for all z in C. the power series diverges for such z also. Now consider z such that jz + ij = 1 then (z + i)m 1m 1 2 = 2 = 2: m m m P1 2 Since m=1 1=m converges. Hence. the series diverges for all z satisfying jz 3j > 0. the given series only converges for z = 3. the power series converges for z = 3. the power series converges for such z also. Now consider z such that jz 3j = 0 then jnn (z 3)n j = nn 0n = 0: P Since 1 n=1 0 converges. the given series converges for all z satisfying jz 2j < 1=2. the series converges for all z satisfying jz + ij < 1 and diverges for all z satisfying jz + ij > 1. (f) We apply the ratio test for ak = 2k =k 2 . Therefore. Hence. the given series converges for all z satisfying jz + ij 1. m m m m 1 j2 (z 2) j = 2 = 1: 2 P Since 1 m=1 1 converges. thus n an nn nn n 1 R = lim = lim = lim n = lim = 0: n!1 an+1 n!1 (n + 1)n+1 n!1 (n + 1) (n + 1) n!1 n+1 n+1 Therefore. the power series diverges for such z also. thus am 1 (m + 1)2 m2 + 2m + 1 1 + 2=m + 1=m2 R = lim = lim 2 = lim = lim = 1: m!1 am+1 m!1 m 1 m!1 m2 m!1 1 hence the radius of convergence is R = 1. thus 27 . (e) We apply the ratio test for an = nn . (d) We apply the ratio test for am = 1=m2 . Hence. the power series converges for such z also. 28 . Hence. ak 2k (k + 1)2 k 2 + 2k + 1 1 + 2=k + 1=k 2 1 R = lim = lim 2 k+1 = lim 2 = lim = : k!1 ak+1 k!1 k 2 k!1 2k k!1 2 2 hence the radius of convergence is R = 1=2. the given series converges for all z satisfying jz 2 ij 1=2. the series con- verges for all z satisfying jz 2 ij < 1=2 and diverges for all z satisfying jz 2 ij > 1=2. Now consider z such that jz 2 ij = 1=2 then n 2n n 2n 1 1 (z 2 i) = 2 = : n2 n 2 n2 P Since 1 2 k=1 1=n converges. Therefore. 3 1 2 3 P L K LLL Find the radius of convergence of the following series. 29 .3. P 1 n (a) z 3 = z + z 3 + z 9 + z 27 + z 81 + P p n=0 (b) z = z 2 + z 3 + z 5 + z 7 + z 11 + p prime Solutions (a) (b) 1 1 R= p = = 1: lim sup k jak j 1 Neither series converges at z = 1.V. so R = 1. V. and the …nite partial sums ! 1 as r ! 1. Show that jf (r )j ! +1 as r ! 1 whenever is a root of unity. Suppose is an N th root of unity. because it is increasing. 30 . Remark. Then n!=N P 1 (r )n! = rn! N = rn! for n > N . Thus f (z) does not extend analytically to any larger open set than the open unit disk.4 1 2 3 P L K P Show that the function de…ned by f (z) = z n! is analytic on the open unit disk fjzj < 1g. N = 1.3. Now rn! " 1 as r " 1. ) jf (r )j ! +1 as r ! 1. Solution Cauchy-Hadamard formula gives R = 1 (radius of convergence) ) f (z) is analytic for jzj < 1. 3. obtain X 1 1 X 1 1 X 1 2 k k 1 z = ) kz = 2 ) k (k 1) z k 2 = : k=0 1 z k=1 (1 z) k=2 (1 z)3 Multiply by z 2 obtain X 1 2z 2 k2 k z2 = : k=2 (1 z)3 We get X 1 X 1 2z 2 z 2z 2 2 k k z = kz k + 3 = z+ k=2 k=2 (1 z) (1 z)2 (1 z)3 31 .V.5 1 2 3 P L K LLL What functions are represented by the following power series? P 1 P 1 (a) kz k (b) k2z k k=1 k=1 Solution (a) P Di¤erentiate the geometric series 1 k k=0 z . obtain X 1 1 X 1 1 k z = ) kz k 1 = : k=0 1 z k=1 (1 z)2 Multiply by z obtain X 1 z kz k = : k=1 (1 z)2 (b) P Di¤erentiate the geometric series 1 k k=0 z twice. 32 .3. and the integrated series k+1 z all have the same radius of convergence. Solution p p Can use Cauchy-Hadamard formula.6 1 2 3 P L K P P Show that the series ak z k . Then lim sup k jak j = p q k!1 ja j lim sup k k jak j = lim sup k k+1k . This largest disk is clearly the same for f . Can also use the characterization of R as the radius of the largest disk to which the function extends and ?????. the di¤erentiated series P ak k+1 kak z k 1 . so all series have the same radius of conver- k!1 k!1 gence. and ????. f 0 . and k k = 1.V. Solution k We apply the Cauchy-Hadamard formula where ak = 2 + ( 1)k . 2 + ( 1)k+1 3k+1 33 . Hence p k 3 if k is even.7 1 2 3 P L K Consider the series X 1 k 2 + ( 1)k zk : k=0 Use the Cauchy-Hadamard formula to …nd the radius of conver- gence of the series. jak j = 1 if k is odd. First observe that r p k 2 + ( 1)k = 2 + ( 1)k k k jak j = Note that 3 if k is even. 2 + ( 1)k = 1 if k is odd. = k+1 = 1 ak+1 if k is odd.V. Hence.3. What happens when the ratio test is applied? Evaluate explicitly the sum of the series. since the lim sup of this sequence is 3 we have by the Cauchy-Hadamard formula 1 1 R= p = : lim sup k jak j 3 Applying the ratio test we have k ak 2 + ( 1)k 3k if k is even. Then we have with jzj < 1=3 X 1 k 2 + ( 1)k zk = k=0 X 1 2n 2n 2n X 1 2n+1 2n+1 = 2 + ( 1) z + 2 + ( 1)2n+1 z = k=0 k=0 X 1 X 1 X 1 X 1 2n 2n 2n+1 2n+1 2 n n = 3 z + 1 z = 9z +z z2 = n=0 n=0 n=0 n=0 1 z = + : 1 9z 2 1 z 2 Note: The series is convergent because j9z 2 j . and that the series have singularities at z = 1=3 and z = 1. in that limk!1 jak j = jak+1 j does not exist. 34 . jz 2 j < 1 since jzj < 1=3. the ratio test is inconclusive. However since jak j = jak+1 j 3.Since the sequence does not converge. it does converge for jzj < 1=3. If jzj > L ". ).7 p Let L = lim sup k jak j. get R 6 L1 . (L + ") jzj < 1. The de…nition p of n ln supnpimplies that given " > 0. and P ak z k diverge. ) R 6 L1 " . then ak z k > (L ")k (L 1")k for in…nitely many k 0 s. then jak j < (L + ")k . Let " ! 0. then ak z k converges.V. 9 > 0. 9N 3 for k > N we have k jak j 6 L+". k z k 6 (L + ")k jzjk = ((L + ") jzj)k converges by comparison with geometric series. 35 . but k jak j > L " for in…nitely P many k 0 s.3. If k > N . If jzj < 1= (L + "). ) R > 1= (L + ") ) jak j > (L ")k . 4). Solution 36 .V.3.8 1 2 3 P L K Write out a proof of the Cauchy-Hadamard formula (3. (d) Log z about z = 1 + 2i. (f) z3 z about z = 2i: Solution (d) Since the values of Log z are unbounded in any neighborhood of the origin it may not be extended to an analytic function on any open disk containing the p origin. (a) z 1 1 about z = i.4. Hence. 1 z i (c) cosh z about z = 0. p p (a) R = 2 (b) R = =2 (c) R = =2 (d) R = 5 (e) R = 3 (f) R = 2 37 . 146.V. the desired radius of convergence is 5 by the second Corollary on p. expanded about the indicated point. (e) z 3=2 about z = 3. But Log z is de…ned and analytic on the disk of radius j1 + p2ij = 5 centered at 1 + 2i.1 1 2 3 P L K LLL Find the radius of convergence of the power series for the following functions. 1 (b) cos z about z = 0. 2 1 2 3 P L K LLL Show that the radius p of convergence of the power series expansion of (z 2 1) = (z 3 1) about z = 2 is 7. 38 .4.V. and (z e2 i=3 )(z+e2 i=3 ) p distance from 2 to nearest singularity is 7. Solution z+1 Rewrite f (z) as . Singularities of f (z) are at e2 i=3 . Show that the radius of convergence of the series is R = 5. through the extension does not coincide with Log z in the part of the disk in the lower half–plane.3 1 2 3 P L K LLL Find the power series expansion of Log z about the point p z = i 2.4.V. 39 . Solution (From Hints and Solutions) p Log z extends to be analytic for jz (i 2)j < 5. Explain why this does not contradict the discontinuity of Log z at z = 2. The radius of convergence of the power series of either branch is 1=4.4 1 2 3 P L K LLL Suppose f (z) is analytic at z = 0 and satis…es f (z) = z + f (z)2 .V. 40 . which is the distance to the singularity at 1=4.4. What is the radius of convergence of the power series expansion of f (z) about z = 0? Solution (From Hints and Solutions) p Near 0 the function coincides with one of the branches of 1 1 4z =2. Solution P 1 in z n P P P 1 ( 1)m z 2m P 1 ( 1)m z 2m+1 eiz = n! = + = (2m)! +i (2m+1)! .4.V. where n = 2m n=0 n even n odd m=0 m=0 in the …rst and n = 2m + 1 in the second series.5 1 2 3 P L K Deduce the identity eiz = cos z + i sin z from the power series expansions. 41 . 6 1 2 3 P L K LLL Find the power series expansions of cosh z and sinh z about z = 0.4. What are the radii of convergence of the series? Solution P 1 z 2n z2 z4 (a) cosh z = cos iz = (2n)! =1+ 2! + 4! + ::: n=0 (iz)3 (iz)5 z3 z5 (b) sinh z = i sin (iz) = ( i) (iz) 3! + 5! ::: = z + 3! + 5! + ::: P 1 z 2n+1 = (2n+1)! n=0 42 .V. V. 43 .7 1 2 3 P L K Find the power series expansion of the principal branch Tan 1 (z) of the inverse tangent function about z = 0. Solution We have d 1 X 1 k 1 Tan z= = z2 .4. dz 1 + z2 k=0 thus X 1 z 2k+1 Tan 1 z= ( 1)k : k=0 2k + 1 And we have R = 1. What is the radius of con- vergence of the series? Hint. Find it by integrating its derivative (a geometric series) term by term. 8 1 2 3 P L K Expand Log (1 + iz) and Log (1 iz) in power series about z = 0.V.4. By comparing power series expansions (see the preceding exercise). establish the identity 1 1 1 + iz Tan z= Log : 2i 1 iz (See Exercise 5 in section I.8) Solution X 1 z k+1 X 1 z k+1 k+1 Log (1 + iz) = ( i) . Log (1 iz) = ik+1 : k=0 k+1 k=0 k+1 Thus 2i X 2j z 2j+1 X 1 1 1 1 + iz z 2j+1 Log = i = ( 1)j : 2i 1 iz 2i j=0 2j + 1 j=0 2j + 1 44 . 1. : : :. Assume a 6= 0. Re z > 0. and m=0 P 1 a0 = a1 = a 0 . am = f m!(1) = a(a 1):::(a m! m+1) = m a .9 1 2 3 P L K Let a be real.V. R = 1 since am+1 am = a m m+1 ! 1 as m=0 m ! 1. 45 . What is the radius of convergence of the series? Write down the series explicitly. 2. Expand z a in a power series about z = 1. and consider the branch of z a that is real and positive on (0.4. Solution f (z) = z a = ea Log z . m > 1. f (z) = a m (z 1)m . f (m) (z) = a (a 1) : : : (a m + 1) z a m . 1). If a is an integer > 0 get R = 1. f (m) (1) = a (a 1) : : : (a m + 1) P 1 (m) f (z) = am (z 1)m . V. 2. Otherwise radius of convergence is 1. 1. For which does the binomial series reduce to a polynomial? Solution Use f (n) = ( 1) : : : ( n + 1) (1 + z) n and the formula for the coe¢ - cient of z n . which is distance to the singularity at 1. The series reduces to a polynomial for = 0. We can obtain the radius of convergence for the series also from the ratio test. the binomial coe¢ cient " choose n" is de…ned by ( 1) ( n + 1) = 1.10 (From Hints and Solutions) 1 2 3 P L K Recall that for a complex number . : : :. and = . 46 .4. n 1: 0 n n! Find the radius of convergence of the binomial series X 1 zn: n n=0 Show that the binomial series represents the principal branch of the function (1 + z) . an an+1 = ( (1):::(1):::( n+1) n+1)( (n+1)! n) n! = n+1n ! 1 as n ! 1. when we divide by 0 above.4. (constant term ????? J0 . de…ne the function Jn (z) by the power series X 1 ( 1)k z n+2k Jn (z) = : k=0 k! (n + k)!2n+2k Show that Jn (z) is an entire function. Solution 1 P ( 1)k z n+2k zn Jn (z) = k!(n+k)!2n+2k = n!2n + O (z n+2 ) k=0 P1 ( 1)k (n+2k)z n+2k 1 zn 1 Jn0 (z) = k!(n+k)!2n+2k = (n 1)!2n + O (z n+2 ) k=0 P1 ( 1)k (n+2k)(n+2k 1)z n+2k 2 zn 2 Jn00 (z) = k!(n+k)!2n+2k = (n 2)!2n + O (z n ) k=0 ( 1)k (n+2k)(n+2k 1) Term multiplying z n+2k in z 2 Jn00 + zJn0 + (z 2 n2 ) Jn is k!(n+k)!2n+2k + ( 1)k (n+2k) ( 1)k n2 k!(n+k)!2n+2k k!(n+k)!2n+2k ( 1)k (k 1)!(n+k 1)!2n+2k 2 = ( 1)k k!(n+k)!2n+2k (n + 2k) (n + 2k 1) + (n + 2k) n2 4k (n + k) =0 Should cheek separately to the sum terms of J0 + J1 .V.11 1 2 3 P L K For …xed n 0. It works. because the case k = 0 we replace 1= (k 1)! by k=k! = 0. This is Bessel’s di¤erential equation. 1 47 . (It’s not clear). ???? work out. Ratio test gives radius of convergence = 1. Show that w = Jn (z) satis…es the di¤erential equation 1 n2 w00 + w0 + 1 w = 0: z z2 Remark. and Jn (z) is Bessel’s function of order n. z term for J1 ). V. then an = 0 for n even. The result for f (z) odd is analogous.12 1 2 3 P L K LLL Suppose P that the analytic function f (z) has power series expansion n an z . Show that if f (z) is an even function. 48 .4. Show that if f (z) is an odd function. If f is even then f (z) f ( z) = 0. and so n=0 2a2n+1 z = 0. then an = 0 for n odd. Solution If f (z) is analytic P1 in Dr n(0) then f ( z) is also analytic in the same region n and P1 f ( z) = 2n+1n=0 ( 1) an z . A power series is 0 i¤ all of its contents are 0 (since (n) an = f (0) =n! which shows an = 0 for n odd. etc. Alternatively. f (k) ( z) = ( 1)k f (k) (z). same argument gives ak = 0 for k even. If f (z) is odd. f 00 ( z) = f 00 (z). f ( z) = f (z). f ( z) = f (z) . apply ever result to zf (z) which is even.f (z) even. ) ak = f (k) (0) =k! for k odd. Chain rule: f 0 ( z) = f 0 (z). 49 . f (k) (0) = ( 1)k f (k) (0) ) f (k) (0) = 0 for k odd. If f (z) and g (z) are ana- lytic. and g (z) is not identically zero. k > N Then the limits are equal in all cases. f (0) 6= 0. f (0) 6= 0. f (z0 ) = g (z0 ) = 0. f (z) = ak z k = f 0 (0) z +O (z 2 ). then f (z) f 0 (z) lim = lim 0 . fg(z) (z) ! fg0 (0) if g 0 (0) 6= 0 ! if g (0) = 0 .4. k = N : N bN bN 0 g (0) = 0 . Solution P 1 Assume z0 = 0 .13 1 2 3 P L K Prove the following version of L’Hospitals’s rule. 50 . g (z) = g 0 (0) z +O (z 2 ) = k=1 8 0 (0) < 1 if g (0) = 0 and f (0) 6= 0 or k < N aN g 0 (0) z+aN z n +O z N +1 .V. k = N : bN 0 if g (0) = 0 . f (0) 6= 0. z!z0 g (z) z!z0 g (z) in the sense that either both limits are …nite and equal. 0 if g 0 (0) 6= 0 8 g (0)+N bN z +O(z ) g (z) < 1 g (0) = 0 and either f (0) 6= 0 or k < N N aN aN ! = g (0) = 0 . or both limits are in…nite. k > N 0 f (z) f ( ) f (z) 0 (0)+ka z k 1 +O z k 0 0 ! fg0 (0) (0) k g 0 (z) = 0 N 1 N . f (0) 6= 0. then Fr (z) = F (rz) has power series P P 1 F (z) = an z k . 51 . To approximate such an F . Show that f can be approximated uniformly on T by a sequence of polynomials in z if and only if f has an extension F that is continuous on the closed disk fjzj 1g and analytic on the interior fjzj < 1g.4. jpn pm j ! 0 uniformly for jzj 6 1. jzj < 1. m ! 1. to F (rz) and F (rz) ! f (z) uniformly for jzj = 1 as v " 1. consider dilates Fr (z) = F (rz). For " > 0. also F (z) is analytic for jzj < 1. then take N such that PN P N an rn z n Fv (z) 6 2" for jzj = 1. pn ! f uniformly for jzj = 1. then jpn pm j ! 0 uniformly for jzj = 1 as n. jzj < 1r . take r < 1 with jFr (z) F (z)j < " for jzj = 1. Conversely if pn are polynomials.V. Solution If f has an analytic extension F . Since pn is analytic for jzj < 1. F (z) = f (z) for jzj = 1.14 1 2 3 P L K Let f be continuous function on the unit circle T = fjzj = 1g. Since pn (z) ! f (z) for jzj = 1. Fr (z) = an rn z k . Then an rn z n F (z) 6 2" + 2" = " n=0 n=0 for jzj = 1. ) fpn g converges uniformly to same function F (z) for jzj 6 1 . so by the maximum principle. Hint. Fr converge uniformly n=j for jzj 6 1. Hence. P 1 n (a) 1 z 2 +1 =1 z2 + z4 ::: = ( 1)n z 2n z12 1+1=z 1 2 = 1 ( 1) z 2 z 2n = n=0 P 1 ( 1)n z 2n+2 n=0 z2 1 1 1 P 1 1 P 1 1 (b) z3 1 = z 1 1=z 3 = z z 3n = z 3n+1 n=0 n=0 1=z 2 P 1 1 1 (c) e = n! z 2n n=0 1 1 1 1 P 1 1 1 (d) z sinh z =z z + 3!z 3 + 5!z 5 + ::: = (2n+1)! z 2n n=0 52 .5.V. we have X 1 3n+1 X 1 1 f (z) = g (1=z) = (1=z) = k=0 k=0 z 3n+1 for all z with jzj > 1.1 1 2 3 P L K LLL Expand the following functions in power series about 1 2 2 (a) z21+1 (b) z3z 1 (c) e1=z (d) z sinh (1=z) Solution (b) Denote the above function by f and de…ne g by (1=w)2 w g (w) = f (1=w) = 3 = for w 6= 0 (1=w) 1 1 w3 and g (0) = 0. Note that g is analytic on the disk D1 (0) with power series representation w X1 n X 1 g (w) = = w w3 = w3n+1 1 w3 k=0 k=0 3 for all w 2 D1 (0) (note that jw j 1 if jwj 1). V.2 1 2 3 P L K Suppose f (z) is analytic at 1.1). With the notation f (1) = b0 and f 0 (1) = b1 . since bk+1 wk ! 0 as w ! 0. z [f (z) f (1)] = f 0 (1)+ zk ! k=0 k=2 k=1 P 1 f 0 (1) as z ! 1.5. k=1 53 . with series expansion (5. show that f 0 (1) = lim z jf (z) f (1)j : z!1 Solution 1 P 0 P 1 P 1 bk+1 f (z) = bk zk = f (1)+ f (1) z + bk zk . V.5. Kumjian) Recall that f is analytic at 1 i¤ there is a function at 0 such that f (z) = g (w) with w = 1=z wherever f (z) makes sense. k=0 with radius of convergence R. Let 0 be the smallest number such that f (z) extends to be analytic for jzj > . We may suppose that f (z) is de…ned for jzj > and that f is analytic at these points.1). Thus. we have X1 bk f (z) = for jzj > . the series converges absolutely for jwj < R and diverges for jwj > R. Solution (A.1) converges absolutely for jzj > and diverges for jzj < . It follows that by substituting w = 1=z. Then setting R = 1= if > 0 and R = 1 if = 0.3 1 2 3 P L K Suppose f (z) is analytic at 1. we have that g is analytic on DR (z0 ) and cannot be extended to an analytic function on any larger open disk centered at z0 . with series expansion (5. g has a power series representation X 1 g (w) = bk w k for jwj < R. Hence. 54 . k=0 zk where the series converges absolutely for jzj > and diverges for jzj < . Show that the series (5. N =0 E 55 . Hint. obtain f (w) = P 1 bn RR n w n+1 . w 2 CnE. and set ZZ dx dy f (w) = . where b n = z dxdy. Solution P n n+1 If jzj 6 M for z 2 E. jwj > R. then 1= (w z) = z =w converges uniformly for z 2 E and jwj > R . Integrate term by term. and …nd a formula for the coe¢ cients of the power series of f (w) at 1 in descending powers of w.4 (From Hints and Solutions) 1 2 3 P L K Let E be a bounded subset of the complex plane C over which area integrals can be de…ned. Show that f (w) is analytic at 1. and R > M . E w z where z = x + iy.5.V. Use a geometric series expansion. Solution To …nd the formula for jwj < 1 .V. Hint. There are two formulae for f (w). RR P RR zn RR If jwj < 1. break the integral into two pieces corre- sponding toZ Zjzj > jwj and Z Z to jzj < jwj. jwj < 1 56 . If jwj < 1. get z dxdy = D w z w z jzj<jwj jzj>jwj | {z } | {z } (1) (2) RR n i(n 1) RR 1 r drd e = 0 . f (w) = RR dxdy dxdy dxdy RR n 1 w z = + = (1)+(2). one valid for jwj 1 and the other for jwj 1.5. in the case that E = fjwj 1g is the unit disk.5 (From Hints and Solutions) 1 2 3 P L K Determine explicitly the function f (w) de…ned in Exercise 4. jwj > 1 f (w) = w. get (1) w1 dxdy 1 z=w = w 1 w n dxdy = 1 w dxdy = w1 jwj2 = jzj<jwj jzj6jwj jzj6jwj w RR dxdy RR 1 dxdy P 1 RR dxdy (2) w z = z 1 w=z = wn z n+1 =0 1>jzj>jwj n=0 =w. f (w) = w z dxdy = w . jwj > 1. Be sure thy agree for jwj = 1. and use geometric series. Solution z2 z4 z6 1 cos z =1+ 2! 4! 6! + O (z 8 ) + (: : :)2 + (: : :)3 + O (z) 2 1 + z2! + a4 z + a6 z 6 + O (z 8 ) 4 2 3 z 4 : 4!1 + 2!1 = 14 24 1 = 24 5 . z6 : 2 2!4! + 2!1 = 18 1 24 = 1 12 sin z z3 z5 z7 z2 5 4 1 6 cos z = z 3! + 5! 7! + ::: 1+ 2 + 24 z 12 z + ::: z:1 z 3 : 21 1 3! = 1 2 1 6 = 31 z 5 : 5!1 1 2 1 3! + 245 1 = 120 1 12 5 + 24 16 = 120 4 = 30 = 2 15 1 5 2 z 7 : 12 24 3! + 215! 7!1 = 7 6 5 2 57! 7+3 7 1 = 420 175+21 1 7! = 265 7! = 7 653 432 57 .6.1 1 2 3 P L K LLL Calculate the terms through order seven of the power series expansion about z = 0 of the function 1= cos z.V. V.6.2 1 2 3 P L K LLL Calculate the terms through order …ve of the power series expansion about z = 0 of the function z= sin z. Solution z z 1 sin z =z z 3 =3!+z 5 =5! ::: = 1 z 2 =3!+z 4 =5! ::: z2 z4 6 =1+ 3! 5! + z7! : : : + (: : :) + (: : :)3 + O (z 2 ) = 2 1 + z 2 =6 + a4 z 4 + a6 z 6 + O (z 7 ) a4 = 5!1 + (3!) 1 2 = 36 1 1 120 = 103603 = 360 7 6(1 14)+4 5 7 a6 = 7!1 = 3!5! 2 + (3!) 1 3 = 3!7! = 1403!7!78 = 3!7! 62 = 32 z 2 7z 4 sin z = 1 + z6 + 360 + O (z 6 ) 58 . since 0! = 1! = 1.1). n 2: 2! 3! n! What is the radius of convergence of the series? Solution We have 1 X 1 X 1 zk = ( 1)k z k for all jzj < 1 and for all z 2 C: 1+z k=0 k=0 k! Since the given function can be expressed as product of these two we have 1 z X 1 ez = e = cn z n for all jzj < 1 1+z 1+z n=0 where ( 1)n ( 1)n 1 ( 1)n 2 ( 1)0 an = + + + + 0! 1! 2! n! by formula (6. c0 = 1.6. c1 = 0 and for n 2 we have.V. ( 1)n 2 ( 1)0 1 1 ( 1)n an = ( 1)n 1 ( 1 + 1)+ + + = ( 1)n + + : 2! n! 2! 3! n! The radius of convergence is 1. 59 . Hence.3 1 2 3 P L K LLL Show that ez 1 1 3 3 4 11 5 = 1 + z2 z + z z + 1+z 2 3 8 30 Show that the general term of the power series is given by n 1 1 ( 1)n an = ( 1) + + . ( 1)n an ! e 1 when n ! 1 . ) a0 = 1 a1 + a2 = 1=2 > > .) 60 . an = 1=n! a n an 1 + an = 1=n! So an = n!1 1 (n 1)! + (n 1 2)! : : : + ( 1)n 0!1 (the two last terms cancel.Solve ez = (1 + z) = a0 + a1 z + a2 z 2 + : : : by multiplying 9 by (1 + z) and com- a0 = 1 > > = Recursively z a 0 + a 1 = 1 paring with the coe¢ cients of e . (In fact.) The radius of convergence is 1 because the function has a pole at 1 . so only even terms appear in the series. What is the radius of convergence of the series? Find the …rst three Bernoulli numbers. Distance from 0 to nearest singularity = 2 = radius of convergence. w cot w = = w cos w sin w 2! w2 4 6! w6 = 1 3! + w5! 7! +::: h 2 4 i 2 2 w4 w2 1 w2! + w4! + O (w6 ) 1 + w6 120 + 6 + O (w6 ) w2 w4 w2 1 1 1 2 + 24 1+ 6 + 36 120 w4 + O (w6 ) = 1 1 7 1 1 1+ 6 2 w2 + 360 12 + 24 w4 + O (w6 ) = 1 2 1 4 6 1 3 w 45 w + O (w ) 2 4 4 cot w = 1 B1 w2! B2 2 4!w ::: ) 2! = 3 .e. at 2 . must keep all powers ????? to and ????? w8 .V. Better to just ????? for B1 and B2 .. i. The function has singularities at z=2 = . Solution z 2 4 6 2 cot (z=2) = 1 B1 z2! B2 z4! B3 z6! : : :. w2 4 w6 1 + w4! +::: Find the two Bernoullinumbers B1 +B2 . B2 = 244!45 . 4! B2 = 45 22 B 1 1 1 24 1 . cot (z=2) = cos(z=2) sin(z=2) is odd. 61 . so z2 cot (z=2) is even. B1 = 6 .4 1 2 3 P L K De…ne the Bernoulli numbers Bn by z z2 z4 z6 cot (z=2) = 1 B1 B2 B3 : 2 2! 4! 6! Explain why there are no odd terms in this series.6. B2 = 30 1 For B3 . V. 4! = 4!1 + 2!1 = 1 4 1 24 = 5 24 . E2 = 1. Solution 1 1 P En n 3 cosh z = n! z . Find the …rst four nonzero Euler numbers. Since cosh z is even. 2 i.6.5 1 2 3 P L K De…ne the Euler numbers En by 1 X En 1 = zn: cosh z n=0 n! What is the radius of convergence of the series? Show that En = 0 for n odd. E4 = 5 E6 1 1 1 6! = 6! + 2 2!4! (2!)3 . E0 = 1 E2 1 E4 2 2! = 2! . En = 0 for n odd. cosh z has zeros at 2 i. 1 1 z2 z4 z6 cosh z = (1+z2 =2!+z 4 =4!+:::) = 1 + E2 2! + E4 4! + E6 6! = 2 z2 z4 z2 z4 z6 1 2! + 4! + ::: + 2! + 4! + 6! ::: (: : :)3 . Distance from 0 to n=0 nearest singularity is =2 = R . : : :. E6 = 61 6! E6 = 1+6 5 33 = 29 6 5 3 = 29 90 = 61 62 . m z .6 1 2 3 P L K Show that the coe¢ cients of a power series "depend continuously" on the function they represent.V. and X 1 X m k fm (z) = ak. 63 . in the following sense. f (z) = ak z k . If ffm (z)g is a sequence of analytic functions that converges uniformly to f (z) for jzj > . k=0 k=0 then for each k 0.6.m ! ak as m ! 1. we have ak. so sin z = zh (z) for some analytic function h with h (0) 6= 0. f (z) = z 3 h (z) and so f has a zero of order three at 0. z0 = k for k 2 Z. Hence. f has a simple zero at z0 = k when k 6= 0. 64 . Thus.7. (d) Note that g (z) = cos z 1 = 0 i¤ z = 2k for k 2 Z. Note that f 0 (z) = 2z sin z + z 2 cos z. Since sin z has a zero of order 1 at 0.V. all zeros are double zeros. Then f has zeros at all integer multiples of i. Hence. we have f 0 (k ) = 2k sin k + (k )2 cos k = ( 1)k k 6= 0: Thus. for any other zero z0 = k where k 2 Zn f0g.e. First observe that g 0 (z) = sin z and so g 0 (2k ) = sin 2k = 0 for all k 2 Z. But g 00 (z) = cos z and so g 0 (2k ) = cos 2k = 1 for all k 2 Z.1 1 2 3 P L K LLL Find the zeros and orders of zeros of the following functions z 2 +1 (a) z2 1 (d) cos z 1 (g) ez 1 (b) 1 z + z5 (e) cos zz 1 1 (h) sinh2 z + cosh2 z (c) z 2 sin z (f) coszz2 1 (i) Log z z (principal value) Solution (c) Let f (z) = z 2 sin z. We will show that all the zeros are double zeros. then cos (n ) cosh (n ) = ( 1)n cosh (n ). e3 i=4 . (d) double zeros at n . n = 1. k 2 Z . at these points. 65 . : : :. 2 . where zj = 1 4 ei( =4+j=4 2 ) . n = 0. : : :. 1 (b) z + z5 = (z + 1) z15 = (z z1 ) (z z2 ) (z z3 ) (z z4 ) z15 . sin (iy) = 2i =i 2 = i sinh y. cos z = 1 ) sin x sinh y = 0 ) y = 0 or x = n . y = 0. (h) simple zeros at i=4 + n i=2.(a) simple zeros at i. thus at z = 1. 2. : : : . Simple poles at z = 0 . 1. we have simple zeros at i2 k. Get x = 0. If y = 0. 2. The zeros are double zeros. : : :. n = 1. At the points dz d = cos zz 1 = z( sin z) z2(cos(z 1)) = cos(2n ) 1 (2n )2 for n 6= 0. : : :. 1. (c) triple zero at 0. since cos (n ) > 1 for n 6= 0. If x = n . Since ez 1 is periodic. This is = 1 only if n = 0. 2 : : :. n = 0. so we have a simple zero at 1. cos (iy) = e 2+e = e y ey ey e y cosh y. has only zeros at z = 2n . simple zeros at n . (i) no zeros. We write 2 3 Log z = Log (1 (1 z)) = 1 z (1 2z) + (1 3z) + : : :. : : :. 2. 2. 2. 1. (b) simple zeros at e i=4 . Since dz (cos z 1) = sin z = 0 at these points. Double zeros at z = 2n . 4 . thus a simple zero at 0. 2 (g) ez 1 = z + z2! + : : :. Other zeros are double. y y (d) cos z = cos (x + iy) = cos x cos (iy) sin x sin (iy). Solutions d are z = 0. then cos x = 1 only at x = 2n . d2 dz 2 (cos z 1) = cos z 6= 0 . (e) cos zz 1 . these four points are simple zeros. n = 0. (i) Log z = 0 when jzj = 1 and Arg z = 0. n = 1. (c) –(i) not analytic at 1. (b) analytic at 1. 66 .7. simple zero. and determine the orders of any zeros at 1.2 (From Hints and Solutions) 1 2 3 P L K Determine which of the functions in the preceding exercise are analytic at 1.V. Solution (a) analytic at 1 . the zeros are simple. 1 < n < 1. 67 .3 1 2 3 P L K LLL Show that the zeros of sin z and tan z are all simple.V.7. since cos (n ) = dz sin zjz=n = 1 6= 0. Solution d Zeros of sin z are at n . w) = 0. Apply theorem. w) = 0 for z. 68 . F (z. w) is entire in z for each …xed w and entire in w for each …xed z. Solution cos (z + w) = cos z cos w + sin z sin w = F (z. w real.4 1 2 3 P L K LLL Show that cos (z + w) = cos z cos w sin z sin w. so cos (z + w) = cos z cos w sin z sin w. Get F (z. E = R. assuming the corresponding identity for z and w real. with D = C.7.V. 5 1 2 3 P L K Show that Z 1 r zt2 +2wt 2 =z e dt = ew . Solution R1 at2 2bt e dt. Compare the result to Exercise IV.1. eat 2bt dt = p1 e x ds. Hint. note the integral is improper. Re a < 0. x real x 1 1 R1 zt2 zbt p 2 =z e e dt = z eb 1 69 . 1 z where we take the principal branch of the square root.7. and evaluate it for z = x >p 0 and w real by making a change of variable and using the known value for z = 1 and w = 0. x > 0. F ( ) = x 1 1 Z1 R1 s2 R1 2 2 (s =2) e e s ds = e (s =2) e =4 ds = e =4 e ds 1 1 1 | {z } R1 s2 p 2 R1 xt2 p b2 =x e e s ds = e =4 . so a limit process may be used 1 R1 2 R1 s2 2b p s for analytic.3. w 2 C. e e 2bt dt = p1 e .V. z. Show that the integral is analytic in z and w. Re z > 0. 6 1 2 3 P L K LLL Suppose f (z) is analytic on a domain D and z0 2 D. Solution Suppose that f (m) (z0 ) = 0 for m and de…ne g on D by g (z) = f (z) f (z0 ). Show that if f (m) (z0 ) = 0 for m 1. Suppose not.V. then since g (m) (z0 ) for m 0.7. then f (z) is constant on D. This result may also be proved using the Uniqueness Principle (the second theorem on page 156). 70 . the power series expansion for g is trivial and therefore g is identically zero on a disk on nonzero radius centered at z0 but this violates the …rst theorem on page 156 that asserts that an analytic function which is not identically zero only has isolated zeros. Then it su¢ ces to show that g is identically zero. then u (x.7.V.7 1 2 3 P L K LLL Show that if u (x. y) vanish at the same point of D. y) is constant on D. Solution 71 . y) is a harmonic function on a domain D such that all the partial derivatives of u (x. Solution 72 .8 1 2 3 P L K With the convention that the function that is identically zero has a zero of in…nite order at each point. show that if f (z) and g (z) have zeros of order n and m respectively at z0 .7. Show that strict inequality can occur here.V. but that equality holds whenever m 6= n. m). then f (z) + g (z) has a zero of order k min (n. then f (z) = g (z)N for some function g (z) analytic near z0 and satisfying g 0 (z) 6= 0. where h (z) has a convergent power series and h (z0 ) 6= 0. Take g (z) = (z z0 ) e(log(h(z)))=N for an appropriate branch of the logarithm.7.9 1 2 3 P L K LLL Show that the analytic function f (z) has a zero of order N at z0 . 73 . Solution (From Hints and Solutions) Write f (z) = (z z0 )N h (z).V. V. Solution f (z)N analytic.10 1 2 3 P L K Show that if f (z) is a continuous function on a domain D such that f (z)N is analytic on D for some integer N . doing circle around z z0 . m is an integral multiple of N . by preceding exercise. except possibly at the isolated points where f (z) = 0. Then h (z) has N th root near z0 . h (z) 6= 0. for this need maximum principle. 74 . Zeros of f (z) are isolated. h (z) analytic.7. (f (z) =g (z))N = (z z0 )m . (z z0 )m=N is continuous in a neighborhood of z0 . then f (z) is analytic on D. f (z)N = g (z)N (z z0 )m . (By the Riemann‘s theorem. f (z) is analytic ????? there but don’t have this yet) Write f (z)N = (z z0 )m h (z). f (z) is analytic. f (z) =g (z) re- turns to original value. If f 0 (z0 ) = 0.11 1 2 3 P L K Show that if f (z) is a nonconstant analytic function on a domain D. f (z) = g (z)N . then the image under f (z) of any open set is open. g 0 (z0 ) 6= 0. h (z0 ) 6= 0. write f (z) = (z z0 )N h (z). so f cover disk centred at 0. then f maps open disks centred at z0 onto open sets. Remark. The proof is easy when f 0 (z) 6= 0. 75 . g covers a ????? disk. Use Exercise 9 to deal with the points where f 0 (z) is zero. Solution If f 0 (z0 ) 6= 0. This is the open mapping theorem for analytic functions. Then f has a N th root near z0 . so f (D) contains a disk centred at f (z0 ). assume f (z0 ) = 0.V. since the Jacobian of f (z) coincides with jf 0 (z)j2 .7. V. 76 .7. Solutions Clearly open mapping theorem ) strict maximum principle for analytic func- tions. since a non constant analytic function can’t attain maximum at z0 and cover a disk centred at z0 . Then strict maximum principle ) maximum principle.12 1 2 3 P L K Show that the open mapping theorem for analytic functions implies the max- imum principle for analytic functions. V. or else f (D) consists of a single point on @D.13 (From Hints and Solutions) 1 2 3 P L K Let fn (z) be a sequence of analytic functions on a domain D such that fn (D) D. Solution f (D) D [ @D.7. 77 . and f (D) cannot contain any point of @D. then f (D) is open. and suppose that fn (z) converges normally to f (z) on D. Show that either f (D) D. If f (z) is not constant. @D) > 1=n. Show that a closed discrete subset of a domain D either is …nite or can be arranged in a sequence fzk g that accumulates only on f1g [ @D.V. En is compact. Solution Let Kn = fz 2 D : d (z. jzj 6 ng.14 1 2 3 P L K A set E is discrete if every point of E is isolated. shading with those in E \ K1 . only …nitely many points of E belongs to Kn ????? points.????? 78 .7. the E \ (K2 nK1 ) . returns to e2 i=3 times initial value. e 2 i=3 at 1. phase change ei =3 at +1. ei =3 at +1. i at +1.V8. phase change ei =3 at +1.1 1 2 3 P L K p Suppose that the principal branch of z 2 1 is continued analytically from z = 2 around the …gure eight path indicated above. 79 . 3 z3 p 1. What is the analytic con- tinuation of the function at the end of the path? Answer the same question 1=3 1=3 form the functions (z 3 1) and (z 6 1) . 1 at 1. returns to initial value. returns to initial value. 3 z 6 1. phase change i at +1. Solution p p z 2 1. ei =3 at +1. 2 1 2 3 P L K Show that f (z) = Log z = (z 1) 12 (z 1)2 + has an analytic contin- it uation around the unit circle (t) = e . How is f2 related to f0 ? Solution P m 1 df dz = z1 = . f (m) (z) = ( 1) zm (m 1)! P 1 f (m) eit ( ) m P 1 itm m Series is m! (z eit ) = f (eit ) + ( 1)m 1 e m (z eit ) m=0 m=1 P 1 itm it m ft (z) = it + ( 1)m 1 e m (z e ) . Determine explicitly the power series ft for each t.V8. get f2 (z) = f0 (z) + 2 i . 0 t 2 . m=1 80 . and show that the radius of convergencepof the power series ft (z) representing the continuation is j (t)j. Solution 81 .8.V. Show that z cannot be analytically along path containing 0.3 1 2 3 P L K p Show that each branch of z can be continued analytically along any path in Cn f0g. Let Z z X1 an F (z) = f ( )d = (z z0 )n+1 z0 n=0 n + 1 be the inde…nite integral of f (z) for z near z0 .8.4 1 2 3 P L K P Let f (z) be analytic on a domain D. What happens in the case D = Cn f0g. and f (z) = 1=z? What happens in the case that D is star-shaped? Solution 82 . …x z0 2 D. and let f (z) = an (z z0 )n be the expansion of f (z) about z0 .V. z0 = 1. Show that F (z) can be continued analytically along any path in D starting at z0 . and that it cannot be extended analytically to any larger open set.V. Observe that f (z) = z +f (z 2 ). and that f (r) ! +1 as r ! 1. Solution 83 .5 1 2 3 P L K Show that the function de…ned by X n f (z) = z2 = z + z2 + z4 + z8 + is analytic on the open unit disk fjzj < 1g.8. Hint. Such a sequence with large gaps between successive nonzero terms is called a lacunary sequence.8. where an = 0 except for n in a sequence nk that satis…es nk+1 =nk 1 + for some > 0. consider g (w) = f (wm (1 + w) =2).6 1 2 3 P L K P Suppose f (z) = an z n . This result is the Hadamard gap theorem. Solution 84 . Show that f (z) does not extend analytically to any point of the unit circle. Remark. There is a slick proof. If f (z) extends ana- lytically across z = 1. Suppose further that the series has radius of convergence R = 1. where m is a large integer. and that this implies that the power series of f (z) converges for jz 1j < ". Show that the power series for g (w) has radius of convergence r > 1.V. Determine the radius of convergence of the power series expansion of f (z) about tei .8.V. Show that there is an angle such that f (z) does not have an analytic continuation along the path (t) = tei . where the series has radius of convergence R < 1. 0 t R. Solution 85 .7 1 2 3 P L K P Suppose f (z) = an z n . Solution 86 . and the radius of convergence of the power series ft (z) tends to 0 as t ! t1 .V. and let (t). be a path such that (a) = z0 . show that there is a parameter value t1 such that there is an analytic continuation ft (z) for a t < t1 .8 1 2 3 P L K Let f (z) be analytic at z0 . a t b. If f (z) cannot be continued analytically along .8. since the satisfy z wn = n 0. w) be a polynomial in z and w. p f (z)) = 0 is called an algebraic function. Solution 87 . the branches of z are algebraic functions. Show that if ft (z) is any analytic continuation of f (z) along any path starting at z0 . f (z)) = 0.V. ft (z)) = 0 for all t. An analytic function f (z) satis…es a polynomial equation P (z. then P (z. For instance. Remark. of degree n in w. Suppose that f (z) is analytic at z0 and satis…es P (z.8.9 1 2 3 P L K Let P (z. 10 1 2 3 P L K Let D be the punctured disk f0 < jzj < "g. Remark. use the fact that any path in D starting at z0 is the composition of a unique path in the half-plane starting at w0 and the exponential function ew . Solution 88 . Show that f (z) has an analytic continuation along any path in D starting at z0 if and only if there is an analytic function g (w) in the half-plane fRe w < log "g such that f (ew ) = g (w) for w near w0 . suppose f (z) is analytic at z0 2 D.8. For the proof. If f (z) does not extend analytically to D but has an analytic continuation along any path in D. we say that f (z) has a branch point at z = 0. and ew0 = z0 .V. VI 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 2 3 4 5 6 7 8 1 . VI.1 1 2 3 P L K 111 LLL Find all possible Laurent expansions centered at 0 of the following functions. we have 1 X 1 X X 1 k 1 2 1 1 1 1 k 2 = 2 1 = 2 = 2 z =[ k 2 = n] = zn: z2 z z 1 z z k=0 z z k=0 n= 1 (b) In the region jzj < 1 we have z 1 2 2 X 1 X 1 =1 =1 =1 2 ( 1)n z n = 1 2 ( 1)n z n : z+1 z+1 1 ( z) n=0 n=1 In the region jzj > 1 we have 2 X ( 1)k 1 z 1 2 2 1 =1 =1 1 =1 = z+1 z+1 z1 z z k=0 z k X1 ( 1)k X1 ( 1)n 1 X1 ( 1)n =1 2 k+1 = [k + 1 = n] = 1 2 =1+2 : k=0 z n=1 zn n=1 z n (c) From computation. (a) z21 z (b) zz+11 (c) (z2 1)(z 1 2 4) Solution (a) In the region 0 < jzj < 1. we have 2 .1. we have 1X k X X 1 1 1 1 1 1 k 1 = = z = z = [k 1 = n] = zn: z2 z z1 z z k=0 k=0 n= 1 In the regionjzj > 1. 1 1 1 1 1 = + (z 2 1) (z 2 4) 3z 2 1 3 z2 4 In the region jzj < 1. we have 1 1 1 1 1 = + = (z 2 1) (z 2 4) 3 z2 1 3 z2 4 1 1 1 1 = + = 3 z 2 1 z12 3 z 2 1 z42 1 X 1 1 X 4n 1 1 = + = 3z 2 n=0 z 2n 3z 2 n=0 z 2n 1X 1 1 X 4n 1X n 1 1 1 2(n+1) = + = (4 1) z = [n + 1 = k] = 3 n=0 z 2(n+1) 3 n=0 z 2(n+1) 3 n=0 1 X 1 1 k = 4 1 z 2k : 3 k= 1 3 . we have 1 1 1 1 1 1 1 1 1 = + 2 = 1 z2 = (z 2 1) (z 2 4) 3z 2 1 3z 4 3z 1 2 z2 34 1 4 1 X 1 1 X z 2n 1 X 2k 1 X z 2n 1 1 1 1 = = z 3z 2 k=0 z 2n 12 n=0 4n 3 k= 1 12 n=0 4n In the region jzj > 2. we have 1 1 1 1 1 1 1 1 1 = + = = (z 2 1) (z 2 4) 3z 2 1 3z 2 4 31 z 2 12 1 z42 1 X 2n 1 X z 2n 1 X 1 1 1 = z = 4 4 n z 2n : 3 n=0 12 n=0 4n 12 n=0 In the region 1 < jzj < 2. 4 .2 1 2 3 P L K 111 LLL For each of the functions in Exercise 1. Determine the largest open set on which each series converges.VI. (b) We have z 1 2 =1 z+1 z+1 The function (z 1) = (z + 1) converges on the set 0 < jz + 1j < 1. where an = 1 z2 z 2n+1 for n 0 k= 1 The function 1= (z 2 z) converges on the set 1 < jz + 1j < 2.1. …nd the Laurent expansion centered at z = 1 that converges at z = 12 . Solution (a) Partial fractions give us 1 1 1 = + z2 z z z 1 Laurent series for 1=z 1 X 1 1 1 1 1 1 = = 1 = = z z+1 1 z + 1 1 z+1 z + 1 k=0 (z + 1)k X 1 1 X1 = k+1 = [k + 1 = n] = (z + 1)n k=0 (z + 1) n= 1 Laurent series for 1=z 1 1 X (z + 1)n 1 X 1 1 1 1 1 1 = = = = (z + 1)n z 1 z+1 2 2 1 z+1 2 2 n=0 2n 2n+1 n=0 Thus we have 1 X 1 1 for n 1 = an (z + 1)n . 5 .(c) Partial fractions give us 1 1 1 1 1 1 1 1 1 = + + = (z 2 1) (z 2 4) 6 z 1 6 z + 1 12 z 2 12 z + 2 1 1 1 1 1 1 1 1 = + + = 6 z + 1 2 6 z + 1 12 z + 1 3 12 z + 1 + 1 1 1 1 1 1 1 1 1 = z+1 + z+1 1 = 12 1 2 6 z + 1 36 1 3 12 (z + 1) 1 + z+1 1 X (z + 1)n 1 1 1 X (z + 1)n 1 X ( 1)k 1 1 1 = + = 12 n=0 2n 6 z + 1 36 n=0 3n 12 k=0 (z + 1)k+1 1 X (z + 1)n 1 1 1 X (z + 1)n 1 X 1 1 n= 1 = + + ( 1)n (z + 1)n = 12 n=0 2n 6z +1 36 n=0 3n 12 1 1 X 1 X 1 1 1 1 1 1 ( 1)n (z + 1)n + (z + 1) 1 + (z + 1)n . 12 n=0 6 12 12 n=0 12 2n 36 3n The function 1= ((z 2 1) (z 2 4)) converges on the set 1 < jz + 1j < 2. tanz2 z 1 z 2 z+ =2 1 4 1 2 z+ =2 . : : : Simple poles at 2 + n lim (z sin z =2) cos z = sin 2 lim cos zz =2 cos =2 = sin =2 sin =2 = 1 z! =2 z! =2 sin z z+ =2 sin( =2) lim (z + =2) cos z = sin 2 lim cos z cos( =2) = sin( =2) = 1 z! =2 z! =2 ) Principal parts at 2 are z 1 =2 . and a0 = a2 = 0. z 2 z 2 z =2 2 z =2 .VI. 1. jz+ =2j=" = 24 2 i 6 . 2. deduce that Z 2 1 Jn (z) = ei(z sin n ) . z 2 C: 2 0 Remark. H H H H a1 = 21 i jzj=3 tan z 2 z dz = 1 2 i jzj=" + jz =2j=" jz+ =2j=" tan z z2 dz tan z sin z 1 H At z 0. ! 0 at ????? ) f (z) = f0 (z) + 6 f1 (z) is the Laurent decomposition for 3 < jzj < 4. so f0 (z) is odd. n 0. converges for jzj > =2.4). establish the Laurent series expansion hw i X1 exp (z 1=z) = Jn (w) z n . z2 . n = 0. Solution (a) f (z) = tan z.3 1 2 3 P L K LLL Recall the powe series for the Bessel function Jn (z). For …xed w 2 C.4.1. z+ 2 +z 2 2z 2z P1 2 1 n (b) f1 (z) = z 2 2 =4 = z 2 2 =4 = z 2 k=0 4 z 2 = P1 2 n 1 2 n=0 4 z 2n+1 . singularities at 2 + n . and de…ne J n (z) = ( 1)n Jn (z).11. 3 < jzj < 4 sin z f (z) = tan z = cos z . jzH =2j=" = 24 2 i At z 2 . jzj=" = 2 i tan z z cos z z 1 1 4 1 H At z 2 . also for 2 < jzj < 32 . given in Exercise V. This Laurent expansion is called Schlömlich formula. 0 < jzj < 1: 2 1 From the coe¢ cient formula (1. ) f0 (z) = f (z) + z 1 =2 + z+1 =2 is analytic for jzj < 32 f1 (z) = z 1=2 z+1 =2 is analytic for jzj > 2 . (c) f (z) is odd. the series converges for jzj < 32 . 7 . otherwise is analytic for jzj < 32 . 3 (d) Since f0 (z) = f (z) f1 (z) has poles at 2 . 4 1 2 3 P L K LLL Suppose that f (z) = f0 (z) + f1 (z) is the Laurent decomposition of an an- alytic function f (z) on the annulus fA < jzj < Bg. Solution H f (z) P 1 Use an = z n+1 dz. ) f0 (z) + f1 (z) have only even ????? ?????. and the Laurent series expansion of f (z) has only even powers of z. or de…nition of f0 (z)+f1 (z). ant the Laurent series expansion of f (z) has only odd powers of z. f even ) an = an for n odd ) an = 0 for n odd.1. f (z) = an z n . then f0 (z) and f1 (z) are even functions. Show that if f (z) is an even function. 8 . Show that if f (z) is an odd function.VI. then f0 (z) and f1 (z) are odd functions. f ( z) = jzj=r 1 P ( 1)n an z n . by the …rst theorem on page 126. Hence. Hence. k6= 1 k6= 1 k6= 1 since z k has a primitive if k 6= 1. Solution By the Laurent Expansion Theorem we have for all z 2 D and r > 0 X 1 Z k 1 f( ) f (z) = ak z where ak = d : k= 1 2 i Cr (0) ( z0 )k+1 1 R Setting c = a 1 = 2 i Cr (0) f ( ) d we have X f (z) c=z = ak z k : k6= 1 R Recall that an analytic function g has a primitive on a region D i¤ g (z) dz = 0 for every piecewise smooth P closed path in D. Show that there is a constant c such that f (z) c=z has a primitive in D.1. Let be such a path in D and observe the series k6= 1 ak z k converges uniformly on any closed annulus fz : r jzj sg with 0 < r < s and hence on .VI. we may integrate the series termwise we obtain: Z Z X X Z X k f (z) c=zdz = ak z = ak z k dz = ak 0 = 0. 9 .5 1 2 3 P L K LLL Suppose f (z) is analytic on the punctured plane D = Cn f0g. Give a formula for c in terms of an integral of f (z). f (z) c=z has a primitive in D. 6 1 2 3 P L K Fix an annulus D = fa < jzj < bg. If f is analytic. its Laurent decomposing. and clearly F n@D = f . so gk ! f uniformly on D. 10 .1. Show that f (z) can be approximated uniformly on @D by polynomials in z and 1=z if and only if f (z) has continuous extension to the closed annulus D [ @D that is analytic on D. V5 14. then by the maximum principle jgk gj j ! 0 uniformly on D. Then f0 + f1 have continuous boundary values. (?????? change of variable w = 1=z ). That does it.VI. Can approximate f0 (z) uniformly by polynomials in z for f1 (z) uniformly by polynomials in 1=z. If f = lim gn (z) uniformly on @D. continuous ????? values. Solution Use the conformity theorem for polynomial approximation. and let f (z) be a continuous function on its boundary @D. unit f = f0 +f1 . ) Solution By III. Use a decomposition of the form u = Re f +c log jzj.4. get for R n 6= 0.3.3. A < jzj < B. ) = 21 c log rd + 2 0 d = 0 + c log r Note: 11 . Z 1 a0 + c log r = u rei d : 2 Hint.VI. (See Exercise III. Show that for each r. bn and c satisfy Z n n 1 an r + a nr = u rei cos (n ) d . n= 1 n6=0 which is uniformly convergent on each circle in the annulus. ) sin n d . n 2 n u (r. n 6= 0. 1 R R If we just integrate. the coe¢ cients an . A < r < B. ) cos n d = n r cos n d + n r cos2 n d = n n ( nr + nr ) R ) n rn + n r n = 1 u (r. n n 1 R The formula for n r nr R = u (r. is similar.7 1 2 3 P L K Show that a harmonic function u on an annulus fA < jzj < Bg has a unique expansion X 1 X u rei = an rn cos (n ) + bn rn sin (n ) + c log r. n 6= 0. where f is analytic on the annulus. n 6= 0.4. ????? useR orthogonality.1. Z n n 1 bn r + b nr = u rei sin (n ) d . we get u (r. n 6= 0. we have u = Re f +c log jzj for some e: Write itsPLaurent series f (z) = 1 k k= a ck z . ) cos n d . Then for Pck = k k . wek have Re f P= 1 k= 1 k Re z + k Im z k u= 1 k k k= 1 k r cos (k ) + k r sin (k ) + c log The series converging uniformly on ????? The term w= 0 ????? Multiply R by cos n . If have strict inequality somewhere.6).1. f ei ie i M. jcj = M . (See i Exercise III.8 (ej) H Multiply f by a unimodular constant. ) f (z) = cz for a constant. di¤erentiate h by hand. VI. The Deriv- R h(z) h(z) ative is H 0 (w) = lim 1w z (w+ w) z w dz w!1 R h(z) w R = lim 1w (z (w+ w))(z w) dz = (zh(z)dz w)2 w!1 12 . ) f ei iei M .9 (ej) For the analyticity. assume jf (z)j 6 M . f (z) iM z.VI.1. R2 then < 2 M . have Re f ei iei d = 2 M. 0 0 Re f ei iei d 6 f ei 6 M . we have Im f ei iei d 0. f (z) dz = jzj=1 R2 R2 f ei d = 2 M . Take real parts. Since f ei iei 6 0 M .1. We conclude that Re f ei iei d M . Similarly.J Groves) z g(z) zez We have that z = 1 is a pole of order 1 since zze 2 1 = z 1 with g (z) = z+1 .2. Determine the order of any pole. Similarly we have.VI. tan z = cos z has simple poles at z = =2 + k for k 2 Z. and …nd the principal part at each pole. (b) (J. Hence.J Groves) We have that z = 1 is a pole of order 2 since (z2 z 1)2 = (zg(z) 1)2 with g (z) = z (z+1)2 and g (z) is analytic near z = 1.J Groves) We have that cos z has simple zeros at z = =2 + k for k 2 Z. Kumjian) 13 . essential. So cos1 z has simple poles at z = =2 + k for k 2 Z. (d) (J. It has an isolated singularity at z = 0. 2 sin z (a) z= (z 2 1) (d) tan z = cos z (g) Log 1 z1 z (b) zze 2 1 (e) z 2 sin z1 (h) (zLog1)z3 (i) e1=(z +1) 2z 2 (c) e 1 z (f) cos z z2 2 =4 Solution (a) (J. and h (z) is analytic near z = 1. z = 1 is a pole h(z) of order 2 since (z z1)2 = (z+1) 2 with h (z) = z (z 1)2 and h (z) is analytic near z = 1. Kumjian) Note that the given function is analytic on the punctured plane D = Cn f0g. (e) (A. the function has a removable singularity at 0. it extends to an analytic function at z = 0. Since sin z is analytic and di¤erent sin z to 0 at these points. and de- termine where they are removable.R. or poles.R. For z 6= 0 we have ! e2z 1 1 X1 (2z)k 1 X 2k z k 1 1 X 2j+1 j 1 = 1+ = = z : z z k=0 k! z k=1 k! z j=0 (j + 1)! Since the function has a power series expansion for all z 6= 0. z = 1 is a pole of order 1 since zez z z2 1 = h(z) z+1 with h (z) = zze 1 . (c) (A.R. and g (z) is analytic near z = 1.1 1 2 3 P L K LLL Find the isolated singularities of the following functions. it follows that Log 1 z1 is analytic on Cn [0. There are no isolated singularities.R. It has an isolated singularity at z = 0. (e) (J. we deduce that our function has a removable singularity at w = 0 and so at z = =2. Set w = z =2.J Groves) The function z 2 ( =2)2 = (z =2) (z + =2) has simple zeros at z = =2 and z = =2. (g) (J. So we suspect a removable singularity.The function is analytic on the punctured plane D = Cn f0g. z n=0 (2n + 1)! z n=0 (2n + 1)! z z = 0 is an essential singularity. (h) (A.R. Since 1 X1 ( 1)n 1 X1 ( 1)n 1 z 2 sin = z2 2n+1 = 2n 1 .R.J Groves) The function Log (z) is analytic on the region Cn ( 1. it also has a removable singularity at z = =2. For z 6= 0 we have 1 X1 ( 1)k 1 2k+1 2 2 z sin =z = z k=0 (2k + 1)! z X1 ( 1)k 1 X1 ( 1)k 1 = 2k 1 =z+ : k=0 (2k + 1)! z k=1 (2k + 1)! z 1 2k Hence z = 0 is an essential singularity for the function since there are an in…- nite number of nonzero terms in the Laurent series with negative exponents. In a similar fashion. Then cos (z) cos (z) cos (w + =2) 1 sin w = = = : z2 ( =2)2 (z =2) (z + =2) w (w + ) w+ w Since we know that sin (w) =w = 1 w2 =3! + : : :. (f) (J.J Groves) The function z 2 sin z1 has an isolated singularity at z = 0. 0]. The function cos z has a simple zero at z = =2. Kumjian) First observe that for z 2 C with jz 1j < 1 we have 1 1 X 1 = = ( 1)k (z 1)k : z 1 + (z 1) k=0 14 . 1]. R. e z2 +1 has an essential singularity at z = i. Similarly. Hence. its power series expansion centered at z0 = 1 my be obtained from that of 1=z by integrating termwise: we obtain (note Log 1 = 0) X1 ( 1)k+1 Log z = (z 1)k . 1 i 1 1 = z2 +1 2 z+i z i so that 1 i 1 i 1 i 1 X1 ( i)n 1 e z2 +1 = e 2 z+i e 2 z i = e2 z+i : n n=0 2 n! (z i)n i 1 1 since e 2 z+i is analytic and non-zero at z = i. (h) (J. The principal part is given by (z 11)2 2(z1 1) . 15 .J Groves) There are isolated singularities at z = i and z = i. it has an essential singularity at z = i. This follows from the fact that 2 3 Log z = (z 1) (z 21) + (z 31) for jz 1j < 1 so that Log z 1 1 1 1 3 = + : (z 1) (z 1)2 2 z 1 3 (i) (J.R. k=1 k for jz 1j < 1.Since Log z is a primitive of 1=z. z = 0 is a pole of order 2.J Groves) We have that z = 1 is a pole of order 2. we have for 0 < jz 1j < 1 Log z 1 X1 ( 1)k+1 k X 1 ( 1)k+1 = (z 1) = (z 1)k 3 = (z 1)3 (z 1)3 k=1 k k=1 k 1 1 X1 ( 1)j = 2 + (z 1)j (z 1) 2 (z 1) j=0 k + 3 Hence. Moreover. so values of e1=(z +1) do not converge.(a) Double poles at 1 . (g) De…ned and analytic for 1 1=z 2 Cn [ 1. 1]. cos z = sin ( =2 + n ) (z =2 n ) + O (z =2 n )3 sin( =2+n )+O((z =2+n )2 ) tan z = sin( =2+n )(z =2 n )+O (:::)3 = z =21 n + O (z =2 n ) Poles ( ) are simple. : : :. z ez ez (b) Singularity at z = 1. (d) tan z = sin z= cos z. 2. No isolated singularities. poles at z = =2 + n . Log z = dw w = 1+w 1 = ( 1)n (w 1)n dw = 1 P 1 n ( 1) (z 1) n+1 (z 1) 2 (z 1) 3 n+1 Log z = (z 1) 2 + 3 ::: n=0 Log z 1 1 (z 1)3 = (z 1)2 z(z 1) + analytic Double pole. 1. z 2 Cn [0. (i) e1=(z +1) . principal part 1= (z =2 n ) . Principal part 1=2 1=2 z 1 at z = 1. 1=z 2 Cn [1. and singularities are essential.????? 2 check e1(1 t ) = f (it) ! +1 as t " 1 or t # 1. singularities 2 at z = i. so no limit at i. singularities at =2 are both removable. 1=z 1 2 Cn [0. Rz R2 dw RP (h) Isolated singularity at z = 1. e1(1 t ) = f (it) ! 0 as 2 2 t # 1 or t " 1. simple poles zze 2 1 = 2 1 z 1 + z+1 . 2 2 (f) cos z= (z =4). z+1 at z = 1. 0). 1). 1). principal part (z 11)2 z(z1 2) at z = 1. Values of 1= (z 2 + 1) approach 1 from all directions as z ! i. n = 0. 16 . principal parts (1=4) (z 1)2 . expanded about the indicated point.VI. 2 . 2 2 (d) z = sin z. R = 3 = ?????? to i (nearest singularity). p (c) removable at z = 0. simple poles at z = 0. R=j i j= 2. R = . 1. (d) sin3 z . about z = 3 + i. 17 .2 1 2 3 P L K LLL Find the radius of convergence of the power series for the following functions. about z = i: Solution (a) Removable singularity at z = 1. (c) sinz z .2. poles at i. (a) zz4 11 . poles at . about z = i. about z = 0. cos z z2 (b) z2 2 =4 . (b) singularities at =2 are removable. principal parts 1= (z =2). By uniqueness. (d) What is the radius of convergence of the power series expansion for f0 (z)? Solution (a) tan z has two poles in the disk jzj < 4. z+ =2+z =2 (b) Use geometric series. then f0 (z) = f (z) f1 (z) is analytic for jzj < 4. (c) Obtain the coe¢ cients a0 . otherwise is analytic for jzj < 3. and determine the largest domain on which it converges. f1 (z) = z2 2 =4 = 2z 2z P 2 1 1 n z2 2 =4 = z2 4 z2 = . (d) Since f0 (z) = f (z) f1 (z) has poles at 3 =2. simple poles at =2. f (z) = f1 (z) + f2 (z) is the Laurent decomposition. the series converges for jzj < 3 =2.VI. and R = 3 =2. so that f0 (z) is analytic for jzj < 4. Converges for jzj > =2. converges for jzj > =2. and f1 (z) is analytic for jzj > 3 and vanishes at 1.2. (b) Write down the series expansion for f1 (z). 18 . a1 = 1 + 8= 2 . k=0 P 1 2 n 1 2 4 z 2n+1 n=0 (c) a0 = a2 = 0. (a) Obtain an explicit expression for f1 (z). If f1 (z) = 1= (z =2) 1 (z + =2). Let f (z) = f0 (z) + f1 (z) be the Laurent decomposition of f (z). and f1 (z) is analytic for jzj > 3 and ! 0 as z ! 1. a1 and a2 of the power series expansion of f0 (z).3 1 2 3 P L K Consider the function f (z) = tan z in the annulus f3 < jzj < 4g. where f1 (z) is the sum of the principal parts of f (z) at its poles Solution Since f1 (z) is a sum of principal parts. By the uniqueness of the Laurent decomposition.2. Show that the Laurent decomposition of f (z) with respect to the annulus fs " < jzj < sg has the form f (z) = f0 (z) + f1 (z). Further f0 (z) = f (z) f1 (z) is analytic at ????? ??????. with only a …nite number of poles in the disk.4 1 2 3 P L K Suppose f (z) is meromorphic on the disk fjzj < sg. hence for jzj < 5. 19 . f is f0 (z) + f1 (z) = f (z). and f1 (z) ! 0 as z ! 1. with poles inside fjzj 6 "g. Use the uniqueness of the Laurent decomposition.VI. f1 (z) is analytic for jzj > ". (z z0 )N f (z) has a removable singularity at z0 . and if (z z0 ) f (z) ! 0 as z ! z0 . prove that if z0 is an isolated singularity of f . we have X 1 X 1 f (z) = ak (z z0 )k N = aj+N (z z0 )j : k=0 j= N It follows that z0 is a pole of order at most N unless ak = 0 for all k = 0. Give a second proof based on Morea’s theorem. : : : . Then by Riemann’s Theorem on removable singularities (see p. 1. the Laurent expansion is of the following form: there is a r > 0 so that N X 1 (z z0 ) f (z) = ak (z z0 )k k=0 for all 0 < jz z0 j < r. 172). N 1.VI. 20 .2. in which case z0 is a removable singularity. then z0 is removable. And hence.5 1 2 3 P L K By estimating the coe¢ cients of the Laurent series. Solution Suppose that z0 is an isolated singularity of f and that (z z0 )N f (z) is bounded near z0 . VI. and if f (z)8 is analytic on D.6 1 2 3 P L K LLL Show that if f (z) is continuous on a domain D.2. Solution 21 . then f (z) is analytic on D. the Laurent expansion is of the following form: there is r > 0 so that N X 1 (z z0 ) f (z) = ak (z z0 )k k=0 for all 0 < jz z0 j < r. (z z0 )N f (z) has a removable singularity at z0 . N 1. then z0 is either removable or a pole of order at most N . 1. 22 . 172). And hence.7 1 2 3 P L K LLL Show that if z0 is an isolated singularity of f (z). and if (z z0 )N f (z) is bounded near z0 . : : : . Solution Suppose that z0 is an isolated singularity of f and that (z z0 )N f (z) is bounded near z0 . we have X 1 k N X 1 f (z) = ak (z z0 ) = aj+N (z z0 )j : k=0 j= N It follows that z0 is a pole of order at most N unless ak = 0 for all k = 0.2.VI. in which case z0 is a removable singularity. Then by Riemann’s Theorem on removable singularities (see p. 2.VI. Show that (a) order (f g. z0 ). z0 ) . z0 )g. (c) order (f + g. (b) order (1=f. order (g. Show that the inequality can occur in (c). z0 ) = order (f. but that equality hold in (c) whenever f and g have di¤erent orders at z0 . z0 ) = order (f. z0 ). Solution 23 . The order of the function 0 is de…ned to be +1. z0 ) + order (g. z0 ) min forder (f.8 1 2 3 P L K A meromorphic function f at z0 is said to have order N at z0 if f (z) = (z z0 )N g (z) for some analytic function g at z0 such that g (z0 ) 6= 0. This shows that the use of notation O (z m ) in section V. Solution 24 .9 1 2 3 P L K Recall that "f (z) = O (h (z)) as z ! z0 " means that there is a constant C such that jf (z)j C jh (z)j for z near z0 .VI.2. Show that if z0 is an isolated singularity of an analytic function f (z). that is the Laurent series of f (z) has the form f (z) = am (z z0 ) + am+1 (z z0 )m+1 + : Remark. then the Laurent coe¢ cients of f (z) are 0 for k < m. and if f (z) = O ((z z0 )m ) as z ! z0 .6 is con- sistent. 10 1 2 3 P L K LLL Show that if f (z) and g (z) are analytic functions that both have the same order N 0 at z0 .2. then f (z) f (N ) (z0 ) lim = (N ) : z!z0 g (z) g (z0 ) Solution 25 .VI. so 0 PN Al P that ak =ak+1 ! z0 as n ! 1. z 1z0 = z10 1 z=z1 0 = 1 z0 z zj . If f (z) = an z n as only one pole on the circle jzj = R. h (z) = bk z k analytic. and suppose that f (z) extends to be meromorphic for jzj < R + ".11 1 2 3 P L K P Suppose f (z) = ak z k is analytic for jzj < R. Solution P If f (z) = an z n is analytic for jzj < R. of ????? N .VI. with only one pole z0 on the circle jzj = R. Show that ak =ak+1 ! z0 as k ! 1. ( 1)d (z(l z1)!)l = 0 0 j=0 1 P 1 j(j 1):::(j l+1)z j l z0 z0l j=l 1 ( 1)l+1 P 1 1 (z z0 )l = (l 1)! z0k+l+1 (k + l) : : : (k + 1) z k k=0 1 ( 1)l+1 P1 1 (z z0 )l = (l 1)! z0k+l+1 (k + l) : : : (k + 1) z k k=0 26 . with polesP of ?????6 N on the circle jzj = R. with A( 1)N +1 kN f (z) = A= (z z0 )N + lower order. then an = O nN 1 . then ak = (N 1)!z N +k+1 (1 + O (1=k)).2. and is meromorphic for jzj < R + ". bk = O 1= jz0 jk . Proof: Write f (z) = (z z0 )l + bk z k P l=0 analytic. We have jbk j 6 1=sk for some s > P 1 j R. Solution Apply Arzela‘-Weierstrass theorem. In any neighbourhood of 0 there are values of ef (z) = ew that are near 0 (e m ) and that are near to (e+m ). 0 essential ) values of f (z) ???? on C as z ! 0. ! values of ef (z) ????? on C as z ! 0.VI. as z ! 0.2. monotone. so ef (z) does not have a limit as z ! 0. 0 a pole ) f (z) ! 1. an isolated singularity of f (z). then z0 is an essential singularity for ef (z) . values of f (z) ?????? of 0 cover the extremum of some ?????? fjwj > Rg. 0 is essential singularity of ef (z) . Suppose z0 = 0. 27 . not removable.12 1 2 3 P L K LLL Show that if z0 is an isolated singularity of f (z) that is not removable. Solution Suppose values of f (z) do not cluster at L as z ! z0 .13 (From Hints and Solutions) 1 2 3 P L K Let S be a sequence converging to a point z0 2 C. to see that g (z) extends to be analytic for jz z0 j < ". then to z0 . Show that either f (z) extends to be meromorphic on some neighborhood of z0 . and f (z) is meromorphic there 28 . or else for any complex number L there is a sequence fwj g such that wj ! z0 and f (wj ) ! L. Then g (z) = 1= (f (z) L) is bounded for jz z0 j < ".VI. Apply Riemann‘s the- orem …rst to the zj `s for j large. and let f (z) be analytic on some disk centered at z0 except possibly at the points of S and z0 . z 6= zj .2. get a m = 0 for m > . with Laurent series as in Exercise 7 of Section 1. Let r ! 0. Solution P 1 P 1 u rei harmonic.14 1 2 3 P L K Suppose u rei is harmonic on a punctured disk f0 < r < 1g. m > 0. 29 . 0 < r < 1. Similarly b m = 0 for m > .2. Suppose > 0 is such that r u rei ! 0 as r ! 0. am r2m + a m = O (r +m ) as r ! 0.VI. Have am rm + a m r m = O (r ). n=0 n6=0 Let m > . u = an rn cos n + bn rn sin n + c log r. Show that an = 0 = bn for n . Consider Laurent series as in 14. 30 .VI. For m > 1. a m = 0 for all m > 0. and u = Re f is analytic at z = 0. c = u(r) Re f (rei ) c log r constant. u rei = Re f rei + c log r. What can you say if you know only that ju (z)j C log (1= jzj) for some …xed constant C and 0 < jzj < ? Solution Assume u (z) = O (log (1= jzj)) R as z ! 0.15 1 2 3 P L K Suppose u (z) is harmonic on the punctured disk f0 < jzj < g. when f is analytic. Show that if u (z) !0 log (1= jzj) as z ! 0. am r2m + a m = O (rm log (1=r)). Have. log(1=r) = log(1=r) + log(1=r) . and b m = 0 for m > 0. c = 0. get u rei d = O (log (1=r)) am rm + a m r m = O (log (1=r)). then u (z) extends to be harmonic at 0.2. and classify them.VI. Determine which have isolated singularities at 1.1 1 2 3 P L K Consider the functions in Exercise 1 of Section 2 above.3. Solution 31 . 2 1 2 3 P L K Suppose that f (z) is an entire function that is not a polynomial. What kind of singularity can f (z) have at 1? Solution 32 .3.VI. VI. then ef (z) has an essential singularity at z = 1. Solution 33 .3 1 2 3 P L K Show that if f (z) is nonconstant entire function.3. 4 1 2 3 P L K Show that each branch of the following functions is meromorphic at 1.VI. 2 5=2 3 (a) (z 1) (b) 3 (z 1) (c) z 2 1=z Solution 34 .3. and obtain the series expansion p for each branch at p 1. 1 1 2 3 P L K LLL Find the partial fractions decompositions of the following functions.4. (a) z21 z (c) (z+1)(z12 +2z+2) z 1 (e) z+1 z 2 +1 1 2 (b) z(z 2 1) (d) (z2 +1)2 (f) zz2 4z+3 z 6 (a) 1 A B = + z2 z z z 1 1 A = lim z f (z) = = 1 z!0 z 1 z=0 1 B = lim (z 1) f (z) = =1 z!1 z z=1 1 1 1 = + z2 z z z 1 (b) z2 + 1 A B C 2 = + + z (z 1) z z 1 z+1 z2 + 1 A = lim z f (z) = = 1 z!0 z2 1 z=0 2 z +1 B = lim (z 1) f (z) = =1 z!1 z (z + 1) z=1 z2 + 1 C = lim (z + 1) f (z) = =1 z! 1 z (z 1) z= 1 z2 + 1 1 1 1 = + + z (z 2 1) z z 1 z+1 (c) 1 A B C = + + (z + 1) (z 2 + 2z + 2) z+1 z+1+i z+1 i 35 .VI. 1 A = lim z f (z) = =1 z! 1 (z 2 + 2z + 2) z= 1 1 1 B = lim (z + 1 + i) f (z) = = z! 1 i (z + 1) (z + 1 i) z= 1 i 2 1 1 C = lim (z + 1 i) f (z) = = z! 1+i (z + 1) (z + 1 + i) z= 1+i 2 1 1 1=2 1=2 = (z + 1) (z 2 + 2z + 2) z+1 z+1+i z+1 i (d) (/snider 106) man så? 1 A B C D 2 = 2 + + 2 + (z 2 + 1) (z i) (z i) (z + i) z i 1 1 A = lim (z i)2 f (z) = 2 = z!i (z + i) z=i 4 d 1 i B = lim (z i)2 f (z) = 2 = z!i dz (z + i) z=i 4 1 1 C = lim (z + i)2 f (z) = 2 = z! i (z i) z= i 4 d 1 i D = lim (z + i)2 f (z) = 2 = z! i dz (z i) z= i 4 1 1=4 i=4 1=4 i=4 = 2 + (z 2 + 1)2 (z i)2 (z i) (z + i) z i (e) z 1 A = +B z+1 z+1 A = lim (z + 1) f (z) = z 1jz= 1 = 2 z! 1 B = lim f (z) = 1 z!1 36 . z 1 2 =1 z+1 z+1 (f) z2 4z + 3 A B = + +C z2 z 6 z+2 z 3 z2 4z + 3 A = lim (z + 2) f (z) = = 3 z! 2 z 3 z= 2 z 2 4z + 3 B = lim (z 3) f (z) = =0 z!3 z (z + 1) z=3 C = lim f (z) = 1 z!1 z2 4z + 3 3 = +1 z2 z 6 z+2 37 . 1 1 A = lim (z 1) f (z) = = z!1 z2 +z+1 z=1 3 1 w B = lim (z w) f (z) = = z!w (z 1) (z e 2 i=3 ) z=e2 i=3 3 1 w C = lim (z w)2 f (z) = 2 i=3 = z!w (z 1) (z e ) z=e 2 i=3 3 z9 + 1 1=3 w=3 w=3 6 = z3 + + + .VI. where w = e2 i=3 z 1 z 1 z w z w 38 . z2 = e2 i=3 . and z3 = e4 i=3 . and we see that z3 = w.2 1 2 3 P L K LLL Use the division algorithm to obtain the partial fractions decomposition of the following functions 3 9 z6 (a) zz2 +1 +1 (b) zz6 +11 (c) (z2 +1)(z 1)2 (a) z3 + 1 z+1 A B = z + = z + + z2 + 1 z2 + 1 z+i z i z+1 1 1 A = lim (z + i) f (z) = = + i z! i z i z= i 2 2 z+1 1 1 B = lim (z i) f (z) = = i z!i z + i z=i 2 2 z3 + 1 z+1 ( 1 + i) =2 (1 + i) =2 2 =z+ 2 =z+ z +1 z +1 z+i z i (b) z9 + 1 3 3 z +1 1 1 A B C 6 = z + 6 = z3+ 3 = z3+ 2 = z3+ + + z 1 z 1 z 1 (z 1) (z + z + 1) z 1 z w z w Because z 3 1 = 0 have tree roots z1 = 1. Set z2 = w.4. Konstigt svar i boken? 1=3 z 1 + zw=3w + zw=3w 1=3 z 1 + w=3 z w w=3 z+w = (c) z6 z6 = (z 2 + 1) (z 2 2z + 1) z4 2z 3 + 2z 2 2z + 1 z6 2z 3 z 2 + 2z 2 = z 2 + 2z + 2 + z4 2z 3 + 2z 2 2z + 1 (z 2 + 1) (z 1)2 2z 3 z 2 + 2z 2 A B C D 2 = + + 2 + (z 2 + 1) (z 1) z + i z i (z 1) z 1 2z 3 z 2 + 2z 2 1 A = lim (z + i)2 f (z) = = z! i (z i) (z 1)2 z= i 4 2z 3 z 2 + 2z 2 1 B = lim (z i) f (z) = 2 = z!i (z + i) (z 1) z=i 4 2z 3 z 2 + 2z 2 1 C = lim (z 1)2 f (z) = 2 = z!1 (z + 1) z=1 2 3 2 d 2z z + 2z 2 5 D = lim (z 1)2 f (z) = = z!1 dz (z 2 + 1) z=1 2 39 . VI.3 1 2 3 P L K LLL Let V be the complex vector space of functions that are analytic on the extended complex plane except possibly at the points 0 and i. Solution 40 .4. where they have poles of order at most two. What is the dimension of V ? Write down explicitly a vector space basis for V . then so do f (z) +g (z) and f (z) g (z). Solution 41 .1 1 2 3 P L K Show that if f (z) and g (z) have period w.VI.5. 2 1 2 3 P L K Expand 1= cos (2 z) in a series of powers of e2 iz that converges in the upper half-plane.VI. Determine where the series converge absolutely and where it converges uniformly.5. Solution 42 . Also …nd an expansion of tan z as in an exponential series that converges in the lower half-plane. that converges in the upper half-plane.VI. Solution 43 . 1 < k < 1.5.3 1 2 3 P L K Expand tan z in a series of powers of exponentials eikz . VI. with real period 2 > 0. for …xed " > 0. C > 0 such that jf (x + iy)j CeAy for y > 0. n A where the series converges uniformly in each half-plane fy "g. Show that X f (z) = an einz= . Suppose that there are A.5. Solution 44 .4 1 2 3 P L K Let f (z) be an analytic function in the upper half-plane that is periodic. 5. Solution 45 .VI. and suppose that there are no periods w of f (z) satisfying 0 < jwj < 1. and sketch the set of periods for each possibility. How many periods of f (z) lie on the unit circle? Describe the possibilities.5 1 2 3 P L K Suppose that 1 are periods of a nonzero doubly periodic function f (z). 2 ). w2 ) = ( 1 . Solution 46 . then 1 and 2 generate the periods of f (z) if and only if there is a 2 2 matrix A with integer entries and with determinant 1 such that A (w1 .VI.6 1 2 3 P L K We say that w1 and w2 generate the periods of a double periodic function if the periods of the function are precisely the complex numbers of the form mw1 + nw2 where m and n are integers.5. and if 1 and 2 are complex numbers. Show that if w1 and w2 generate the periods of a doubly periodic function f (z). Show that the series X 1 1 m. whose periods are generated by w1 and w2 .VI.n= 1 (z (mw1 + nw2 ))k converges uniformly on any bounded subset of the complex plane to dou- ble periodic meromorphic function f (z). Let k 3.5.7 1 2 3 P L K Let w1 and w2 be two complex numbers that do lie on the same line through 0. Solution 47 . Show that the number of periods in any annulus fN jzj N + 1g is bounded by CN for some constant C. Strategy. The cosine series is 4 1 1 j j= cos + 2 cos 3 + 2 cos 5 + : 2 3 5 Solution 48 . Find the i complex Fourier series f e and show that it can be expressed as a cosine series. . Sketch the graphs of the …rst three partial sums of the cosine series.6.1 1 2 3 P L K Consider the continuous function f ei = j j.VI. Discuss the convergence of the series. Does it converge uniformly? Partial answer. < (the principal value of the argument).VI.6. Solution 49 .2 1 2 3 P L K Let f ei . while the sine series converges at . Di¤erentiate the complex Fourier series term by term and determine where the di¤erentiated series converges. Show that the complex Fourier series diverges at = . Find the complex Fourier series of f ei and the sine series of f ei . Find the i complex Fourier series of f e and show that it can be expressed as a cosine series. show that 2 1 1 1 =1 + + : 12 22 32 42 Solution 50 . Discuss the convergence of the series.6.VI. .3 1 2 3 P L K Consider the continuous function f ei = 2 . Does it converge uniformly? By substituting = 0. . Solution 51 . Find i the complex Fourier series of f e and show that it can be expressed as a cosine series.VI. Relate the Fourier series to the series of the function in Exercise 3.4 1 2 3 P L K Consider the continuous function f ei = 4 2 2 2 .6. then ck e converges absolutely for each . Solution 52 . and the series converges uniformly for .VI.5 1 2 3 P L K P P ik Show that if ck converges absolutely.6. VI.6. where g (z) and h (z) are continuous functions on the unit circle that extend continuously to be analytic on the open unit disk.6 1 2 3 P L K Show that any function f ei on the unit circle with absolutely convergent Fourier series has the form f ei = g ei + h (ei ). Solution 53 . 7 1 2 3 P L K P If f ei ck eik . and the series converges uniformly to f ei . Solution 54 . Formula (6.6. This is called Parseval’s identity. then Z X1 2 d f e i = jck j2 : 2 k= 1 Remark.6) show Parseval’s identity holds for a function f ei if and only if the partial sums of the Fourier series of f ei converge to f ei in the sense of "mean-square" or "L2 approximation".VI. 6.8 1 2 3 P L K By applying Parseval’s identity to the picewise constant function with series (6. show that 2 1 1 1 =1+ 2 + 2+ 2+ : 8 3 5 7 Use this identity and som algebraic manipulation to show that 2 1 1 1 =1+ 2 + 2+ 2+ : 6 2 3 4 Solution 55 .5).VI. VI.9 1 2 3 P L K By applying Parseval’s identity to the function of Exercise 1. show that 4 1 1 1 =1+ 4 + 4+ 4+ : 96 3 5 7 Use this identity and som algebraic manipulation to show that 4 1 1 1 =1+ 4 + 4+ 4+ : 90 2 3 4 Solution 56 .6. with Laurent expansion ak z k .10 1 2 3 P L K If f (z) is analytic P in some annulus containing the unit circle jzj = 1.6.VI. then I X 1 1 jf (z)j2 jdzj = jak j2 : 2 jzj=1 k= 1 Solution 57 . Solution 58 . with Fourier series P ck eik .6. Show that f ei extends to be analytic on some annulus con- taining the unit circle if and only if there exist r < 1 and C > 0 such that jck j Crjkj for 1 < k < 1.VI.11 1 2 3 P L K Let f ei be a continuous function on the unit circle. 12 1 2 3 P L K Using the convergence theorem for Fourier series. Solution 59 . First approximate f eik by a smooth function. 1 < k < 1. by …nite linear com- binations of exponentials eik . prove that every continti- nously function on the unit circle in the complex plane can be approximated uniformly there by trigonometric polynomials. Strategy. that is.6.VI. sot the P boundary is given by iks a smooth periodic function (s). with equality if and only if D is a disk. (a) P 2 Show that k jck j2 = 1. Solution 60 . Hint.4. This proves the isoperimetric theorem.1. (b) P Show that the area of D is k jck j2 . (c) Show that the area of D is . the curve that surrounds the largest area is a circle. Hint.6.VI. Remark. Use Exercise IV. Among all smooth closed curves of a given length. Let ck e be the Fourier series of (s). 0 s 2 . Apply Parseval’s identity to 0 (s) and use j 0 (s)j = 1 for a curve parameterized by arc length.13 1 2 3 P L K Let D be a domain bounded by a smooth boundary curve of length 2 . We parametrize the boundary of D by arc length s. VI. for …xed m and n. k= m 2 k= m 2 k= m for any choice of complex numbers bk . is the corresponding partial sum of the Fourier series. This i shows Pn that the best mean-square approximate to f e by exponential sums ik m bk e .14 1 2 3 P L K Show that Z X n 2 Z X n 2 X n d d f e i ik bk e = f e i ik ck e + jbk ck j2 . Solution 61 . m k n. Remark.1. Solution 62 .1.VI.15 1 2 3 P L K Show that a continously di¤erentiable function on the unit circle has an absolutely convergent Fourier series. where ak = 1=ik and bk is the Fourier coe¢ cient of the derivative. Strategy. qPUse Bessel’ qP s inequality and the Cauchy-Schwarz inequality P 2 k k j kj j k j2 . Write the Fourier coe¢ cients ck of f ei as ak bk . at each of which the derivative has limits from the left and from the right. whose Fourier series is absolutely convergent. Strategy. Suppose that f ei is piecewise continuously di¤erentiable.VI.16 1 2 3 P L K Let f ei be a continuous function on the unit circle. in the sense that it has a continuous derivative except at a …nite number of points. Solution 63 . Cancel the discontinuities of the derivative using translates of the function in Exercise 3. Show that the Fourier series f ei is absolutely convergent.6. 17 1 2 3 P L K Let f ei be a piecewise continuously di¤erentiable. Strategy. where f1 e satis…es the hypotheses of Exercise 15.VI. Show that i i i f e = f1 e + bj hj e . and otherwise to the average of the limits of f ei from P the ileft and from the right. and each hj ei is obtained from the function of Exercise 2 by change of variable 7 ! j. to f ei if the function is continuous at ei .6. Show that the Fourier series of f ei converges at each point. Solution 64 . at each of which both the function and its derivative have limits from the left and from the right. in the sense that it is continuously di¤erentiable except at a …nite number of points. VII 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 X X X X 2 X X X X X X X X X X X X 3 X X X X X X X 4 X X X X X X X X X X 5 X X X X X X 6 X X X X X X X 7 X X X 8 X 1 . 1 (f) Res [cot z. 0 = lim = 1: z z!0 z e) By rule 2. h i (a) Res z21+4 .1 1 cos z ez (b) Res z 2 +4 .0 z 5n .1 Evaluate the following residues. 1 1 1 i Res . 1 1 1 i Res . wk Solution a) By rule 4. 2i = = = : z2 +4 2z z=2i 4i 4 b) By rule 4. 2i (e) Res (h) Res z2 .0 = = 1: sin z cos z z=0 2 .1 = = : z5 1 5z 4 z=1 5 d) By rule 1. 0] = Res . h cos z i d Res . 0] (i) Res zn 1 .0 (g) Res z Log z . 0 = lim cos z = sin zjz=0 = 0: z2 z!0 dz f) By rule 3. 2i (d) Res sin z2 z . 1 1 1 Res . h cos z i cos z Res [cot z. sin z sin z Res 2 . 2i = = = : z2 +4 2z z= 2i 4i 4 c) By rule 4.VII.1.0 1 z +1 (c) Res z5 1 . z5 z5 z 5 1!z 4 2!z 3 3!z 2 4!z hence ez 1 1 Res 5 . where k = 0. wk = = n 1 = = = : zn 1 nz n 1 z=wk nwk nwkn 1 n n 3 . z z Res .0 = = : z 4! 24 i) We have poles of the form wk = e2 ik=n .1 = = 1: Log z 1=z z=1 h) By Laurent expansion z z2 z3 z4 ez 1+ 1! + 2! + 3! + 4! + o (z 5 ) 1 1 1 1 1 = = + + + + + o (1) . : : : n 1. by rule 3.g) By rule 3. 1. zn + 1 zn + 1 wkn + 1 2 2wk 2e2 ik=n Res . 2. remark that wkn = 1. VII.1.2 Calculate the residue at each singularity in the complex plane of the following functions. z (a) e1=z (b) tan z (c) (z2 +1)2 (d) z21+z Solution (a) The function e1=z has a singularity at z = 0, by Laurent expansion 1 1 e1=z = 1 + + + :::: z 2!z 2 Thus by de…nition Res e1=z ; 0 = 1: (b) sin z The function tan z = cos z has isolated singularities at z = 2 + n, 1< n < 1, they are simple poles. By rule 3, sin z sin z Res [tan z; =2 + n ] = Res ; =2 + n = = 1: cos z sin z z= =2+n (c) z The function (z 2 +1)2 has isolated singularities at z = i, they are double poles. By rule 2, z d z (z + i)2 2z (z + i) Res 2 ;i = = = 0; (z 2 + 1) dz (z + i)2 z=i (z + i)4 z=i z d z (z i)2 2z (z i) Res 2; i = = = 0: 2 (z + 1) dz (z i)2 z= i (z i)4 z= i (d) 4 1 The function z(z+1) has isolated singularities at z = 0 and z = 1 they are simple poles. By rule 1, 1 1 1 Res ; 0 = lim = = 1; z (z + 1) z!0 z + 1 z+1 z=0 1 1 1 Res ; 1 = lim = = 1: z (z + 1) z! 1 z z z= 1 5 VII.1.3 Evaluate R the followingR integrals using Rresidue theorem. (a) jzj=1 sin z2 z dz (c) jzj=2 cosz z dz (e) jz 1j=1 z81 1 dz R z R z4 R (b) jzj=2 z2e 1 dz (d) jzj=1 sin z dz (f) jz 1=2j=3=2 tanz z dz Solution (a) Using the residue theorem and rule 1, Z sin z sin z sin z 2 dz = 2 i Res 2 ; 0 = 2 i lim = 2 i (1) = 2 i: jzj=1 z z z!0 z (b) Using the residue theorem and rule 3, Z ez dz = jzj=2 z2 1 ez ez = 2 i Res ; 1 + Res ; 1 = z2 1 z2 1 z e ez e1 e 1 =2 i + =2 i = 2z z=1 2z z= 1 2 = 2 i sinh 1: (c) Using the residue theorem and rule 3, Z h z i h z i z dz = 2 i Res ; + Res ; = jzj=2 cos z cos z 2 cos z 2 ! z z =2 i + =2 i = sin z z= =2 sin z z= =2 2 2 = 2 2 i: (d) Using the residue theorem and rule 3, 6 Z z4 z4 z4 dz = 2 i Res ;0 = 2 i = 0: jzj=1 sin z sin z cos z z=0 (e) VII.1.3e |z - 1| = 1 Using the residue theorem with the contour in Figure VII.1.3e and rule 4, Z 1 1 1 1 8 dz = 2 i Res ; 1 + Res ; ei =4 + Res ;e i =4 = jz 1j=1 z 1 z8 1 z8 1 z8 1 1 1 1 i =2 i 7 + 7 + = 1 + e i=4 + e i=4 = 8z z=1 8z z=ei =4 8z 7 z=e i =4 4 i 1 i 1+i i p = 1+ p + p = 1+ 2 : 4 2 2 4 (f) The function tanz z = sin z=z cos z has removable singularity at z = 0, and a isolated singularity at z = 2 inside the domain jz 1=2j = 3=2. Using the residue theorem and rule 3, 7 Z tan z dz = jz 1=2j=3=2 z ! sin z=z sin z=z sin z=z 2 = 2 i Res ; =2 i lim =2 i = (2 i) = cos z 2 z! =2 sin z sin z z= =2 = 4i: 8 VII.1.4 1 2 3 P L K Suppose P (z) and Q (z) are polynomials such that the zeros of Q (z) are simple zeros at the points z1 ; : : : ; zm ; and deg P (z) < deg Q (z). Show that the partial fractions decomposition P (z) =Q (z) is given by P (z) X P (zj ) 1 m = 0 (z ) z : Q (z) j=1 Q j zj Solution Since the zeros of Q (z) are simple, the polynomial quotient P (z) =Q (z) has at most simple poles. By Rule 3, the residue at zj is P (z) =Q0 (z) : Hence the principal part of P (z) =Q (z) at zj is P (zj ) 1 : Q0 (zj ) z zj Since deg P (z) < deg Q (z), then P (z) !0 Q (z) as z ! 1, and P (z) =Q (z) has no principal part at 1. By partial fraction decomposition our polynomial quotient P (z) =Q (z) or can be written as the sum of its principal parts at its pole, P (z) X P (zj ) 1 m = 0 (z ) z : Q (z) j=1 Q j zj 9 VII.1.5 1 2 3 P L K Let f (z) be a meromorphic function on the complex plane that is doubly periodic, and suppose that none of the poles of f (z) lie on the boundary of the period parallelogram P constructed in Section VI.5. By integrating f (z) around of P , show that the sum of the residues at the poles of f (z) in P is zero. Conclude that there is no doubly periodic meromorphic function with only one pole, a simple pole, in the period parallelogram. Solution R Follow directions. @P = 0, the curve integrals over opposite sides cancel. Thus the sum of the residues is zero. In particular, there can not be a single nonzero residue. 10 VII.1.6 Consider the integral Z i(z 1=2)2 e 2 iz dz; @DR 1 e where DR is the parallelogram with vertices 12 (1 + i) R. (a) p Use the residue theorem to show that the integral is (1 + i) = 2. (b) By parameterizing the sides of the parallelogram, show that the integral tends to Z 1 2 (1 + i) e 2 t dt 1 as R ! 1. (c) Use (a) and (b) to show that Z 1 s2 p e ds = : 1 Solution (a) VII.1.6 γ2 γ3 z1 γ1 γ4 Set 11 Z i(z 1=2)2 e I= 2 iz dz; @DR 1 e i.e, we integrate 2 e i(z 1=2) f (z) = 1 e 2 iz along the parallelogram contour in Figure VII.1.6. The parallelogram has vertices at 1=2 (1 + i) R. Residue at a simple pole at z1 = 0, where by Rule 3, " 2 # 2 e i(z 1=2) e i(z 1=2) e i=4 Res ; 0 = = : 1 e 2 iz 2 ie 2 iz 2 i z=0 Using the Residue Theorem, we obtain that Z i(z 1=2)2 e e i=4 2 iz dz = 2 i ; @DR 1 e 2 i i.e., Z i(z 1=2)2 e 1+i 2 iz dz = p : @DR 1 e 2 (b) 1 Integrate along 1 parametrized by z = 2 + (1 + i) t with R t R, and let R ! 1. This gives Z Z R i(z 1=2)2 e z = 12 + (1 + i) t f (z) dz = dz = = 1 R 1 e 2 dz = (1 + i) dt iz Z R 2 Z R 2 e i((1+i)t) (1 + i) e 2 t = 1 dt = (1 + i) 2 i(1+i)t dt ! R 1 e 2 i( 2 +(1+i)t) R 1+e Z 1 2 e 2 t ! (1 + i) 2 i(1+i)t dt: 1 1+e 1 Now integrate along 2 parameterized by z = t+ 2 + (1 + i) R with 0 t 1, and let R ! 1. 12 Z f (z) dz = 2 Z i(z 1=2)2 e z = t + 12 + (1 + i) R = dz = = 2 1 e 2 iz dz = dt Z Z e (t 2t)i e 2 (R Rt) 2 2 1 i( t+(1+i)R)2 1 e = 2 i( t+1=2+(1+i)R) dt = dt 0 1 e 0 1 e (2t 1 2R)i e2 R e (t 2t)i e 2 (R Rt) (R2 Rt) 2 2 2 e sup 1 sup = 0 t 1 1 e (2t 1 2R)i e2 R 0 t 1 j1 je (2t 1 2R)i e2 R jj e 2 (R Rt) (R2 Rt) 2 2 e = sup = sup ! 0: 0 t 1 j1 e2 Rj 0 t 1 e2 R 1 1 Integrate along 3 , parameterized by z = 2 + (1 + i) t with R t R, and let R ! 1. This gives Z Z R i(z 1=2)2 1 e z = 2 + (1 + i) t f (z) dz = dzdz = = 3 R e1 2 dz = (1 + i) dt iz Z R 2 Z R 2 i(1+i)t 2 t2 e i( 1+(1+i)t) (1 + i) dt e e dt = 1 = (1 + i) 2 i(1+i)t ! R 1 e 2 i( 2 +(1+i)t) R 1+e Z 1 2 i(1+i)t 2 t2 e e dt ! (1 + i) 2 i(1+i)t dt: 1 1+e 1 Now integrate along 4 parameterized by z = t 2 (1 + i) R with 0 t 1, and let R ! 1. 13 Z f (z) dz = 4 Z i(z 1=2)2 e z = t 12 (1 + i) R = dz = = 4 1 e 2 iz dz = dt Z 1 (1+i)R)2 Z 1 (2R 2Rt+t2 2t+1)i 2 (R2 +R Rt) e i(t 1 e e = dz = dz = 0 1 e 2 i(t 1=2 (1+i)R) 0 1 e (2t 1 2R)i e 2 R e (2R 2Rt+t2 2t+1)i e 2 (R2 +R Rt) e 2 (R2 +R Rt) sup (2t 1 2R)i e 2 R 1 sup (2t 1 2R)i e 2 R jj 0 t 1 1 e 0 t 1 j1 je e 2 (R2 +R Rt) sup 2 Rj ! 0: 0 t 1 j1 e Using the Residue Theorem and letting R ! 1, we obtain that Z 1 2 t2 Z 1 2 i(1+i)t 2 t2 e e e dt (1 + i) 2 i(1+i)t dt + 0 + (1 + i) 2 i(1+i)t dt + 0 = 1 1+e 1 1+e e i=4 = 2 i ; 2 i and therefore Z 2 t2 1 e 1 + e 2 i(1+i)t 1+i (1 + i) 2 i(1+i)t dt = p ; 1 1+e 2 i.e., Z 1 2 t2 1+i (1 + i) e dt = p : 1 2 (c) Comparing real parts thus yields, Z 1 1 2 t2 e dt = p ; 1 2 and changing variables as follows gives the desired result 14 Z 1 p Z 1 1 2 t2 s = p2 t 1 s2 p = e dt = =p e ds; 2 1 ds = 2 dt 2 1 i.e., Z 1 s2 p e dt = : 1 15 VII.2.1 1 2 3 P L K Show using residue theory that Z 1 dx 2 2 = ; a > 0: 1 x +a a Remark. Check the result by evaluating the integral directly, using the arctangent function. Solution VII.2.1 γ2 z1 -R γ1 R z2 Set Z 1 dx I= ; 1 x 2 + a2 and integrate 1 1 f (z) = = z2 + a2 (z ia) (z + ia) along the contour in Figure VII.2.1. Residue at a simple pole at z1 = ia, where by Rule 2, 1 1 i Res ; ia = = : z2 +a 2 2z z=ia 2a 16 Integrate along 1 , and let R ! 1. This gives Z Z R Z 1 dx dx f (z) dz = 2+a 2 ! 2 + a2 = I: 1 R x 1 x Integrate along 2, and let R ! 1. This gives Z 1 f (z) dz 2 R ! 0: 2 R a2 R Using the Residue Theorem and letting R ! 1, we obtain that i I +0=2 i ; 2a and therefore Z 1 dx = ; a > 0: 1 x2 + a2 a Remark. Check of the result by evaluating the integral directly gives, because the integrand is positive the integral can be computed taking a single limit, namely, Z 1 Z b b dx dx 1 x = lim = lim arctan = ; a > 0: 1 x + a2 b!1 2 b 2 x +a 2 b!1 a a b a 17 VII. Solution VII.2. where by Rule 2. ia = lim 2 = = : (z 2 + a2 ) z!ia dz (z + ia) (z + ia)3 z=ia 4a3 18 .2 1 2 3 P L K Show using residue theory that Z 1 dx = 3: 2 2 2 2a 1 (x + a ) Remark. Residue at a double pole at z1 = ia. 1 d 1 2 i Res 2 . 1 (x2 + a2 )2 and integrate 1 1 f (z) = = (z 2 + a2 )2 (z ia) (z + ia)2 2 along the contour in Figure VII.2. Check the result by di¤erentiating the formula in the pre- ceding exercise with respect to the parameter.2.2.2 γ2 z1 -R γ1 R z2 Set Z 1 dx I= . and the primitive of the derivative R is uniformly bounded by a a independent function h (x) such that h (x) < 1 on every such like interval. This gives Z Z R Z 1 dx dx f (z) dz = 2 ! = I: (x 2 + a2 ) (x 2 +a2 )2 1 R 1 Integrate along 2. we obtain that that i I +0=2 i .1 we have the formula Z 1 1 (1) 2 2 dx = . and let R ! 1.Integrate along 1. From Exercise VII.2. We start by di¤erentiate the left hand side in the formula with respect to the parameter a Z 1 Z 1 Z 1 d 1 2a dx (2) dx = dx = 2a . we have that 19 . This gives Z 1 f (z) dz R ! 0: 2 (R2 a2 )2 R3 Using the Residue Theorem and letting R ! 1. both the integrand and its derivative are con- tinuous. 4a3 and therefore Z 1 dx = : 1 (x2 + a2 )2 2a3 Remark. because for every compact interval with a 6= 0. da 1 x + a2 2 1 (x + a2 )2 2 1 (x2 + a2 )2 and di¤erentiate the right hand side in the formula in the same way d (3) = : da a a2 By (1) . a > 0: 1 x +a a It is allowed to di¤erentiate both sides in the formula.(3).and let R ! 1. Z 1 dx = : 1 (x2 + a2 )2 2a3 20 . z2 d z2 2z (z + i)2 2z 2 (z + i) i Res .2. Check the result by combining the preceding two exercises. and let R ! 1. 1 (x2 + 1)2 and integrate z2 z2 f (z) = = (z 2 + 1)2 (z i)2 (z + i)2 along the contour in Figure VII. This gives 21 .2. i = lim = = : (z 2 + 1)2 z!i dz (z 2 + i)2 (z + i)4 z=i 4 Integrate along 1.VII. Residue at a double pole at z1 = i.2. Solution VII. where by Rule 2.3.3 Show using residue theory that Z 1 x2 dx 2 = : 2 1 (x + 1) 2 Remark.3 γ2 z1 -R γ1 R z2 Set Z 1 x2 dx I= . and let R ! 1. Z Z R Z 1 x2 dx x2 dx f (z) dz = ! = I: 1 R (x2 + 1)2 1 (x2 + 1)2 Integrate along 2. we obtain that i I +0=2 i . and put a = 1 in the formulas from the preceding two exercises Z 1 Z 1 Z 1 x2 dx dx dx 2 = 2 2 = = : 1 x +1 2 2 2 2 1 (x + 1) 1 (x + 1) 22 . We split the integrand using partial fraction. This gives Z R2 f (z) dz R ! 0: 2 (R2 1)2 R Using the Residue Theorem and letting R ! 1. 4 and therefore Z 1 x2 dx = : 1 (x2 + 1)2 2 Remark. 2.2. 23 .4 Using residue theory.4. show that Z 1 dx 4 =p : 1 x +1 2 Solution VII.VII. 1 x4+1 and integrate 1 1 f (z) = = z4 +1 z 1+i p z 1+i p z 1 i p z 1p i 2 2 2 2 along the contour in Figure VII. 2 p 1 1+i 1 2 Res 4 . Residue at a simple pole at z1 = 1+i p .2.4 γ2 z2 z1 -R γ1 z3 z4 R Set Z 1 dx I= . p = 3 = ( 1 i) z +1 2 4z z=(1+i)=p2 8 1+i Residue at a simple pole at z2 = p 2 . where by Rule 2. where by Rule 2. and let R ! 1. 8 8 and therefore Z 1 dx =p : 1 x4+1 2 24 . p 1 1+i 1 2 Res 4 . p = 3 p = (1 i) z +1 2 4z z=( 1+i)= 2 8 Integrate along 1. and let R ! 1. we obtain that p p ! 2 2 I +0=2 i ( 1 i) + (1 i) . This gives Z Z R Z 1 dx dx f (z) dz = 4 ! 4 = I: 1 R x +1 1 x +1 Integrate along 2. This gives Z 1 f (z) dz 4 R ! 0: 2 R 1 R3 Using the Residue Theorem and letting R ! 1. 5 γ2 z2 z1 -R γ1 z3 z4 R Set Z 1 Z 1 x2 x2 I= dx ) 2I = dx 0 x4 + 1 1 x4 + 1 and integrate z2 z2 f (z) = 4 = z +1 z 1+i p z 1+i p z 1 i p z 1p i 2 2 2 2 along the contour in Figure VII.5 Using residue theory. Residue at a simple pole at z1 = 1+i p .2.2. show that Z 1 x2 dx = p : 0 x4 + 1 2 2 Solution VII. 2 p z2 1+i 1 2 Res 4 . 25 . where by Rule 3.2. p = p = (1 i) : z +1 2 4z z=(1+i)= 2 8 1+i Residue at a simple pole at z2 = p 2 . where by Rule 3.5.VII. and let R ! 1. and let R ! 1. we obtain that p p ! 2 2 2I + 0 = 2 i (1 i) + ( 1 i) . p z2 1+i 1 2 Res 4 . This gives Z R2 f (z) dz R ! 0: 2 R4 1 R Using the Residue Theorem and letting R ! 1. This gives Z Z R Z 1 x2 x2 f (z) dz = 4 dx ! 4 dx = 2I: 1 R x +1 1 x +1 Integrate along 2. 8 8 and therefore Z 1 x2 dx = p : 0 x4 + 1 2 2 26 . p = p = ( 1 i) : z +1 2 4z z=( 1+i)= 2 8 Integrate along 1. VII.2. 27 .6.6 γ2 z1 z2 -R γ1 R z3 z4 Set Z 1 x I= dx 1 (x2 + 2x + 2) (x2 + 4) and integrate z z f (z) = = (z 2 + 2z + 2) (z 2 + 4) (z ( 1 + i)) (z 2i) (z ( 1 i)) (z + 2i) along the contour in Figure VII.2. z z 1 1 Res .2.6 Show that Z 1 x dx = : 1 (x2 + 2x + 2) (x2 + 4) 10 Solution VII. 2i = 2 = i: (z 2 2 + 2z + 2) (z + 4) (z + 2z + 2) 2z z=2i 20 10 Residue at a simple pole at z2 = 1 + i. where by Rule 3. where by Rule 3. Residue at a simple pole at z1 = 2i. we obtain that 1 1 1 3 I +0=2 i i+ + i 20 10 20 20 and therefore Z 1 x dx = : 1 (x2 + 2x + 2) (x2 + 4) 10 28 . z z 1 3 Res . and let R ! 1. This gives Z Z R x f (z) dz = dx ! R (x2 + 2x + 2) (x2 + 4) 1 Z 1 x ! 2 2 dx = I: 1 (x + 2x + 2) (x + 4) Integrate along 2 . 1 + i = = + i: (z 2 + 2z + 2) (z 2 + 4) (2z + 2) (z 2 + 4) z= 1+i 20 20 Integrate along 1. This gives Z R f (z) dz 2 R ! 0: 2 (R 2R 2) (R2 4) R2 Using the Residue Theorem and letting R ! 1. and let R ! 1. 7. a > 0: 1 x4 + 1 2 2 2 Solution VII.2. 1 x4 + 1 and integrate eiaz eiaz f (z) = = z4 + 1 z 1+i p z 1+i p z 1 i p z 1p i 2 2 2 2 along the contour in Figure VII.7 γ2 z2 z1 -R γ1 z3 z4 R Set Z 1 cos (ax) I= dx a > 0. where by Rule 3.2.VII. Residue at a simple pole at z1 = 1+i p .2. 2 29 .7 1 2 3 P L K LLL Show that Z 1 p cos (ax) a= 2 a a dx = p e cos p + sin p . p = = z4 + 1 2 4z 3 z=(1+i)=p2 p a=p2 p p a=p2 p p p a= 2 p p p a= 2 p 2e cos a= 2 2e sin a= 2 2e cos a= 2 2e sin a= 2 = i i+ 8 8 8 8 1+i Residue at a simple pole at z2 = p 2 . = z4 + 1 2 4z 3 p z=( 1+i)= 2 p a= 2 p p p p a= 2 p p p a= 2 p p p a= 2 p 2e cos a= 2 2e sin a= 2 2e cos a= 2 2e sin a= 2 = i i 8 8 8 8 Integrate along 1. eiaz 1+i eiaz Res p =. 1 x4 + 1 4 4 and therefore Z 1 p cos (ax) a= 2 a a 4 dx = p e cos p + sin p . and let R ! 1. a > 0: 1 x +1 2 2 2 30 . eiaz 1 + i eiaz Res . where by Rule 3. This gives Z Z R Z 1 Z 1 Z 1 eax cos (ax) sin (ax) sin (ax) f (z) dz = dx ! dx+i dx = I+i dx 1 R x4 + 1 1 x4 + 1 1 x4 + 1 1 x4 + 1 Integrate along 2 . this gives Z 1 f (z) dz 4 R ! 0: 2 R 1 R3 Using the Residue Theorem and letting R ! 1. we obtain that Z 1 p p a= 2 p p p a= 2 p ! sin (ax) 2e cos a= 2 2e sin a= 2 I+i dx = 2 i i i . Because jeiaz j 1 in the upper half plane if a > 0. and let R ! 1. 2.8 Show that Z 1 cos x dx = : 1 (x2 + 1)2 e Solution VII.8. where by Rule 2. Residue at a double pole at z1 = i.2.VII.8 γ2 z1 -R γ1 R z2 Set Z 1 cos x I= dx 1 (x2 + 1)2 and integrate eiz eiz f (z) = = (z 2 + 1)2 (z i)2 (z + i)2 along the contour in Figure VII. eiz d eiz ieiz (z + i)2 2eiz (z + i) e 1 Res . This gives 31 .2. i = lim == = i (z 2 + 1)2 z!i dz (z + i)2 (z + i)4 z=i 2 Integrate along 1. and let R ! 1. Z f (z) dz = 1 Z 1 Z 1 Z 1 eix cos x sin x = dx ! dx + i dx = 1 (x2 + 1)2 1 (x2 + 1)2 1 + 1)2 (x2 Z 1 sin x =I +i 2 dx: 2 1 (x + 1) Integrate along 2. this gives Z 1 f (z) dz R ! 0: 2 (R2 1)2 R3 Using the Residue Theorem and letting R ! 1. 2 1 (x + 1) 2 and therefore Z 1 cos x dx = : 1 (x2 + 1)2 e 32 . Because jeiz j 1 in the upper half plane. we obtain that Z 1 sin x e 1 I +i 2 dx + 0 = 2 i i . and let R ! 1. 1 ei2z 1 ei2z 1 2 1 Res .i = = ie i 2 (z 2 + 1) 4z z=i 4 4 Integrate along 1 .2.VII.2. Residue at a simple pole at z1 = i. and let R ! 1.9 Show that Z 1 sin2 x 1 dx = 1 : 1 x2 + 1 2 e2 Solution VII. where by Rule 3.2. This gives 33 .9.9 γ2 z1 -R γ1 R z2 Set Z 1 Z 1 sin2 x 1 cos 2x I= dx = dx 1 x2 + 1 1 2 (x2 + 1) and integrate 1 ei2z 1 ei2z f (z) = = 2 (z 2 + 1) 2 (z i) (z + i) along the contour in Figure VII. Because jeiz j 1 in the upper half plane. this gives Z 2 f (z) dz R ! 0: 2 2(R2 1) R Using the Residue Theorem and letting R ! 1. and let R ! 1. 1 2 (x + 1) 4 4 and therefore Z 1 sin2 x 1 dx = 1 : 1 x2 + 1 2 e2 34 . we obtain that Z 1 sin 2x 1 2 1 I +i 2 dx + 0 = 2 i ie i . Z f (z) dz = 1 Z R Z 1 Z 1 1 ei2x 1 cos 2x sin 2x = dx ! dx + i dx = R 2 (x2 + 1) 1 2 (x2 + 1) 2 1 2 (x + 1) Z 1 sin 2x =I +i 2 dx 1 2 (x + 1) Integrate along 2. 35 .10a z1 γ1 -R R z2 γ2 Set Z 1 cos ax I= dx. It depends analytically on a in the intervals 1 < a < 0 and 0 < a < 1.VII.10a. and Re b 6= 0 or cos (iab) = 0. b > 0 1 x 2 + b2 and integrate eiaz eiaz f (z) = = z 2 + b2 (z ib) (z + ib) along the contour in Figure 7.10 Show that Z 1 cos (ax) e jajb dx = . a < 0.2. 1 < a < 1. where by Rule 3. Case 1: a < 0 VII. b > 0: 1 x 2 + b2 b For which complex values of the parameters a and b does the in- tegral exist? Where does the integral depend analytically on the parameters? Solution The integral exists only for a real. It depends analytically on b for Re b 6= 0. (The lower half-plane) Residue at a simple pole at z2 = ib.2.2. Because jeiaz j 1 in the lower half plane if a < 0. eiaz eiaz eab Res . we obtain that eab I +0= 2 i i . This gives Z f (z) dz = 1 Z R Z 1 Z 1 eiax cos (ax) sin (ax) = dx ! dx + i dx = R x 2 + b2 1 x 2 + b2 2 1 x +b 2 Z 1 sin (ax) =I +i 2 2 dx 1 x +b Integrate along 2 and let R ! 1. a < 0: b Case 2: a > 0 36 . ib = = i z 2 + b2 2z z= ib 2b Integrate along 1 and let R ! 1. 2b and hence eab (1) I= . this gives Z 1 f (z) dz R ! 0: 2 R 2 b2 R Using the Residue Theorem and letting R ! 1. (The upper half-plane) Residue at a simple pole at z1 = ib.2. VII. This gives Z f (z) dz = 3 Z R Z 1 Z 1 eiax cos (ax) sin (ax) = dx ! dx + i dx = R x 2 + b2 1 x 2 + b2 2 1 x +b 2 Z 1 sin (ax) =I +i 2 2 dx 1 x +b 37 . a > 0.2. 1 x2 + b2 and integrate eiaz eiaz f (z) = = z 2 + b2 (z ib) (z + ib) along the contour in Figure VII. eiaz eiaz e ab Res .10b γ4 z1 -R γ3 R z2 Set Z 1 cos ax I= dx. where by Rule 3. ib = = i z 2 + b2 2z z=ib 2b Integrate along 3 and let R ! 1. b > 0.10b. we obtain that e ab I +0=2 i i . this gives Z 1 f (z) dz R ! 0: 4 R 2 b2 R Using the Residue Theorem and letting R ! 1. Because jeiaz j 1 in the upper half plane if a > 0. 1 < a < 1. a > 0: b By (1) and (2) we can concluede that Z 1 cos (ax) e jajb 2 2 dx = . 2b and hence e ab (2) I= . b > 0: 1 x +b b 38 .Integrate along 4 and let R ! 1. The integrand.e. so that ia are zeros of cos x. but they are not same zeros. b 2 = ( 1. and get other case in this limit. b2 62 ( 1. So we may as well assume 0 < a < b. we can assume that a. 0] i. ib are zeros of cos x. and extend the formula by analyticity. a. is an even function of a and b. b in the right half-plane. For this.11 γ2 z2 z1 -R γ1 R z3 z4 R1 The integral 1 (x2 +acos x 2 2 2 )(x2 +b2 ) dx converges absolutely if a . a. so it su¢ ces to evaluate it for a. 0]. Solution The integral converges absolutely if a2 . and get other case in this limit. (Problem should may be simpli…ed to avoid thus points. It also converges absolutely for certain points on the imaginary axis.11 1 2 3 P L K Evaluate Z 1 cos x dx: 1 (x2 + a2 ) (x2 + b2 ) Indicate the range of the parameters a and b. b > 0. b in the right half-plane. VII. b 2= iR. It also converges absolutely for certain points on imaginary axis. We can also assume that a 6= b. We can assume that a 6= b.) The integrand. ib are zeros of cos x. but they are not the same zeros. 39 . is an even function of a and b. b 62 iR. so it su¢ ces to evaluate it for a.2. so that ia are zeros of cos x. we can assume that a. b > 0. So we may as well assume that 0 < a < b.VII. For this.2. and extend the formula by analyticity. Because jeiz j 1 in the upper half plane. ib = = (z 2 + a2 ) (z 2 + b2 ) 2z (z 2 + a2 ) z=bi 2b (b2 a2 ) i Integrate along 1 and let R ! 1. a 6= b) Set Z 1 cos x I= dx 1 (x2 + a2 ) (x2 + b2 ) and integrate eiz eiz f (z) = = (z 2 + a2 ) (z 2 + b2 ) (z ia) (z ib) (z + ia) (z + ib) along the contour in Figure VII. 2a (b2 a2 ) i 2b (b2 a2 ) i 40 . eiz eiz e b Res . where by Rule 3.Case 1: (a. ia = = (z 2 + a2 ) (z 2 + b2 ) 2z (z 2 + b2 ) z=ai 2a (b2 a2 ) i Residue at a simple pole at z1 = ib. this gives Z 1 f (z) dz R ! 0: 2 (R2 a2 ) (R2 b2 ) R3 Using the Residue Theorem and letting R ! 1. we obtain that a b e e I +0=2 i .2. eiz eiz e a Res . This gives Z Z R eix f (z) dz = 2 2 2 2 dx ! R (x + a ) (x + b ) 1 Z 1 Z 1 cos x sin x ! 2 2 2 2 dx + i 2 2 2 2 dx = 1 (x + a ) (x + b ) 1 (x + a ) (x + b ) Z 1 sin x =I +i 2 2 2 2 dx 1 (x + a ) (x + b ) Integrate along 2 and let R ! 1. where by Rule 3. b > 0.11. Residue at a simple pole at z1 = ai. we have Z 1 cos x dx = . 1 (x2 + 1)2 e which thus yields the answer from Exercise 8. b > 0. we get Z 1 cos x dx = 3 (1 + a) : 1 (x2 + a2 )2 2a For a = 1. a = b) From Case 1 we have Z 1 cos x be a ae b (b a) e a +a e a e b dx = = : 1 (x2 + a2 ) (x2 + b2 ) ab (b2 2 a) ab (b + a) (b a) If we let b ! a. 41 .and therefore Z 1 cos x be a ae b dx = : 1 (x2 + a2 ) (x2 + b2 ) ab (b2 a2 ) Case 2: (a. X f (z) Res . Suppose there is b < m 1 such that jf (z)j jzjb for z in the upper half-plane. z2 . and let f (z) be a function that is analytic in the upper half- plane and across the real line. zj : Q (z) 42 . : : : . 1 Q (x) Q (z) summed over the zeros zj of Q (z) in the upper half-plane. zj in the upper half plane.12 γ2 zj .12. Solution VII.2. zj .2.1 z3 zj z2 z1 -R γ1 R Set Z 1 f (x) I= dx 1 Q (x) and integrate f (z) f (z) = Q (z) along the contour in Figure VII. Show that Z 1 X f (x) f (z) dx = 2 i Res . The sum of the residues at the poles z1 .VII.12 Let Q (z) be a polynomial of degree m with no zeros on the real line.2. jzj > 1. and let R ! 1. zj Q (z) and therefore Z 1 X f (x) f (z) dx = 2 i Res . This gives Z Z z = Rei f Rei f Rei f (z) dz = d R 2 0 0 jQ (Rei )j jQ (Rei )j Rb ! 0: Rm (C + O (1=R)) R (C + O (1=R)) CR Using the Residue Theorem and letting R ! 1. and let R ! 1. This gives Z Z R Z 1 f (x) f (x) f (z) dz = dx ! dx = I: 1 R Q (x) 1 Q (x) Integrate along 2.E.Integrate along 1. zj 1 Q (x) Q (z) Q. we obtain that X f (z) I +0=2 i Res .D. 43 . 3. Residue at a simple pole at z1 = 0.1.3.VII. z2 + 1 z2 + 1 Res . 0 = = 1: z 3 + 4z 2 + z 3z 2 + 8z + 1 z=0 44 . where by Rule 3.3.1 Show using residue theory that Z 2 cos 2 d =2 1 p : 0 2 + cos 3 Solution VII.1 |z| = 1 z3 z2 z1 Z 2 cos I= d : 0 2 + cos A change of variables as in book page 203-204 gives Z I 1 I 2 cos 2 z + z1 dz 1 z2 + 1 I= d = = dz: 0 2 + cos jzj=1 2 + 12 z + 1 z iz i 3 2 jzj=1 z + 4z + z Integrate z2 + 1 z2 + 1 f (z) = = p p z 3 + 4z 2 + z z z 2 3 z 2+ 3 along the unit circle contour in Figure VII. p Residue at a simple pole at z2 = 2+ 3. where by Rule 3. p z2 + 1 p z2 + 1 2 3 Res 3 . we thus obtain p ! 1 2 3 I=2 i 1 . 2+ 3 = 2 = : z + 4z 2 + z 3z + 8z + 1 p z= 2+ 3 3 Using the Residue Theorem. i 3 and therefore Z 2 cos 2 d =2 1 p : 0 2 + cos 3 45 . 46 . where by Rule 3. b = 6) |z | = 1 z1 z2 Set Z 2 1 I= d : 0 a + b sin A change of variables as in book page 203-204 gives Z 2 I I 1 1 dz 2 I= d = 1 1 = dz 0 a + b sin jzj=1 a+b 2i z z iz jzj=1 bz 2 + 2azi b Integrate 2 2 f (z) = = p p bz 2 + 2azi b b z a a2 b2 i z a+ a2 b2 i b b along the unit circle contour in Figure p VII.2 (a = 7.2.3.2 Show using residue theory that Z 2 d 2 =p .3.VII. a > b > 0: 0 a + b sin a 2 b2 Solution VII.3. a+ a2 b2 Residue at a simple pole at z1 = b i. p 2 a+ a2 b2 2 i Res 2 . a > b > 0: 0 a + b sin a2 b2 47 . a2 b2 and therefore Z 2 1 2 d =p . i = p = p : bz + 2azi b b 2bz + 2ai z= a+ a2 b2 i a2 b2 b Using the Residue Theorem. we thus obtain 1 I=2 i p i . 3 (a = 2) |z| = 1 z3 z2 z1 Set Z Z sin2 =2 t sin2 (2 t) I= d = = dt = 0 a + cos d = dt 2 a + cos(2 t) Z 2 Z 2 sin2 (2 t) sin2 t = dt = dt: a + cos(2 t) a + cos t We have Z 2 sin2 2I = d : 0 a + cos A change of variables as in book page 203-204 gives Z 2 I 1 1 2 I 2 sin2 2i z z dz 1 (z 2 1) 2I = d = 1 1 = dz: 0 a + cos jzj=1 a+ 2 z+ z iz 2i jzj=1 z 2 (z 2 + 2az + 1) 48 .3 Show using residue theory that Z h p i sin2 d = a a2 1 .VII. a > 1: 0 a + cos Solution VII.3.3. 3. Residue at a double pole at z1 = 0 . 0 = lim = 2a: z (z + 2az + 1) z!0 dz (z 2 + 2az + 1) p Residue at a simple pole at z2 = a + a2 1.3. " 2 # 2 2 2 (z 2 1) =z p (z 2 1) =z p Res 2 . where by Rule 2. 2i and therefore Z p sin2 d = a a2 1 . " # 2 2 (z 2 1) d (z 2 1) Res 2 2 . where by Rule 3. a > 1: 0 a + cos 49 . we thus obtain 1 p 2I = 2 i 2a + 2 a2 1 . a + a2 1 = = 2 a2 1: z + 2az + 1 2z + 2a p z= a+ a2 1 Using the Residue Theorem.Integrate 2 2 (z 2 1) (z 2 1) f (z) = 2 2 = p p z (z + 2az + 1) z2 z a + a2 1 z a a2 1 along the unit circle contour in Figure VII. VII.3.4 Show using residue theory that Z p d = 2: 1 + sin2 Solution VII.3.4 |z | = 1 z4 z3 z1 z2 Set Z 1 = t I= d = = 1 + sin2 d = dt Z 0 Z 2 1 1 = 2 dt = dt: 2 1+ sin ( t) 0 1+ sin2 t We have Z 2 1 I= d : 0 1 + sin2 A change of variables as in book page 203-204 gives Z 2 I I 1 1 dz 4 z I= d = = dz: 0 1 + sin2 jzj=1 1+ 1 z 1 2 iz i jzj=1 z4 6z 2 + 1 2i z 50 . 2 1 = 3 = p : z4 6z 2 + 1 4z 12z z= p ( 2 1) 8 2 Using the Residue Theorem. we thus obtain 4 1 1 I=2 i p p .Integrate z f (z) = = z4 6z 2 + 1 z = p p p p z 2 1 z 2+1 z 2+1 z 2 1 along the unit circle contour in Figure VII. i 8 2 8 2 and therefore Z p 1 d = 2: 1 + sin2 51 .2 = 2 1 . z p z 1 Res .3.4 p Residues at a simple poles at z1. where by Rule 3. 3. 0 r < 1: 1 2r cos + r 2 Solution VII.5 Show using residue theory that Z 1 r2 d 2 = 1.5 (r = 1/2) |z | = 1 z2 z1 Set Z 1 r2 d = t I= = = 1 2 2r cos + r 2 d = dt Z 0 1 r2 dt = 2 = 2 1 2r cos ( t) + r 2 Z 2 1 r2 dt = 2 0 1 + 2r cos t + r 2 A change of variables as in book page 203-204 gives 52 .VII.3. 1 r2 1 r2 Res .4 Residue at a simple pole at z1 = r. Z 2 1 r2 dt I= = 0 1 + 2r cos t + r2 2 I 1 1 r2 dz = = 2 jzj=1 1 + 2r 21 z + z1 + r2 iz I 1 1 r2 = dz: 2 i jzj=1 (r + z) (rz + 1) Integrate 1 r2 1 r2 f (z) = = (r + z) (rz + 1) r (z + r) (z + 1=r) along the unit circle contour in Figure VII. we thus obtain 1 I=2 i (1) . 0 r < 1: 1 2r cos + r 2 53 . r = = 1: r (z + r) (z + 1=r) 2rz + r2 + 1 z= r Using the Residue Theorem.3. 2 i and therefore Z 1 r2 d 2 = 1. where by Rule 3. VII. 2 0 1 + cos w w2 where the integrand in the left side in the identity can be rewritten as 1 X 1 k cos k = ( 1) : 1 + cos w k=0 wk Now. ( 1) ( 1) ( 2) (1 + x) = 1+ x+ x2 + x3 + + xn + .6 By expanding both sides of the identity Z 2 1 d 1 =p 2 0 w + cos w 2 1 in a power series at 1. we work on the left side of the identity Z 2 X Z 1 1 k cos k 1 X1 ( 1)k 2 I= ( 1) d = cosk d : 2 0 k=0 wk 2 k=0 w k 0 R2 Because 0 cosk d = 0 if k odd.3. we use the power series expansion (MH p. we have X1 Z 2 1 1 (2) I =1+ cos2k d : k=1 2 0 w2k Now. show that Z 2 1 (2k)! cos2k d = . 192. 2! 3! n we have 54 . k 0: 2 0 22k (k!)2 Solution Rewrite both sides in the given identity Z 2 1=2 1 d 1 (1) I= = 1 .) for real and 1< x < 1. k 0: 2 0 22k (k!)2 55 .. 1=2 1 1 = w2 1 1 1 3 2 1 3 5 3 2 1 2 2 1 2 2 2 1 =1+ + + + 1! w2 2! w2 3! w2 1 3 5 7 4 1 3 5 7 1 2k k 2 2 2 2 1 2 2 2 2 2 1 + + + + = 4! w2 k! w2k X1 1 3 (2k 1) 1 =1+ = k=1 2k k! w2k X1 1 2 3 4 5 : : : (2k 1) 2k 1 =1+ = k=1 2k k! 2 4 6 : : : 2k w2k X1 1 2 3 4 5 : : : (2k 1) 2k 1 1+ = k=1 2k k! 2k (1 2 3 : : : k) w2k X1 (2k)! 1 X1 (2k)! 1 =1+ = 1 + .e. k k k=1 2 k!2 k! w 2k k=1 22k (k!)2 w 2k i. 1 1=2 X1 (2k)! 1 (2) 1 =1+ : w2 k=1 22k (k!)2 w2k Comparing the two power series expressions (2) and (3) of the both sides in the identity (1) gives us together with the uniqueness theorem for the power series Z 2 1 (2k)! cos2k d = . 3.7.7 (w = 2) |z| = 1 z2 z1 Set Z 2 d I= 0 (w + cos )2 A change of variables as in book page 203-204 gives Z 2 Z 2 I d 1 dz 4 z I= = = 2 dz: 0 (w + cos )2 0 w+ 1 2 z+ 1 z 2 iz i 2 jzj=1 (z + 2wz + 1) 4z Integrate f (z) = (z 2 +2wz+1)2 along the unit circle contour in Figure VII.3. Check you answer by di¤erentiating the integral of 1= (w + cos ) with respect to the parameter w. Solution VII. 1] : 0 (w + cos ) (w2 1)3=2 Specify carefully the branch of the power function.3. It su¢ ces to check the identity for w > p 1. where by Rule 2.VII. 1] that is positive for w 2 (1. p We choose branch of w 2 1 on Cn [ 1. 56 .7 Show using residue theory that Z 2 d 2 w 2 = . w 2 C n [ 1. 1). Residue at a double pole at z1 = w + w2 1. both the integrand and its derivative are continuous. w 2 C n [ 1. because for every compact interval with w 2 C n [ 1. i 4 (w2 1)3=2 and therefore Z 2 d 2 w 2 = .3. w + w2 1 = (z 2 + 2wz + 1) d z = lim p p 2 = z! w+ w2 1 dz z w w2 1 p 2 p z w w2 1 2z z w w2 1 = p 4 = z w w2 1 p z= w+ w 1 2 p z w w 2 1 2z = p 3 = z w w2 1 p z= w+ w2 1 1 w = : 4 (w 2 1)3=2 Using the Residue Theorem. 1] : 0 (w + cos ) (w2 1)3=2 Check.6 we have the formula Z 2 1 d 1 (1) =p 2 0 w + cos w 2 1 It is allowed to di¤erentiate both sides in the formula. We start by di¤erentiating the left hand side in the formula with respect to the parameter w 57 . and that the primitive of the derivative R is uniformly bounded by an w independent funtion h ( ) such that h ( ) d < 1 on every such like interval on w. z p Res 2 . we thus obtain ! 4 1 w I=2 i . From Exercise VII. 1]. we thus have that Z 2 d 2 w 2 = : 0 (w + cos ) (w 2 1)3=2 58 . dw 2 0 w + cos 2 0 (w + cos )2 and di¤erentiate the right hand side in the formula in the same way d 1 w (3) p = : dw w 2 1 (w2 1)3=2 By (1) .(3). Z 2 Z 2 d 1 d 1 d (2) = . 1 = jzj e z= 1 =e : 1+z 59 . Residue at a simple pole at z1 = 1.4.4. 0 < a < 1: 0 1+x sin ( a) Solution VII.1.1 γ2 γ1 z1 -R γ4 γ3 R Set Z 1 x a I= dx 0 1+x and integrate z a jzj a e ia arg z f (z) = = 1+z 1+z a along the keyhole contour in Figure VII. a jzj e ia arg z a ia arg z ia Res .VII.1 By integrating around the keyhole contour. show that Z 1 x a dx = . where 0 < arg z < 2 .4. where by Rule 3. We make a branchcut for z along the positive real axis. This gives then 0 < a < 1 Z " a f (z) dz 2 " 2 "1 a ! 0: 4 1 " Using the Residue Theorem and letting R ! 1 and " ! 0+ . 60 .Integrate along 1. and let R ! 1 and " ! 0+ . and let " ! 0+ . This gives Z f (z) dz = 3 Z a Z " jzj e ia arg z z = xe2 i x a e 2 ia = dz = = dx ! 3 1+z dz = dx R 1+x Z 0 a 2 ia Z 1 x e x a ! dx = e 2 ia dx = e 2 ia I: 1 1 + x 0 1+x Integrate along 4. and hence solving for I. we obtain that 2 ia ia I +0 e I +0=2 i e : ia Multiplying with e . and let R ! 1 and " ! 0+ . This gives Z f (z) dz = 1 Z a Z R jzj e ia arg z z = xe0i x a dz = = dx ! 1+z dz = dx " 1+x 1 Z 1 x a ! dx = I: 0 1+x Integrate along 2 and let R ! 1. This gives then 0 < a < 1 Z R a 2 f (z) dz 2 R ! 0: 2 R 1 Ra Integrate along 3. this yields ia ia e e I = 2 i. e. e e sin ( a) i.. 0 < a < 1: 0 1+x sin ( a) 61 . Z 1 x a dx = . 2 i I= ia ia = . Check the result by changing variable and comparing with Exercise 1. where by Rule 3. show that Z 1 dx = . 0 1 + xb and integrate 1 1 f (z) = = 1+z b 1 + jzjb eib arg z along the pie-slice of aperture 2 =b contour in Figure VII. Residue at a simple pole at z1 = e i=b .4. We make a branch cut for z b along the negative imaginary axis. 1 i=b 1 1 i=b Res .2 By integrating around the boundary of a pie-slice domain of aper- ture 2 =b.VII.2 (b = 12) γ3 γ2 z1 ε γ1 R Set Z 1 dx I= . where =2 < arg z < 3 =2. Solution VII.e = = e : 1 + zb bz b 1 z=e i=b b 62 .2.4.4. b > 1: 0 1 + xb b sin ( =b) Remark. this yields i=b i=b 2 i e e I= . and let R ! 1 and " ! 0+ . This gives then b > 1 Z 1 2 R 2 f (z) dz ! 0: 2 Rb 1 b Rb 1 Integrate along 3. and let " ! 0+ . This gives then b > 1 Z 1 2 " 2 " f (z) dz b ! 0: 4 1 " b b Using the Residue Theorem and letting R ! 1 and " ! 0+ . and let R ! 1 and " ! 0+ . This gives Z f (z) dz = 3 Z Z " 1 z = xe2 i=b 1 = dz = = e2 i=b dx ! 3 1+ jzjb eib arg z dz = e2 i=b dx R 1 + xb Z 0 Z 1 1 1 ! e2 i=b dx = e 2 i=b dx = e2 i=b I: 1 1 + xb 0 1 + xb Integrate along 4. This gives Z f (z) dz = 1 Z Z R 1 z = xe0i 1 = dz = = dx ! 1+ jzjb eib arg z dz = dx " 1+x b 1 Z 1 1 ! dx = I: 0 1 + xb Integrate along 2 and let R ! 1.Integrate along 1. b and hence solving for I. we obtain that 1 I +0 e2 i=b I +0=2 i e i=b : b i=b Multiplying with e . 63 . b > 1: 0 1 + xb b sin ( =b) Remark. 2 i I= i=b i=b ) = (e e b sin ( =b) i. Z 1 dx = .) 2 3 Z 1 xb = t dx I= dx = 4 x = t1=b 5= 0 1 + xb 1 1=b 1 dx = b t dt Z 1 1=b 1 Z 1t 1 1 t (1 1=b) = dt = dt = 0 b 1+t b 0 1+t 1 = = : b sin ( (1=b 1)) b sin ( =b) 64 .4. (Because b > 1. then 0 < 1 1=b < 1 and the restriction that 0 < a < 1 for the integral in Exercise VII. we make the substitution xb = t and use the integral from Exercise VII.1 is satis…ed. We begin with the …rst integral I..4.1 where we choose a = 1 1=b.e. 3.VII.3 γ2 γ1 z1 -R γ4 γ3 R Set Z 1 log x I= . 0 < a < 1: 0 a x (x + 1) sin2 ( a) Remark. Residue at a simple pole at z1 = 1. 65 .4.3 By integrating around the keyhole contour. 0 xa (x + 1) and integrate log x log jzj + i arg z f (z) = = a ia arg z xa (x + 1) jzj e (z + 1) along the keyhole contour in Figure VII. so that 0 < arg z < 2 . We make a branch cut for log z along the positive real axis. where the cimicircle 4 have radius ". Solution VII. show that Z 1 2 log x cos ( a) dx = .4.4. where by Rule 3. Check the result by di¤erentiating the identity in Exercise 1. 1 = = ie : jzj e (z + 1) jzja eia arg z z= 1 Integrate along 1. This gives then 0 < a < 1 q Z log2 R + (2 )2 2 log R f (z) dz a (R 2 R ! 0: 2 R 1) R Integrate along 3. we obtain that 66 . and let R ! 1 and " ! 0+ . log jzj + i arg z log jzj + i arg z ia Res a ia arg z . This gives then 0 < a < 1 Z p log2 " + 2 f (z) dz a 2 " 2 "1 a jlog "j ! 0: 4 " (1 ") Using the Residue Theorem and letting R ! 1 and " ! 0+ . and let R ! 1 and " ! 0+ . This gives Z f (z) dz = 1 Z ZR log jzj + i arg z z = xe0i log x = dz = = dx ! jzja eia arg z (z + 1) dz = dx a " x (x + 1) 1 Z 1 log x ! a dx = I: 0 x (x + 1) Integrate along 2and let R ! 1.and let " ! 0+ . This gives Z f (z) dz = 3 Z Z " log jzj + i arg z z = xe2 i log x + 2 i = a ia arg z dz = = ! 3 jzj e (z + 1) dz = dx R x e a 2 ia (x + 1) Z 0 log x + 2 i ! a 2 ia (x + 1) dx = 1 x e Z 1 Z 1 2 ia log x 2 ia dx = e a dx 2 ie a = 0 x (x + 1) 0 x (x + 1) = e 2 ia I 2 ie 2 ia J Integrate along 4. da 0 1+x 1 1+x 1 xa (x + 1) 67 .1 we have the formula Z 1 x a (1) dx = . both the integrand and its derivative is continuous. we get simultaneous equations 2 2 sin ( a) J = 2 2 sin ( a) I 2 cos ( a) J = 0. We begin to di¤erentiate the left side in the formula with respect to the parameter a Z 1 Z 1 Z 1 d x a log x x a log x (2) dx = dx = dx.e. And that the primitive of the derivative is uniformly limited by a constant on every compact interval on a. and hence the solutions Z 1 2 Z 1 log x cos ( a) dx I= dx = . J= = . Z 1 2 log x cos ( a) dx = . 0 < a < 1: 0 a x (x + 1) sin2 ( a) Remark. because for every compact interval with 0 < a < 1. this yields ia ia ia e e I 2 ie J= 2 2. 0 a x (x + 1) sin2 ( a) 0 xa (x + 1) sin ( a) i. there 0 < a < 1.4. 0 < a < 1: 0 1+x sin ( a) It is allowed to di¤erentiate both sides in the formula. 2 ia 2 ia ia I +0 e I 2 ie J +0=2 i ie : ia Multiplying with e . From the Exercise VII. and thus ia 2i sin ( a) I 2 ie J= 2 2: Separating real and imaginary parts. and di¤erentiate the right hand side in the formula in the same way d cos ( a) (3) = : da sin ( a) sin2 ( a) By (1) .(3). we have that Z 1 2 log x cos ( a) dx = : 0 xa (x + 1) sin2 ( a) 68 . 4. where 0 < arg z < 2 . 69 . show that by integrating around the keyhole con- tour that Z 1 x a a (a + 1) (a + m 2) m dx = .4. We make a branchcut for z along the positive real axis.4 γ2 γ1 z1 -R γ4 γ3 R Set Z 1 x a I= dx 0 (1 + x)m and integrate z a jzj a e ia arg z f (z) = = (1 + z)m (1 + z)m a along the keyhole contour in Figure VII. Solution VII.VII.4.4.4 For …xed m 2. 1 m < a < 1: 0 (1 + x) (m 1)! sin ( a) Remark. The result can be obtained also by integrating the formula in Ex- ercise 1 by parts. where m 2 Z R a 2 f (z) dz 2 R ! 0: 2 (R 1)m Rm+a 1 Integrate along 3. z0 ] = lim [(z z0 )m f (z)] : z!z0 (m 1)! dz m 1 Residue at a pole of order m at z1 = 1 z a 1 dm 1 Res m . and let R ! 1 and " ! 0+ . 1 = lim z a = (1 + z) z! 1 (m 1)! dz m 1 1 = lim ( a) ( a 1) ( a m + 2) z ( a (m 1)) = z! 1 (m 1)! 1 = lim ( 1)m 1 a (a + 1) (a + m 2) jzj( a (m 1)) ei( a (m 1)) arg z = z! 1 (m 1)! 1 = ( 1)m 1 a (a + 1) (a + m 2) j 1j( a (m 1)) ei( a (m 1)) = (m 1)! 1 = ( 1)m 1 a (a + 1) (a + m 2) ( 1)m 1 e ia = (m 1)! a (a + 1) (a + m 2) = e ia : (m 1)! Integrate along 1. This gives 70 . This gives then 1 m < a < 1. and let R ! 1 and " ! 0+ . This gives Z Z Z jzj a e ia arg z z = xe0i R x a f (z) dz = dz = = dx ! (1 + z)m dz = dx " (1 + x)m 1 1 Z 1 x a ! dx = I: 0 (1 + x)m Integrate along 2 and let R ! 1.From Sa¤/Snider page 310 we have 1 dm 1 Res [f (z). where m 2 Z a " f (z) dz 2 " 2 "1 a ! 0: 4 (1 ")m Using the Residue Theorem and letting R ! 1 and " ! 0+ . Set (this is our induction hypothesis) Z 1 x a a (a + 1) (a + m 2) Im (a) = m dx = 0 (1 + x) (m 1)! sin ( a) 71 . this yields ia ia a (a + 1) (a + m 2) e e I=2 i . Z 1 x a a (a + 1) (a + m 2) m dx = . (m 1)! and hence solving for I. and let " ! 0+ .e. This gives then 1 m < a < 1. (e e (m 1)! (m 1)! sin ( a) i. we obtain that 2 ia a (a + 1) (a + m 2) ia I +0 e I +0=2 i e : (m 1)! ia Multiplying with e . 2i a (a + 1) (a + m 2) a (a + 1) (a + m 2) I= ia ia ) = . 1 m < a < 1: 0 (1 + x) (m 1)! sin ( a) Remark. Z f (z) dz = 3 Z Z " a 2 ia jzj a e ia arg z z = xe2 i x e m dz = = m dx ! 3 (1 + z) dz = dx R (1 + x) Z 0 a 2 ia Z 1 x e 2 ia x a ! m dx = e m dx = e 2 ia I: 1 (1 + x) 0 (1 + x) Integrate along 4. b (b + 1) (b + p 2) = (p 1)! sin ( b) Z 1 Z 1 x b 1 = p dx = x b dx = 0 (1 + x) 0 (1 + x)p 1 Z 1 b+1 x b+1 1 x p = p dx = 1 b (1 + x) 0 0 1 b (1 + x)p+1 Z 1 p x b+1 = dx: 1 b 0 (1 + x)p+1 Set b = a + 1 and solve for the integral in the left side in the inequality above 72 .From Exercise VII. thus our induction hypothesis is valid for m = 2.4. Now suppose that our formula is valid for m = p.1 we have Z 1 x b dx = . 0 < b < 1: 0 1+x sin ( b) Use Exercise VII. 0 (1 + x) sin ( (a + 1)) sin ( a) that is the same as I2 (a).4. and integrate by parts.1 and integrate by part Z 1 Z 1 x b 1 = dx = x b dx = sin ( b) 0 1+x 0 1+x 1 Z 1 b+1 x b+1 1 x 1 = dx = 1 b 1+x 0 0 1 b (1 + x)2 Z 1 1 x b+1 = dx: 1 b 0 (1 + x)2 Set b = a + 1 and solve for the integral in the left side in the inequality above Z 1 x a a 2 dx = a = . We have showed that the formula Im (a) is valid for all m > 2. 73 . p! sin ( a) that is the same as Ip+1 (a). thus our induction hypothesis is valid for m = p + 1. Z 1 x a dx = 0 (1 + x)p+1 a (a + 1) (a + 2) ((a + 1) + p 2) = = p (p 1)! sin ( (a + 1)) a (a + 1) (a + 2) (a + (p + 1) 2) = . that was to be shown. 4. where 0 < arg z < 2 .4. 74 . b = 2) γ2 γ1 z2 z1 -R γ4 γ3 R Set Z 1 log x I= dx 0 (x + a) (x + b) and integrate (log z)2 (log jzj + i arg z)2 f (z) = = (z + a) (z + b) (z + a) (z + b) around the keyhole contour in Figure VII.5 (a = 1. a 6= b: 0 (x + a) (x + b) 2 (a b) Solution VII.VII. Residue at a simple pole at z1 = a. a. show that Z 1 log x (log a)2 (log b)2 dx = . where by Rule 3.5. We make a branchcut for log z along the positive real axis. b > 0.4.5 (log z)2 By integrating a branch of (z+a)(z+b) around the keyhole contour. " # (log jzj + i arg z)2 (log jzj + i arg z)2 (log b + i )2 Res . and let R ! 1 and " ! 0+ . " # (log jzj + i arg z)2 (log jzj + i arg z)2 (log a + i )2 Res . and let R ! 1 and " ! 0+ . where by Rule 3. This gives Z f (z) dz = 1 Z Z (log jzj + i arg z)2 z = xe0i R (log x)2 = dz = = dx ! 1 (z + a) (z + b) dz = dx " (x + a) (x + b) Z 1 (log x)2 ! dx = J: 0 (x + a) (x + b) Integrate along 2 and let R ! 1. This gives Z f (z) dz = 3 Z Z " (log jzj + i arg z)2 z = xe2 i (log x + 2 i)2 = dz = = dx ! 3 (z + a) (z + b) dz = dx R (x + a) (x + b) Z 0 Z 1 (log x + 2 i)2 (log x + 2 i)2 ! dx = dx = 1 (x + a) (x + b) 0 (x + a) (x + b) Z 1 Z 1 Z 1 (log x)2 log x 4 2 = dx 4 i + dx = 0 (x + a) (x + b) 0 (x + a) (x + b) 0 (x + a) (x + b) = J 4 iI+K 75 . b = = : (z + a) (z + b) (z + a) a b z= b Integrate along 1. This gives Z log2 R + (2 )2 4 log2 R f (z) dz 2 R ! 0: 2 (R a) (R b) R Integrate along 3. a = = : (z + a) (z + b) (z + b) b a z= a Residue at a simple pole at z1 = b. This gives Z log2 " + (2 )2 2 " log2 " f (z) dz 2 " ! 0: 4 (a ") (b ") ab Using the Residue Theorem and letting R ! 1 and " ! 0+ . Z 1 log x (log a)2 (log b)2 dx = . b > 0. and let " ! 0+ . a. b a a b and hence. (log a)2 (log b)2 I= . 2 (a b) i. a 6= b: 0 (x + a) (x + b) 2 (a b) 76 .e. we obtain that ! (log jbj + i )2 (log jaj + i )2 J +0 J 4 iI+K + 0 = 2 i + . setting imaginary parts equal.Integrate along 4. where 0 < arg z < 2 .VII.4. show that Z 1 a x log x sin ( a) a 2 cos ( a) dx = .4.6.6 Using residue theory. We make a branchcut for z a and log z along the real axis.6 γ2 γ1 z1 -R γ4 γ3 R Set Z 1 xa log x I= dx 0 (1 + x)2 and integrate z a log z jzja eia arg z (log jzj + i arg z) f (z) = = (1 + z)2 (1 + z)2 along the keyhole contour in Figure VII. Residue at a double pole at z1 = 1. where by Rule 3. 1 < a < 1: 0 (1 + x)2 sin2 ( a) Solution VII. 77 .4. jzja eia arg z (log jzj + i arg z) Res . This gives Z f (z) dz = 1 Z Z jzja eia arg z (log jzj + i arg z) z = xe0i R xa log x = dz = = 2 dx ! (1 + z)2 dz = dx " (1 + x) 1 Z 1 xa log x ! dx = I: 0 (1 + x)2 Integrate along 2 and let R ! 1. and let R ! 1 and " ! 0+ . and let R ! 1 and " ! 0+ . This gives 78 . 1 = (1 + z)2 d a az a log z z a = (z log z) = + = dz z= 1 z z z= 1 a z = (a log z + 1) = z z= 1 jzja eia arg z = (a (log jzj + i arg z) + 1) = z z= 1 ia =e ( 1 ai) : Integrate along 1. This gives q Z Ra log2 R + (2 )2 2 log R f (z) dz 2 R ! 0: 2 (R 1)2 R1 a Integrate along 3. Z f (z) dz = 3 Z Z " a 2 ia jzja eia arg z (log jzj + i arg z) z = xe2 i x e (log jxj + 2 i) = 2 dz = = dx ! 3 (1 + z) dz = dx R (1 + x)2 Z 0 a 2 ia Z 1 a x e (log jxj + 2 i) 2 ia x (log jxj + 2 i) ! 2 dx = e dx = 1 (1 + x) 0 (1 + x)2 Z 1 a Z 1 2 ia x log x 2 ia xa dx = e dx 2 ie = 0 (1 + x)2 0 (1 + x)2 = e2 ia I 2 ie2 ia J Integrate along 4. and let " ! 0+ . this yields ia ia ia e e I 2 ie J = 2 i( 1 ai) . 0 (1 + x)2 sin2 ( a) 0 (1 + x) sin ( a) 79 . and hence the solutions Z 1 Z 1 xa log x sin ( a) a 2 cos ( a) xa dx a I= = . we get simultaneous equations 2 sin ( a) J = 2 2 a 2 sin ( a) I 2 cos ( a) J = 2 . J= 2 = . we obtain that I +0 e2 ia I 2 ie2 ia J +0=2 i e ia ( 1 ai) . and thus ia 2i sin ( a) I 2 ie J = 2 2a 2 i: Separating real and imaginary parts. ia multiplying with e . This gives q Z " log2 " + (2 )2 a f (z) dz 2 " 2 "1+a jlog "j ! 0: 4 (1 ")2 Using the Residue Theorem and letting R ! 1 and " ! 0+ . i. Z 1 xa log x sin ( a) a 2 cos ( a) dx = .e. 1 < a < 1: 0 (1 + x)2 sin2 ( a) 80 . Residue at a simple pole at z1 = e i=b .7. We make a branch cut for z a 1 along the negative imaginary axis .4. where by Rule 3. 0 1 + xb and integrate za 1 jzja 1 ei(a 1) arg z f (z) = = 1 + zb 1 + jzjb eib arg z along the pie-slice.7 (b = 12) γ3 γ2 z1 ε γ1 R Set Z 1 xa 1 I= .7 Show that Z 1 xa 1 dx = . where =2 < arg z < 3 =2. Where does the integral depend analytically on the parameter a? Solution VII.4.VII. and evaluate it.4. of aperture 2 =b. 0 < a < b: 0 1 + xb b sin ( a=b) Determine for which complex values of the parameter a the integral exists (in the sense that the integral of the absolute value is …nite). contour in Figure VII. 81 . and let " ! 0+ . This gives Z f (z) dz = 1 Z Z jzja 1 i(a 1) arg z e z = xe0i R xa 1 = dz = = dx ! 1 1 + jzjb eib arg z dz = dx " 1 + xb Z 1 xa 1 ! dx = I: 0 1 + xb Integrate along 2 and let R ! 1. This gives then 0 < a < b Z "a 1 2 " 2 "a f (z) dz ! 0: 4 1 "b b b Using the Residue Theorem and letting R ! 1 and " ! 0+ . and let R ! 1 and " ! 0+ . we obtain that 1 I +0 e2 ia=b I +0=2 i e ia=b . and let R ! 1 and " ! 0+ . This gives then 0 < a < b Z Ra 1 2 R 2 f (z) dz ! 0: 2 Rb 1 b bRb a Integrate along 3. b ia=b and multiplying with e . this yields 82 . za 1 i=b za 1 1 ia=b Res .e = b = e : 1 + zb bz 1 z=e i=b b Integrate along 1. This gives Z f (z) dz = 3 Z Z jzja 1 i(a 1) arg z e z = xe2 i=b " xa 1 e2 i(a 1)=b = = = e2 i=b dx ! 3 1+ jzjb eib arg z dz = e2 i=b dx R 1 + xb Z 0 Z 1 xa 1 e2 i(a 1)=b 2 i=b 2 ia=b xa 1 ! e dx = e dx = e2 ia=b I: 1 1 + xb 0 1 + xb Integrate along 4. e. ia=b ia=b 2 i e e I= b and hence solving for I. Z 1 xa 1 dx = . 2 i I= ia=b ia=b ) = . b (e e b sin ( a=b) i.. 1 + xb 1 + xb 1 + xb thus Re a 1 > 1 and Re a 1 b < 1. 83 . Thus the integrand depends analytically on the parameter a when 0 < Re a < b. 0 < a < b: 0 1 + xb b sin ( a=b) Now we suppose a and b complex and rewrite the nominator in the integrand xa 1 = e(a 1) log x = e(Re a 1) log x i Im a log x e = xRe a 1 ei Im a log x : The absolute value of the integrand is xa 1 xRe a 1 ei Im a log x xRe a 1 = = . of aperture 2 =3. show that Z 1 Z 1 log x 2 2 1 2 dx = .8 By integrating a branch of (log z) = (z 3 + 1) around the boundary of an indented sector of aperture 2 =3. J= dx 0 x3 + 1 0 x3 +1 and integrate log z log jzj + i arg z f (z) = = z3 + 1 z e i=3 (z e i ) z e5 i=3 along the indented sector. We make a branchcut for log z along the negative imaginary axis.4.8. dx = p : 0 x3 + 1 27 0 x3 + 1 3 3 Remark.4.VII. Compare the results with those of Exercise 3 (after changing vari- ables) and Exercise 2.4. where by Rule 3. Solution VII. Residue at a simple pole at z1 = e i=3 . 84 . contour in Figure VII.8 γ2 γ3 z1 γ4 z2 ε γ1 R z3 Set Z 1 Z 1 log x 1 I= dx. where =2 < arg z < 3 =2. This gives Z f (z) dz = 3 Z Z " log jzj + i arg z z = xe2 i=3 log x + 2 i=3 2 i=3 = dz = 2 i=3 = e dx ! 3 3 z +1 dz = e dx R x3 + 1 Z 0 Z 1 log x + 2 i=3 2 i=3 log x + 2 i=3 2 i=3 ! e dx = e dx = 1 x3 + 1 0 x3 + 1 Z 1 Z 2 i=3 log x 2 i 2 i=3 1 1 2 i 2 i=3 = e 3 dx e 3 dx = e2 i=3 I e J: 0 x +1 3 0 x +1 3 Integrate along and let " ! 0+ . and let R ! 1 and " ! 0+ . we obtain that 85 . q Z log2 " + 23 2 2 " 2 " jlog "j f (z) dz 3 ! 0: 4 1 " 3 3 Using the Residue Theorem and letting R ! 1 and " ! 0+ . log jzj + i arg z i=3 log jzj + i arg z i 2 i=3 Res . and let R ! 1. This gives 4. This gives q Z log2 R + 23 2 2 R 2 log R f (z) dz 3 ! 0: 2 R 1 3 3R2 Integrate along 3. and let R ! 1 and " ! 0+ . This gives Z f (z) dz = 1 Z Z R log jzj + i arg z z = xe0i log x = dz = = dx ! z3 + 1 dz = dx 3 " x +1 1 Z 1 log x ! dx = I: 0 x3 + 1 Integrate along 2.e = = e : z3 + 1 3z 2 z=e i=3 9 Integrate along 1. 3. 2 i 2 i I + 0 e2 i=3 I e i=3 J +0=2 i e 2 i=3 : 3 9 i=3 Multiplying with e . where we choose a = 2=3.2. and hence the solutions Z 1 Z 1 log x 2 2 1 2 I= dx = . J= dx = p : 0 x3 + 1 27 0 x3 +1 3 3 Remark. where we choose b = 3. 3 9 and thus p p i 2 23 3Ii + J= : J 3 3 9 Separating real and imaginary parts. we get the simultaneous equations p 2 3 p 3 J = 29 3I 3 J = 0.4. this yields 2 2 i=3 2 2i sin ( =3) I ie J= . 2 3 Z 1 x3 = t log x I= dx = 4 x = t1=3 5= 0 x3 + 1 1 2=3 dx = 3 t dt Z 1 1 log t 1 2 cos 23 2 2 = dt = = : 9 0 t2=3 (t + 1) 9 sin2 23 27 For the second integral J. we make the substitution x3 = t and use the integral from Exercise VII. We begin with the …rst integral I. Z 1 dx 2 J= b = = p : 0 1+x 3 sin ( =3) 3 3 86 .4. we use the integral from Exercise VII. where by Rule 3.9.4. where =2 < arg z < 3 =2.9 By integrating around an appropriate contour and using the results of Exercise 8.4.9 γ2 γ3 z1 γ4 z2 ε γ1 R z3 Set Z 1 (log x)2 I= dx 0 x3 + 1 Integrate (logjzj+i arg z)2 (log z)2 f (z) = = z3 + 1 z e i=3 (z e i) z e5 i=3 along the pie-slice contour with 0 < arg z < 2 =3 in Figure VII.e = = e : z3 + 1 3z 2 i=3 27 z=e 87 . show that Z 1 (log x)2 10 3 dx = p : 0 x3 + 1 81 3 Solution VII. Residue at a simple pole at z1 = e i=3 . We make a branchcut for log z along the negative imaginary axis.4.VII. " # (log jzj + i arg z)2 i=3 (log jzj + i arg z)2 2 2 i=3 Res . and let R ! 1. and let R ! 1 and " ! 0+ .Integrate along 1. and let R ! 1 and " ! 0+ . This gives Z f (z) dz = 1 Z Z (log jzj + i arg z)2 z = xe0i R (log x)2 dz = = dx ! 1 z3 + 1 dz = dx " x3 + 1 Z 1 (log x)2 ! dx = I: 0 x3 + 1 Integrate along 2. This gives Z 2 log2 R + 23 2 R 2 log2 R f (z) dz ! 0: 2 R3 1 3 3R2 Integrate along 3. This gives Z f (z) dz = 3 Z Z " (log jzj + i arg z)2 z = xe2 i=3 (log x + 2 i=3)2 2 i=3 = dz = = e dx ! 3 z3 + 1 dz = e2 i=3 dx R x3 + 1 Z 0 Z 1 (log x + 2 i=3)2 2 i=3 (log x + 2 i=3)2 2 i=3 ! e dx = e dx = 1 x3 + 1 0 x3 + 1 Z 1 Z Z 2 i=3 (log x)2 4 i 2 i=3 1 log x 4 2 2 i=3 1 1 = e dx e dx+ e dx = 0 x3 + 1 3 0 x3 + 1 9 0 x3 + 1 Z Z 2 i=3 4 i 2 i=3 1 log x 4 2 2 i=3 1 1 = e I e dx+ e : 3 0 x3 + 1 9 0 x3 + 1 Integrate along 4. and let " ! 0+ . we obtain that Z 1 Z 1 2 i=3 4 i 2 i=3 log x 4 2 2 i=3 1 2 2 i=3 I+0 e I e dx+ e +0 = 2 i e : 3 0 x3 + 1 9 0 3 x +1 27 88 . This gives Z 2 log2 " + 23 2 " 2 " log2 " f (z) dz ! 0: 4 1 "3 3 3 Using the Residue Theorem and letting R ! 1 and " ! 0+ . 4 i i=3 2 2 4 2 i=3 2 2 3 2i sin ( =3) I e + e p = i: 3 27 9 3 3 27 And hence. this yields Z 1 2 Z 1 i=3 i=3 4 i log x i=3 4 1 i=3 2 3 e e I e dx+ e dx = i. p 4 3 12 3 2 3 3I+ + = . 81 81 27 and therefore Z 1 (log x)2 10 2 dx = p : 0 x3 + 1 81 3 89 . i=3 multiplying with e . 3 0 x3 + 1 9 0 3 x +1 27 and thus. setting imaginary parts equal. substituting the values for the integrals from Exercise 9. log jzj + i arg z log jzj + i arg z Res .VII. 90 .10 γ4 z2 γ6 γ2 -R γ5 γ1 z 1 γ3 R z3 Set Z 1 log x I= dx 0 x3 1 Integrate log z log jzj + i arg z f (z) = = z3 1 (z 1) z e2 i=3 z e4 i=3 along the indented half-disk contour in Figure VII.10 By integrating a branch of (log z) = (z 3 1) around the boundary of an indented half-disk and using the result of Exercise 8. Residue at a simple pole at z1 = 1. where by Rule 3. where by Rule 3. where =2 < arg z < 3 =2.4.10. show that Z 1 log x 4 2 dx = : 0 x3 1 27 Solution VII.1 = = 0: z 3 1 3z 2 z=1 Residue at a simple pole at z2 = e2 i=3 .4.4. We make a branchcut for log z along the negative imaginary axis. and let " ! 0+ . log jzj + i arg z 2 i=3 log jzj + i arg z 2 i 4 i=3 Res . This gives Z Z f (z) dz + f (z) dz = 1 Z 3 Z log jzj + i arg z log jzj + i arg z = dz + dz = 1 z3 1 3 z3 1 z = xe0i = = dz = dx Z 1 " Z R log x log x = 3 dx + 3 dx ! " x 1 1+" x 1 Z 1 log x ! dx = I: 0 x3 1 Integrate along 2. and let R ! 1 and " ! 0+ . and let R ! 1. This gives Z f (z) dz ! i ( 0) 0 = 0: 2 Integrate along 4. and let R ! 1 and " ! 0+ . This gives Z p log2 R + 2 log R f (z) dz 3 R ! 0: 4 R 1 R2 Integrate along 5.e = = e : z3 1 3z 2 z=e2 i=3 9 Integrate along 1 and 3. This gives 91 . 2 2 2 4 2 4 i=3 I i p = e : 27 3 3 9 And hence. Z p log2 " + f (z) dz " " jlog "j ! 0: 6 1 "3 Using the Residue Theorem and letting R ! 1 and " ! 0+ . 2 2 4 2 I+ = . 27 18 and therefore Z 1 log x 4 2 dx = : 0 x3 1 27 92 . 0 x3 + 1 0 x3 +1 27 and thus. setting imaginary parts equal. This gives 6. Z f (z) dz = 5 Z Z " log jzj + i arg z z = xe i log x + i = dz = = e id ! 5 z 3 1 dz = e i dx R x 3 1 Z 0 Z 1 log x + i log x + i ! 3 e i dx = dx = 1 x 1 0 x3 + 1 Z 1 Z 1 log x 1 = dx i dx = 0 x3 + 1 0 3 x +1 Z 1 Z 1 log x 1 = 3 dx i 3 dx: 0 x +1 0 x +1 Integrate along and let " ! 0+ . we obtain that Z 1 Z 1 log x 1 2 i 4 i=3 I +0+0 dx i dx + 0 = 2 i e . substituting the values for the integrals from Exercise 9. VII. 1) from above. It extends analytically to (0.1 Use the keyhole contour indented on the lower edge of the axis at x = 1 to show that Z 1 log x 2 2 dx = . Thus f (z) is analytic on the keyhole domain. We make a branchcut for log z along the positive imaginary axis.5.5. 0 < a < 1: 0 xa (x 1) 1 cos (2 a) Solution VII.5.1. and the 93 . where 0 < arg z < 2 .1 γ2 γ1 -R γ5 z γ3 R γ6 γ4 Set Z 1 log x I= dx 1 xa (x 1) and integrate log z (log jzj + i arg z) f (z) = = a(logjzj+i arg z) za (z 1) e (z 1) along the contour in Figure VII. 1) from below has a simple pole at z = 1. the extension to (0. This gives Z f (z) dz = 1 Z ZR (log jzj + i arg z) z = xe0i log x = dz = = dx ! ea(logjzj+i arg z) (z 1) dz = dx a " x (x 1) 1 Z 1 log x ! dx = I: 0 xa (x 1) Integrate along and let " ! 0+ . q Z log2 R + (2 )2 2 log R f (z) dz a 2 R ! 0: 2 R (R 1) Ra Integrate along 3 and 5. 1 = = 2 ie : ea(logjzj+i arg z) (z 1) ea(logjzj+i arg z) z=1 Integrate along 1. Residue at a simple pole at z = 1.apparent singularity at z = 1 is removable. This gives Z Z f (z) dz+ f (z) dz = 3 Z 5 Z (log jzj + i arg z) (log jzj + i arg z) = a(logjzj+i arg z) dz+ a(logjzj+i arg z) (z dz = 3 e (z 1) 5 e 1) Z 1+" Z " z = xe2 i 2 ia (log x + 2 i) 2 ia (log x + 2 i) = =e dx+e dx ! dz = dx R xa (x 1) 1 " x (x a 1) Z 0 Z 1 2 ia (log x + 2 i) 2 ia (log x + 2 i) !e a dx = e dx = 1 x (x 1) 0 xa (x 1) Z 1 Z 1 2 ia log x 2 ia 1 = e a dx 2 ie a dx = e 2 ia I 2 ie 2 ia J: 0 x (x 1) 0 x (x 1) Integrate along 4. (log jzj + i arg z) (log jzj + i arg z) 2 ia Res . This gives 94 . This gives when 0 < a < 1 2. However. where by Rule 3. use fractional residue formula and let " ! 0+ . and let R ! 1 and " ! 0+ . and let R ! 1 and " ! 0+ . 0 < a < 1: 0 x (x 1) 1 cos (2 a) 95 . This gives 0 < a < 1 6. we obtain that 2 ia 2 ia I e I 2 ie J + 2 2e 2 ia + 0 = 0: Multiplying with e2 ia . q Z log2 " + (2 )2 f (z) dz a (1 2 " 2 "1 a jlog "j ! 0: 6 " ") Using the Residue Theorem and letting R ! 1 and " ! 0+ . Z 2 ia f (z) dz ! i( 0) 2 ie = 2 2e 2 ia : 4 Integrate along and let " ! 0+ . this yields Ie2 ia I 2 iJ + 2 2 = 0: We equate real parts 2 (cos (2 a) 1) I + 2 = 0. and therefore Z 1 log x 2 2 a dx = . where by Rule 3. 96 . Solution VII. and integrate around the boundary of a half- disk indented at z = 0.2.2 γ4 z1 γ2 -R γ1 z2 γ3 R z3 Set Z 1 sin ax I= dx 1 x (x2 + 1) and integrate eiaz eiaz f (z) = = z (z 2 + 1) z (z i) (z + i) along the contour in Figure VII.VII. Replace sin (az) by eiaz .5.0 = 2 = 1: z (z 2 + 1) 3z + 1 z=0 Residue at a simple pole at z2 = i.5. eiaz eiaz Res .2 Show using residue theory that Z 1 sin (ax) a 2 dx = 1 e . a > 0: 1 x (x + 1) Hint.5. Residue at a simple pole at z1 = 0. where by Rule 3. we obtain that Z 1 cos ax e a 2 dx + iI i + 0 = 2 i . this gives Z 1 f (z) dz R ! 0: 4 R (R2 1) R2 Using the Residue Theorem and letting R ! 1 and " ! 0+ .. 97 . I= e a. and let R ! 1 and " ! 0+ . 1 x (x + 1) 2 and hence. and let R ! 1. This gives Z Z f (z) dz + f (z) dz = 1 3 Z Z eiaz eiaz z = x = dz + dz = = 1 2 z (z + 1) 3 2 z (z + 1) dz = dx Z " Z R eiax eiax = 2 dx + 2 dx ! R x (x + 1) " x (x + 1) Z 1 Z 1 Z 1 eiax cos ax sin ax ! 2 dx = 2 dx + i 2 dx = 1 x (x + 1) 1 x (x + 1) 1 x (x + 1) Z 1 cos ax = 2 dx + iI: 1 x (x + 1) Integrate along 2. Because jeiaz j 1 in the upper half plane if a > 0. eiaz eiaz e a Res .e. This gives Z f (z) dz ! i ( 0) 1 = i: 2 Integrate along 4 . i = = : z (z 2 + 1) 3z 2 + 1 z=i 2 Integrate along 1 and 3. use fractional residue formula and let " ! 0+ . setting imaginary parts equal. i. a > 0: 1 x (x2 + 1) 98 .Z 1 sin ax a dx = 1 e . 3 γ8 γ2 γ4 γ6 -R γ1 z3 γ3 z1 γ5 z2 γ7 R Set Z 1 sin (ax) I= dx 1 x ( 2 a2 x 2 ) and integrate eiaz eiaz f (z) = = z ( 2 a2 z 2 ) a2 z (z =a) (z + =a) along the contour in Figure VII.3.5. eiaz eiaz 1 Res . where by Rule 3. = = : z ( 2 a2 z 2 ) a 2 3a2 z 2 z= =a 2 2 99 . Residue at a simple pole at z1 = 0. a > 0: 1 x( ax) Solution VII.0 = = : z ( 2 a2 z 2 ) 2 3a2 z 2 z=0 2 Residue at a simple pole at z2 = =a.3 Show using residue theory that Z 1 sin (ax) 2 2 2 2 dx = .VII.5. eiaz eiaz 1 Res .5. where by Rule 3. 5 and 7. and let R ! 1 and " ! 0+ . This gives Z 1 1 f (z) dz ! i ( 0) 2 = i: 2 2 2 Integrate along 4. eiaz eiaz 1 Res . use fractional residue formula and let " ! 0+ . where by Rule 3. = = : z ( 2 a2 z 2 ) a 2 3a2 z 2 z= =a 2 2 Integrate along 1. This gives Z 1 1 f (z) dz ! i ( 0) 2 = i: 4 100 .Residue at a simple pole at z3 = =a. use fractional residue formula and let " ! 0+ . This gives Z Z Z Z f (z) dz + f (z) dz + f (z) dz + f (z) dz = 1 3 5 7 Z Z eiaz eiaz = dz + dz+ 1 z ( 2 a2 z 2 ) 3 z ( 2 a2 z 2 ) Z iaz Z e eiaz + 2 dz + dz = 5 z( a2 z 2 ) 7 z ( 2 a2 z 2 ) z = x = = dz = dx Z =a " Z " eiax eiax = dx + dx+ R x ( 2 a2 x 2 ) =a+" x ( 2 a2 x 2 ) Z =a " Z R eiax eiax + dx + dx ! " x ( 2 a2 x 2 ) =a+" x ( 2 a2 x2 ) Z 1 eiax ! 2 dx = 1 x( a2 x 2 ) Z 1 Z 1 cos (ax) sin (ax) = 2 2 2 dx + i 2 dx = 1 x( ax) 1 x( a2 x 2 ) Z 1 cos (ax) = 2 dx + iI: 1 x( a2 x 2 ) Integrate along 2. 3. Z 1 sin (ax) 2 2 2 2 dx = .. 2 2 i. setting imaginary parts equal 1 1 1 I = 0. 1 x (x + 1) 2 2 and hence. we obtain that Z 1 cos ax 1 1 1 2 dx + iI i i i = 0. This gives Z 1 1 f (z) dz ! i ( 0) 2 = i: 6 2 2 Integrate along 8 . a > 0: 1 x( ax) 101 . this gives Z 1 f (z) dz R ! 0: 2 R (a R2 2 2) a2 R 2 Using the Residue Theorem and letting R ! 1 and " ! 0+ .Integrate along 6.use fractional residue formula and let " ! 0+ .e. Because jeiaz j 1 in the upper half plane if a > 0. and let R ! 1. 5. 1 eiz d Res .4.4 Show using residue theory that Z 1 1 cos x dx = : 0 x2 2 Solution VII.5. This gives 102 .5. 0 = lim 1 eiz = ieiz z=0 = i z2 z!0 dz Integrate along 1 and 3.VII. where by Rule 2. Residue at a double pole at z1 = 0. and let R ! 1 and " ! 0+ .4 γ4 γ2 -R γ1 z1 γ3 R Set Z 1 Z 1 1 cos x 1 cos x I= dx ) 2I = dx 0 x2 1 x2 and integrate 1 eiz f (z) = z2 along the contour in Figure VII. . use fractional residue formula and let " ! 0+ . This gives Z f (z) dz ! i ( 0) ( i) = : 2 Integrate along 4 . Z 1 1 cos x dx = : 0 x2 2 103 . we obtain that Z 1 sin x 2I i 2 dx + 0 = 0.e. setting imaginary parts equal 2I = 0. this gives Z 2 2 f (z) dz 2 R ! 0: 4 R R Using the Residue Theorem and letting R ! 1 and " ! 0+ . Because jeiaz j 1 in the upper half plane if a > 0. and let R ! 1. Z Z f (z) dz + f (z) dz = 1 3 Z Z 1 eiz 1 eiz = dz + dz = 1 z2 3 z2 z = x = = dz = dx Z " Z R Z 1 1 eix 1 eix 1 eix dx + dx ! dx = R x2 " x2 1 x2 Z 1 Z 1 Z 1 1 cos x sin x sin x 2 dx i 2 dx = 2I i 2 dx: 1 x 1 x 1 x Integrate along 2. i. 1 x and hence. show that Z 1 sin2 x 2 dx = : 1 x Solution VII.VII.5. and let R ! 1 and " ! 0+ .5. This gives 104 . Residue at a double pole at z1 = 0.5 γ2 γ4 -R γ3 z1 γ1 R Set Z 1 Z 1 sin2 x 1 cos 2x I= dx = dx 1 x2 1 2x2 and integrate ei2z 1 f (z) = z2 along the contour in Figure VII.5 By integrating (e 2iz 1) =z 2 over appropriate indented contours and using Cauchy´s theorem.5. where by Rule 2.5. ei2z 1 d i2z Res . 0 = lim e 1 = 2iei2z z=0 = 2i z2 z!0 dz Integrate along 1 and 3. and let R ! 1. 1 2x and hence. this gives Z 2 2 f (z) dz 2 R ! 0: 4 R R Using the Residue Theorem and letting R ! 1 and " ! 0+ . This gives Z f (z) dz ! i ( 0) 2i = 2 : 2 Integrate along 4 . i. 2I + 2 = 0. Z Z f (z) dz + f (z) dz = 1 3 Z Z ei2z 1 ei2z 1 z = x = dz + dz = = 1 z2 3 z2 dz = dx Z " i2x Z R i2x Z 1 i2x e 1 e 1 e 1 = dx + dx ! dx = R x2 " x2 1 x2 Z 1 Z 1 Z 1 cos 2x 1 sin 2x sin 2x = dx + i dx = 2I + i dx: 1 x2 1 x2 1 x2 Integrate along 2. setting real parts equal. use fractional residue formula and let " ! 0+ .. Because jeiaz j 1 in the upper half plane if a > 0.e. we obtain that Z 1 sin 2x 2I i 2 dx + 2 + 0 = 0. Z 1 sin2 x dx = : 1 x2 105 . We make a branchcut for log z along the negative imaginary axis.10. show that Z 1 log x 4 2 dx = : 0 x3 1 27 Remark. 106 . See also Exercise 4. log jzj + i arg z log jzj + i arg z Res .VII.6 By integrating a branch of (log z) = (z 3 1) around the boundary of an indented sector of aperture 2 =3. where =2 < arg z < 3 =2. Residue at a simple pole at z1 = 1. Solution VII.5.1 = = 0: z 3 1 3z 2 z=1 Residue at a simple pole at z2 = e2 i=3 .5.6.6 γ5 γ4 γ6 z2 γ7 γ8 γ2 γ1 z1 γ3 R z3 Set Z 1 log x I= dx 0 x3 1 and integrate log z log jzj + i arg z f (z) = = z3 1 (z e0i ) z e2 i=3 z e4 i=3 along the contour in Figure VII.5. where by Rule 3. where by Rule 3. This gives Z Z f (z) dz+ f (z) dz = 1 3 Z Z log jzj + i arg z log jzj + i arg z z = xe0i dz+ dz = = 1 z3 1 3 z 3 1 dz = dx Z 1 " Z R Z 1 log x log x log x = 3 dx+ 3 dx ! dx = I: " x 1 1+" x 1 0 x3 1 Integrate along 2. use fractional residue formula and let " ! 0+ . log jzj + i arg z 2 i=3 log jzj + i arg z 2 i 2 i=3 Res . use fractional residue formula and let " ! 0+ . and let R ! 1 and " ! 0 . and let R ! 1 and " ! 0+ . This gives Z Z f (z) dz + f (z) dz = 5 7 Z Z log jzj + i arg z log jzj + i arg z z = xe2 i=3 = dz+ dz = = 5 z3 1 7 z3 1 dz = e2 i=3 dx Z 1+" Z " log x + 2 i=3 2 i=3 log x + 2 i=3 2 i=3 = e dx + e dx ! R x3 1 1 " x3 1 Z 0 Z 1 Z 2 i=3 log x + 2 i=3 2 i=3 log x 2 i 2 i=3 1 1 !e dx = e dx e dx = 1 x3 1 0 x3 1 3 0 x3 1 Z 2 i=3 2 i 2 i=3 1 1 = e I e dx: 3 0 x3 1 Integrate along 6. This gives 107 . and let R ! 1.e = = e : z3 1 3z 2 z=e2 i=3 9 + Integrate along 1 and 3. This gives q Z log2 R + 23 2 2 R 2 log R f (z) dz 3 ! 0: 4 R 1 3 R2 Integrate along 5 and 7. This gives Z f (z) dz ! i ( 0) 0 = 0: 2 Integrate along 4. this yields p ! Z 1 3 2 i 1 1 2 2 I i I dx + = 0: 2 2 3 0 x3 1 9 We equate real parts 3 2 2 I+ = 0. we obtain that Z 2 i=3 2 i 2 i=3 1 1 2 2 2 i=3 I +0 e I e dx + e = 0: 3 0 x3 1 9 2 i=3 Multiplying with e . Z 2 2 i 2 i=3 2 f (z) dz ! i( 0) e = e2 i=3 : 6 9 9 Integrate along and let " ! 0+ . 2 9 and hence Z 1 log x 4 2 dx = : 0 x3 1 27 108 . q Z log2 " + 23 2 2 " 2 " jlog "j f (z) dz 3 ! 0: 8 1 " 3 3 Using the Residue Theorem and letting R ! 1 and " ! 0+ . This gives 8. 1 Integrate 1= (1 x2 ) directly. and show that Z 1 dx PV = 0: 0 1 x2 Show that Z 1 Z 1 dx dx = +1. = 1: 0 1 x2 1 1 x2 Solution Partial fractions gives 1 1 1 1 = + 1 x2 2 1+x 1 x Z 1 Z 1 " Z 1 dx dx dx PV = lim + = 0 1 x2 "!0 0 1 x2 1+" 1 x2 1 1 = lim [log (1 + x) log (1 x)]10 " + [log j1 + xj log j1 xj]1 1+" = "!0 2 2 ! 1 " 1 1 1+x 1 1+x = lim log + log = "!0 2 1 x 0 2 1 x 1+" 1 2 " 2+" = lim log log = "!0 2 " " 1 2 " = lim log = 0: "!0 2 2+" Inspection gives Z 1 Z 1 " dx dx 1 = lim = lim+ [log (1 + x) log (1 x)]10 " = 0 1 x2 "!0+ 0 1 x 2 "!0 2 1 " 1 1+x 1 2 " = lim+ log = lim+ log = +1 : "!0 2 1 x 0 "!0 2 " 109 .6. using partial fractions.VII. and the second integral Z 1 Z 1 dx dx 1 = lim = lim+ [log (1 + x) log (1 x)]1 1+" = 1 1 x2 "!0+ 1+" 1 x 2 "!0 2 1 1 1+x 1 2+" = lim+ log = lim+ log ! 1: "!0 2 1 x 1+" "!0 2 " 110 . For a linear change of variables.VII. C" z2 1 as " ! 0.4) Z 1 Z 1 " Z 0 Z log x log jxj + i log jxj + i log z dx+ dx+ dx+ dz = 0: 0 x2 1 1 x2 1 1+" x2 1 C" z 2 1 Taking imaginary part of (5. x ! x. we can take the limit either before or after change of variable. 111 . Otherwise we can not do this. and by using ( ) we get Z 1 " Z 0 x = x dx + dx = = 1 x2 1 1+" x2 1 dx = dx Z 1 Z 1 " dx + dx ! 0 as " ! 0.211 in the preceding section and making a change of variable Solution Identity (5.4). by estimation on p. get Z 1 " Z 0 Z log z dx + dx + Im dz = 0: 1 x2 1 1+" x 2 1 C" z2 1 Note Z log z ( ) Im dz ! 0.4) p. 211.6. Making the change of variables.2 Obtain the principal value in Exercise 1 by taking imaginary parts of the identity (5. 1+" x2 1 0 x2 1 which gives that Z 1 dx PV = 0: 0 1 x2 Note. show that Z 1 1 a PV 2 dx = 2 . 1 1 1 Res .3 By integrating around the boundary of an indented half-disk in the upper half-plane.VII. Residue at a simple pole at z1 = a.a = = : (z 2 + 1) (z a) z2 +1 z=a a2 +1 Residue at a simple pole at z1 = i.3 (a = 2) γ4 z2 γ2 -R γ1 z1 γ3 R z3 Set Z 1 1 I = PV dx 1 (x2 + 1) (x a) and integrate 1 1 f (z) = = (z 2 + 1) (z a) (z i) (z + i) (z a) along the contour in Figure VII.6. where by Rule 3.3. 1 < a < 1: 1 (x + 1) (x a) a +1 Solution VII. where by Rule 1. 112 .6.6. .e. This gives Z 1 f (z) dz ! i( 0) = i: 2 a2 +1 a2 + 1 Integrate along 4. Z 1 1 a PV dx = : 1 (x2 + 1) (x a) a2 +1 113 . we obtain that 1 a I i=2 i + i . and let " ! 0+ . This gives Z Z f (z) dz + f (z) dz = 1 3 Za " ZR 1 1 = 2 dx + dx ! (x + 1) (x a) (x2 + 1) (x a) R a+" Z 1 1 ! dx = I: 1 (x2 + 1) (x a) Integrate along 2. 1 1 1 a Res . This gives Z 1 f (z) dz R ! 0: 4 (R2 1) (R jaj) R2 Using the Residue Theorem and letting R ! 1 and " ! 0+ . and let R ! 1. a2 + 1 2 (a2 + 1) 2 (a2 + 1) i.i = 2 = + i: (z 2 + 1) (z a) 3z 2az + 1 z=i 2 (a2 + 1) 2 (a2 + 1) Integrate along 1 and 3. and let R ! 1 and " ! 0+ . 1 γm a1 a2 a m .k6=j (aj ak ) 114 .4.6.6.VII. By integrating around the boundary of an indented half-disk in the upper half-plane. Residue at a simple pole at zj = aj . show that Z 1 1 PV dx = 0: 1 (x a1 ) (x a2 ) (x am ) Solution VII. aj = Qm = Qm : k=1 (z ak ) k=1. 1 j m.4 γR γ1 γ2 γ m . see Figure VII.4 Suppose m 2 and a1 < a2 < < am .k6=j (z ak ) z=aj k=1.6. 1 1 1 Res Qm . where by Rule 3.1 am -R R Set Z 1 1 I = PV dx 1 (x a1 ) (x a2 ) (x am ) and integrate 1 1 f (z) = = Qm (z a1 ) (z a2 ) (z am ) k=1 (z ak ) along the semicircular contour indented at z = aj with small semicircles of radii ". This gives m Z ! X X m 1 f (z) dz ! i( 0) Qm : j=1 j j=1 k=1.k6=j (aj ak ) Integrate along R. This gives Z Z a1 " X1 Z ak+1 " m Z R f (z) dz = f (z) dz+ f (z) dz+ f (z) dz = L R k=1 ak +" am +" Z a1 " m 1 Z X ak+1 " Z R 1 1 1 = Qm dz+ Qm dz+ Qm dz = R k=1 (z ak ) k=1 ak +" k=1 (z ak ) am +" k=1 (z ak ) z = x = = dz = dx Z a1 " X1 Z ak+1 " m Z R 1 1 1 = Qm dx+ Qm dx+ Qm dx ! R k=1 (x ak ) k=1 ak +" k=1 (x ak ) am +" k=1 (x ak ) Z 1 1 ! Qm dx = I: 1 k=1 (x ak ) Integrate along 1. and let R ! 1 and " ! 0+ . we obtain that X m 1 I i Qm = 0: j=1 k=1. this gives Z 1 f (z) dz Qm R ! 0: R k=1 (z jak j) Rm 1 Using the Residue Theorem and letting R ! 1 and " ! 0+ . note that the the sums of the residues at a simple pole at zj = aj . and let R ! 1. 1 j m.Integrate along the union of line segments on real axis denoted by L. and let " ! 0+ . 2. we have Z 1 1 PV dx = 0: 1 (x a1 ) (x a2 ) (x am ) 115 . and identifying the real part. are real.k6=j (aj ak ) Now. : : : m. za 1 jzja 1 ei(a 1) arg z 1 Res . Residue at a simple pole at z1 = 1.5 (b = 3) γ5 γ4 γ6 z2 γ7 γ8 γ2 γ1 z1 γ3 R z3 Set Z 1 xa 1 I = PV dx 1 xb 1 and integrate za 1 jzja 1 ei(a 1) arg z f (z) = b = z 1 jzjb eib arg z 1 along the contour in Figure VII. 0 < a < b: 1 xb 1 b b Hint.5 Show that Z 1 xa 1 a PV dx = cot . where =2 < arg z < 3 =2.6. We make a branchcut for z a 1 along the negative imaginary axis. indented at z = 1 and z = e2 i=b . where by Rule 3. For b > 1 one can integrate a branch of z a 1 = z b 1 around a sector of aperture 2 =b.6. 1 = = : zb 1 bz b 1 b z=1 116 .VII. Solution VII.5.6. and let R ! 1 and " ! 0+ . where by Rule 3. za 1 2 i=b jzja 1 i(a 1) arg z e e2 i=b 2 i(a 1)=b e Res . and let " ! 0+ . and let R ! 1. and let R ! 1 and " ! 0+ .e = = : zb 1 bz b 1 b z=e2 i=b Integrate along 1 and 3. This gives Z Z f (z) dz+ f (z) dz = 1 3 Z Z jzja 1 i(a 1) arg z e jzja 1 i(a 1) arg z e z = xe0i = dz + dz = = 1 jzjb eib arg z 1 jzjb eib arg z 1 3 dz = dx Z 1 " Z R a 1 Z 1 a 1 xa 1 x x = b dx + b dx ! dx = I: " x 1 1+" x 1 0 xb 1 Integrate along 2. This gives when 0 < a < b Z Ra 1 2 R 2 f (z) dz ! 0: 4 Rb 1 b bRb a Integrate along 5 and 7.Residue at a simple pole at z2 = e2 i=b . This gives Z Z f (z) dz+ f (z) dz = 5 7 Z Z jzja 1 i(a 1) arg z e jzja 1 i(a 1) arg z e z = xe2 i=b dz + = dz = 5 jzjb eib arg z 1 b ib arg z 7 jzj e 1 dz = e2 i=b dx Z 1+" a 1 2 i(a 1)=b Z " a 1 2 i(a 1)=b x e 2 i=b x e = b e dx + b e2 i=b dx! R x 1 1 " x 1 Z 0 a 1 2 i(a 1)=b Z 1 a 1 x e 2 i=b 2 i=b 2 i(a 1)=b x ! b e dx = e e dx = 1 x 1 0 xb 1 = e2 i=b e2 i(a 1)=b I 117 . This gives Z 1 i f (z) dz ! i( 0) = : 2 b b Integrate along 4. .e. we obtain that i e2 i=b 2 i(a 1)=b e + 0 e2 I i=b 2 i(a 1)=b e I i = 2 i 0: b b Hence solving for I. and let " ! 0+ .Integrate along 6. b 1 e2 i=b e2 i(a 1)=b b 1 e2 ia=b b sin b b b i. and let " ! 0+ . a i 1 + e2 i=b 2 i(a 1)=b e i 1 + e2 ia=b cos b a I= = = a = cot . 0 < a < b: 1 xb 1 b b 118 . This gives Z e2 i=b 2 i(a 1)=b e e2 i=b 2 i(a 1)=b e f (z) dz ! i( 0) = i : 6 b b Integrate along 8. This gives when 0 < a < b Z "a 1 2 " 2 "a f (z) dz ! 0: 8 1 "b b b Using the Residue Theorem and letting R ! 1 and " ! 0+ . Z 1 xa 1 a PV dx = cot . VII.6.6 By integrating a branch of (log z) = z b 1 around an indented sec- tor of aperture 2 =b, show that for b > 1, Z 1 2 Z 1 log x 1 dx = ; PV dx = cot ( =b) : 0 xb 1 b2 sin2 ( =b) 0 xb 1 b Solution VII.6.6 (b = 3) γ5 γ4 γ6 z2 γ7 γ8 γ2 γ1 z1 γ3 R z3 Set Z 1 Z 1 log x 1 I= dx J = PV dx 0 xb 1 0 xb 1 and integrate log z log jzj + i arg z f (z) = = z b 1 jzjb eib arg z 1 along the contour in Figure VII.6.5. We make a branchcut for log z along negative imaginary axis, where =2 < arg z < 3 =2. Residue at a simple pole at z1 = 1, where by Rule 3, log z log jzj + i arg z Res b ;1 = = 0: z 1 bz b 1 z=1 Residue at a simple pole at z2 = e2 i=b , where by Rule 3, 119 log z 2 i=b log jzj + i arg z 2 ie2 i=b Res ;e = = : zb 1 bz b 1 z=e2 i=b b2 Integrate along 1 and 3, and let R ! 1 and " ! 0+ . This gives Z Z f (z) dz+ f (z) dz = 1 3 Z Z log jzj + i arg z z = xe0i log jzj + i arg z = dz + dz = = 1 jzjb eib arg z 1 b ib arg z 3 jzj e 1 dz = dx Z 1 " Z R Z 1 logx logx logx = b dx + b dx ! dx = I: " x 1 1+" x 1 0 xb 1 Integrate along 2, and let " ! 0+ . This gives Z f (z) dz ! i ( 0) (0) = 0: 2 Integrate along 4, and let R ! 1. This gives when b > 1 q Z log2 R + 2b 2 2 R 2 log R f (z) dz b ! 0: 4 R 1 b bRb 1 Integrate along 5 and 7, and let R ! 1 and " ! 0+ . This gives Z Z f (z) dz+ f (z) dz = 5 7 Z Z log jzj + i arg z log jzj + i arg z z = xe2 i=b = dz + dz = = b ib arg z 5 jzj e 1 b ib arg z 7 jzj e 1 dz = e2 i=b dx Z 1+" Z " logx + 2 i=b 2 i=b logx + 2 i=b 2 i=b = b e dx + e dx ! R x 1 1 " xb 1 Z 0 Z 1 Z 1 logx + 2 i=b 2 i=b 2 i=b logx 2 i 2 i=b 1 ! b e dx = e b dx e dx = 1 x 1 0 x 1 b 0 xb 1 2 i 2 = e2 i=b I e i=b J: b 120 Integrate along 6, and let " ! 0+ . This gives Z 2 ie2 i=b 2 2 e2 i=b f (z) dz ! i( 0) = : 6 b2 b2 Integrate along and let " ! 0+ . This gives when b > 1 8, q Z log2 " + 2b 2 2 " 2 " jlog "j f (z) dz b ! 0: 8 1 " b b Using the Residue Theorem and letting R ! 1 and " ! 0+ , we obtain that 2 i 2 2 2 e2 i=b I +0 e2 i=b I e i=b J+ = 2 i 0: b b2 2 i=b Multiplying with e , this yields 2 i 2 i=b2 2 e J + 2 = 0: 1 I b b Separating real and imaginary parts, we get simultaneous equations 2 2 cos b 1 I + 2b2 = 0 sin 2b I 2b J = 0; i.e., Z 1 2 Z 1 log x 1 I= dx = ; J = PV dx = cot ( =b) : 0 xb 1 b2 sin2 ( =b) 0 xb 1 b 121 VII.6.7 Suppose that P (z) and Q (z) are polynomials, deg Q (z) deg P (z)+2, and the zeros of Q (z) on the real axis are all simple. Show that Z 1 X X P (z) P (z) P (z) PV dx = 2 i Res ; zj + i Res ; xk ; 1 Q (z) Q (z) Q (z) summed over the poles zj of P (z) =Q (z) in the open upper half- plane and the poles xk of P (z) =Q (z) on the real axis. Remark. In other words, the principal value of the integral is 2 i times the sum of the residues in the upper half-plane, where we count the poles on the real axis as being half in and half out of the upper half-plane. Solution VII.6.7 γR zj - 1 z3 zj z2 z1 γ1 γ2 γ m - 1 γm a1 a2 a m - 1 am -R R Use a semicircular contour indented at ak = xk with small semicircles of radii ", see Figure VII.6.7. Get Z P (z) P (z) P (z) ! i( 0) Res ; xk = i Res ; xk k Q (z) Q (z) Q (z) The condition deg Q deg P + 2 guarantees that the integral converges at 1: Otherwise we would have to use a residue at 1 (in this case deg Q = deg P + 1). 122 Use the Residue Theorem, take limit, and the result follows Set Z 1 P (x) I = PV dx 1 Q (x) and integrate P (z) f (z) = Q (z) along the contour in Figure VII.6.7. The sum of the residues at the simple poles ak = xk ; 1 k m on the real axis is X m P (z) Res ; xk k=1 Q (z) The sum of the residues at the simple poles zk ; 1 k j inside the integra- tion contour is j X P (z) Res ; zj k=1 Q (z) Integrate along segments on real axis, and let R ! 1 and " ! 0+ . This gives Z Z a1 " X1 Z ak+1 " m Z R f (z) dz = f (z) dz+ f (z) dz+ f (z) dz = R k=1 ak +" am +" Z a1 " X1 Z ak+1 " m Z R P (z) P (z) P (z) = dz+ dz+ dz = R Q (z) k=1 ak +" Q (z) am +" Q (z) z = x = = dz = dx Z a1 " X1 Z ak+1 " P (x) m Z R P (x) P (x) = dx+ dx+ dx ! R Q (x) k=1 ak +" Q (x) am +" Q (x) Z 1 P (x) ! dx = I: 1 Q (x) 123 Integrate along R, where we have deg Q (z) deg P (z) + 2 and let R ! 1. This gives Z Rdeg P f (z) dz R ! 0: R Rdeg Q R Integrate along 1; 2; : : : m, and let " ! 0+ . This gives m Z X X m P (z) f (z) dz ! i( 0) Res ; xk : k=1 k k=1 Q (z) Using the Residue Theorem and letting R ! 1 and " ! 0+ , we obtain that j ! Xm P (z) X P (z) I i Res ; xk =2 i Res ; zk ; k=1 Q (z) k=1 Q (z) i.e., Z 1 Xm Xm P (x) P (z) P (z) PV dx = 2 i Res ; zj + i Res ; xk : 1 Q (x) j=1 Q (z) k=1 Q (z) 124 VII.7.1 1 2 3 P L K LLL Show that Z 1 jsin xj dx = +1: 0 x Hint: Show that the area under the mth arc of jsin xj =x is 1=m. Solution VII.7.1 π 2π 3π 4π 5π For x in the interval (m 1) + 4 x m 4 , we have 1 jsin xj sin =p ; 4 2 and 1 1 : x m We estimate the integrand jsin xj 1 1 p : x 2m We have Z m =4 jsin xj 1 1 C dx p = : (m 1) + 4 x 2m m Hence 125 Z m m Z X k =4 jsin xj jsin xj 1 1 dx dx = C 1 + + : : : + ! +1 0 x k=1 (k 1) + 4 x 2 m as m ! 1. We have used the fact that the harmonic series 1 + 12 + 13 + 14 + + m1 tends R m+1 dx to in…nity as as m ! 1, by taking the oversum for the integral 1 x , Z m+1 1 1 1 dx 1+ + + + = log (m + 1) ! +1 2 3 m 1 x as m ! 1. 126 VII.7.2 Show that Z R x3 sin x lim dx = : R!1 R (x2 + 1)2 2e Solution VII.7.2 γ2 z1 -R γ1 R z2 Set Z 1 x3 sin x I= dx 1 (x2 + 1)2 and integrate z 3 eiz z 3 eiz f (z) = = (z 2 + 1)2 (z i)2 (z + i)2 along the contour in Figure VII.7.2. Residue at a double pole at z1 = i, where by Rule 2, z 3 eiz d z 3 eiz 1 Res 2 ; i = lim 2 = : (z 2 + 1) z!i dz (z + i) 4e Integrate along 1 , and let R ! 1. This gives 127 Z f (z) dz = 1 Z Z R z 3 eiz z = x x3 eix = dz = = 2 dx ! 1 (z 2 + 1)2 dz = dx 2 R (x + 1) Z 1 Z 1 3 Z 1 3 x3 eix x cos x x sin x ! 2 dx = 2 dx + i 2 dx = 1 (x2 + 1) 2 1 (x + 1) 2 1 (x + 1) Z 1 3 x cos x = 2 dx + iI: 2 1 (x + 1) Integrate along 2 , and let R ! 1. Use Jordan’s Lemma this gives Z Z R3 iz R3 f (z) dz e jdzj < ! 0: 2 (R2 1)2 2 (R2 1)2 R Using the Residue Theorem and letting R ! 1, we obtain that Z 1 3 x cos x 1 2 dx + iI = 2 i ; 2 1 (x + 1) 4e and hence, setting imaginary parts equal, i.e., Z 1 3 x sin x 2 dx = : 2 1 (x + 1) 2e 128 VII.7.3 Evaluate the limits Z R x sin (ax) lim dx; 1 < a < 1: R!1 R x2 + 1 Show that they do not depend continuously on the parameter a. Solution VII.7.3a z1 γ1 -R R z2 γ2 Case 1 a < 0: Set Z 1 x sin (ax) I= dx 1 x2 + 1 and integrate zeiaz zeiaz f (z) = = z2 + 1 (z i) (z + i) along the contour in Figure VII.7.3a. Residue at a simple pole at z2 = i, where by Rule 3, zeiaz zeiaz ea Res ; i = = : z2 + 1 2z z= i 2 Integrate along 1, and let R ! 1. This gives 129 Z Z Z R zeiaz z = xe0i xeiax f (z) dz = dz = = dx ! 1 1 z2 + 1 dz = dx R x2 + 1 Z 1 Z 1 Z 1 xeiax x cos (ax) x sin (ax) ! 2 dx = dx + i dx = 1 x +1 1 x2 + 1 2 1 x +1 Z 1 x cos (ax) = dx + iI: 1 x2 + 1 Integrate along 2 , and let R ! 1. Use Jordan’s Lemma this gives Z Z R R f (z) dz 2 eiaz jdzj < 2 ! 0: 2 R 1 2 R 1 R Using the Residue Theorem and letting R ! 1, we obtain that Z 1 x cos (ax) ea dx + iI + 0 = 2 i ; 1 x2 + 1 2 and hence, setting imaginary parts equal, I= ea : Case 2 a = 0: Z R Z R x sin (ax) 0 lim dx = lim =0 R!1 R x2 + 1 R!1 R x2 +1 Case 3 a > 0: 130 VII.7.3b γ4 z1 -R γ3 R z2 Set Z 1 x sin (ax) J= dx 1 x2 + 1 and integrate zeiaz zeiaz f (z) = = z2 + 1 (z i) (z + i) along the contour in Figure VII.7.3b Residue at a simple pole at z1 = i, where by Rule 3, zeiaz zeiaz e a Res ; i = = : z2 + 1 2z z=i 2 Integrate along 3, and let R ! 1. This gives Z Z Z R zeiaz z = xe0i xeiax f (z) dz = dz = = dx ! 3 3 z2 + 1 dz = dx R x2 + 1 Z 1 Z 1 Z 1 xeiax x cos (ax) x sin (ax) ! 2 dx = dx + i dx = 1 x +1 1 x2 + 1 2 1 x +1 Z 1 x cos (ax) = dx + iJ: 1 x2 + 1 Integrate along 4, and let R ! 1. Use Jordan’s Lemma this gives 131 Z Z R R f (z) dz eiaz jdzj < ! 0: 4 R2 1 4 R2 1 R Using the Residue Theorem and letting R ! 1, we obtain that Z 1 x cos (ax) e a dx + iI + 0 = 2 i ; 1 x2 + 1 2 and hence, setting imaginary parts equal, I = e a: and we have the solution 8 Z 1 < ea ; a < 0; x sin (ax) dx = 0 a=0 x2 + 1 : 1 e a; a > 0: Not continuous at a = 0 becauce sin 0 = 0. 132 Solution VII.VII. R!1 0 Z R lim xa 1 sin xdx = (a) sin ( a=2) . show that Z R lim xa 1 cos xdx = (a) cos ( a=2) . 0 < a < 1. integrate by parts.7. and integrate 133 .4 By integrating z a 1 eiz around the boundary of a domain in the …rst quadrat bounded by the real and imaginary axes and a quater- circle. R!1 0 where (a) is the gamma function de…ned by Z 1 (a) = ta 1 e t dt: 0 Remark: The formula for the sine integral holds also for 1 < a < 0.7.4 γ3 γ2 γ1 R Set Z R Z R a 1 I = lim x cos xdx J = lim xa 1 sin xdx R!1 0 R!1 0 where 0 < a < 1. To see this. 0 < a < 1. This gives Z f (z) dz = 1 Z Z R z = xe0i a 1 i(a 1) arg z iz = jzj e e dz = = xa 1 eix dx ! dz = dx 0 1 Z 1 Z 1 Z 1 a 1 ix a 1 ! x e dx = x cos xdx + i xa 1 sin xdx = I + iJ: 0 0 0 Integrate along 2 . Mathematics Handbook page 278 134 . Use Jordan’s Lemma this gives Z Z a 1 f (z) dz R eiz jdzj < Ra 1 ! 0: 2 2 R1 a Integrate along 3. This gives Z f (z) dz = 3 Z Z R z = xe i=2 a 1 i(a 1) arg z iz = jzj e e dz = i=2 = xa 1 e ia=2 e x dx ! dz = e dx 0 3 Z 1 Z 1 ! xa 1 e ia=2 e x dx = e ia=2 xa 1 e x dx = e ia=2 (a) = 0 0 = (a) cos( a=2) (a) sin ( a=2) : Using the Residue Theorem and letting R ! 1. and let R ! 1.7. and let R ! 1 and " ! 0+ . f (z) = z a 1 eiz = jzja 1 i(a 1) arg z iz e e along the contour in Figure VII.4 Integrate along 1 . and let R ! 1 and " ! 0+ . we obtain that I + iJ (a) cos( a=2) (a) sin ( a=2) = 0: We equate real and imaginary parts I= (a) cos( a=2) J= (a) sin ( a=2) : Formulas for the Gamma function. (z + 1) = z (z) Z R Z R a 1 a 1 R x cos xdx = x sin x 0 (a 1) xa 2 sin xdx 0 0 Z R Z R a 2 a 1 R (a 1) x sin xdx = x sin x 0 xa 1 cos xdx 0 0 Z R Z R a 2 a 1 (a 1) x sin xdx = R sin R xa 1 cos xdx 0 0 Z R Z R a 2 Ra 1 sin R 1 x sin xdx = xa 1 cos xdx 0 a 1 a 1 0 Z R sin R 1 xa 2 sin xdx = (a) cos ( a=2) 0 (1 a) R1 a a 1 Z R (a + 1) lim xa 1 sin xdx = cos (a + 1) = R!1 0 a 2 = (a) [cos ( a=2) cos ( =2) sin ( a=2) sin ( =2)] = = (a) sin ( a=2) : 135 . 5 Show that Z R Z R p 2 2 lim sin x dx = lim cos x dx = p . R!1 0 R!1 0 2 2 by integrating eiz around the boundary of the pie-slice domain determined by 0 < arg z < =4 and jzj < R. and integrate 2 f (z) = eiz along the contour in Figure VII. Solution VII.VII.7.5 Integrate along 1 . This gives 136 . and let R ! 1. Remark.7. The identities can also be deduced from the preceding exercise by changing variable.7. These improper integrals are called the Fresnel integrals.5 γ3 γ2 γ1 R Set Z R Z R 2 I = lim sin x dx J = lim cos x2 dx R!1 0 R!1 0 where 1 < a < 0. Z Z Z R z = xe0i iz 2 2 f (z) dz = e dz = = eix dx ! dz = dx 0 1 Z 11 Z 1 Z 1 2 ! eix dx = cos x2 dx + i sin x2 dx = J + iI: 0 0 0 Integrate along 2 as we can parametrize as z = Reit where 0 t =4. this gives Z Z 2 z = Reit f (z) dz = eiz dz = = 2 2 dz = iReit dt Z =4 Z =4 2 e2it 2 = eiR iReit dt R eiR (cos(2t)+i sin(2t)) dt = 0 0 Z =4 Z =4 h i =4 R2 sin 2t 4R2 t= 4R2 t= =R e dt R e ds = e = 0 0 4R 0 R2 = 1 e ! 0: 4R 4R i=4 Integrate along 3 as we can parametrize as z = xe where 0 x R. Use that 4x= sin 2x for 0 x =4. This gives Z Z Z R iz 2 z = xe i=4 x2 i=4 f (z) dz = e dz = = e e dx ! dz = e i=4 dx 0 3 3 Z 1 Z 2 1 + i 1 x2 ! e xe i=4 dx = p e dx = 0 2 Z 1 0 Z 1 1 x2 1 x2 = p e dx i p e dx: 2 0 2 0 We will use integral (41) from Mathematics Handbook page 177 with a = 1. Z 1 r ax2 1 e dx = : 0 2 a and hence 137 . and let R ! 1. and let R ! 1. Z Z 1 Z 1 p p 1 x2 1 x2 f (z) dz = p e dx ip e dx = p i p : 3 2 0 2 0 2 2 2 2 Using the Residue Theorem and letting R ! 1. we obtain that p p J + iI + 0 p i p = 0: 2 2 2 2 We equate real and imaginary parts p p I= p J= p : 2 2 2 2 138 . one for each branch of the square root. (z 2 + 1)2 z4 hence 1 Res . 1. z2 1 z 1 z2 z z 2 z hence z Res . 1 = 0: (z 2 + 1)2 c) z3 + 1 z (z 2 1) + z + 1 z 1 2 = 2 =z+ 2 + 2 = z 1 z 1 z 1 z 1 1 1 1 1 1 1 1 =z+ 1 + 2 =z+ 1 + 2 + 4 + ::: + 2 .1 1 2 3 P L K Evaluate the residue at 1 of the following functions. There are two possibilities for (f).1 = 1: z2 1 139 . a) z 1 1 1 1 1 = 1 = 1+ + 4 + ::: . 3 (a) z2z 1 (c) zz2 +11 z n e1=z . z 1 z2 z 1 z z z z 1 hence z3 + 1 Res .1 = 1: z2 1 b) 1 1 =O as z ! 1. : : : (e) q 1 9 (b) (z2 +1)2 (d) zz6 +11 (f) zz ab Note.VII.8. n = 0. 1 = 0: z6 1 e) 1 1 1 1 1 z n e1=z = z n 1 + + 2 + 3 + 3 + + ::: = z 2!z 3!z 3!z 4!z 4 1 1 1 = zn + zn 1 + zn 2 + zn 3 + zn 4 +: 2! 3! 4! 1 We can see that the coe¢ cient of z is 1 . : : : : (n + 1)! f) r s r z a 1 a=z 1 aw = = .d) z9 + 1 z 3 (z 6 1) + z 3 + 1 3 z3 + 1 = = z + . 1 = . 1. n = 0. z6 1 z6 1 z6 1 hence z9 + 1 Res . z b 1 b=z 1 bw analytic at w = 0: 1 1 aw 1=2 a(1 bw)+b(1 aw) 1 p1 coe¢ cient of w is g 0 (0) = 2 1 bw (1 bw)2 = 2 1 (b a) = w=0 b a 2 : 140 . (n + 1)! hence 1 Res z n e1=z . 2 γ1 γ2 γ4 γ3 Set Z 1 x4 I= p dx 0 x (1 x) and integrate z4 z4 f (z) = p =p p z (1 z) jzjei arg z=2 j1 zjei arg0 (1 z)=2 along the contour in Figure VII.8. 141 .2 1 2 3 P L K Show by integrating around the dogbone contour that Z 1 x4 35 p dx = : 0 x (1 x) 128 Solution VII.7.8.8.2. 05 = w2 w w2 1 1 1 w w 1 1 1 d4 1 = Res p = lim p = w5 w 1 w!0 4! dz 4 w 1 35 35 35 35 = 9=2 = 9=2 i9 arg(w 1)=2 = e i=2 = i 128 (w 1) 128 (jw 1j) e 128 128 142 . VIII. 2 3 1 1 1 1 Res [f (z) .2 arg z arg0 z y y π π π 0 0 0 π π π 0 0 0 -2 -1 1 2 x -2 -1 1 2 x −π −π −π 0 0 0 π π π 2π 2π 2π arg z arg0 (1 z) y y π π π 0 0 0 2π 2π 2π 2π π π -2 -1 1 2 x -2 -1 1 2 x −π −π −π 0 0 0 0 0 0 0 π π 4 The argument for the function f (z) = p pz jzjei arg z=2 j1 zjei arg0 (1 z)=2 y 3π/2 3π/2 3π/2 π π/2 π/2 -2 -1 1 2 x −π/2 −π/2 −π/2 0 π/2 π/2 Residue at a simple pole at z1 = 1 .8.13 and Rule 1. 1] = Res f .0 = Res 4 q w4 . where by result from Exercise VII.8. This gives Z "4 f (z) dz p 2 " 2 "9=2 ! 0: 4 " (1 ") Using the Residue Theorem and letting R ! 1 and " ! 0+ . 128 i.Integrate along 1. This gives Z f (z) dz = 1 Z 1 " Z 1 " z4 x4 = p p dz = p p dx = " jzjei arg z=2 j1 zjei arg0 (1 z)=2 " jxjei 0=2 j1 xjei 0=2 Z 1 " Z 1 x4 x4 = p dx ! p dx = I: " x (1 x) 0 x (1 x) Integrate along 2. and let " ! 0+ .. This gives Z f (z) dz = 3 Z " Z " z4 1 = p p dz = p p dx = 1 " jzjei arg z=2 j1 zjei arg0 (1 z)=2 1 " jxjei 0=2 j1 xjei2 =2 Z 1 " Z 1 x4 x4 = p dx ! p dx = I: " x (1 x) 0 x (1 x) Integrate along 4. and let " ! 0+ . Z 1 x4 35 p dx = : 0 x (1 x) 128 143 . and let " ! 0+ . This gives Z (1 + ")4 f (z) dz p 2 " 2 "9=2 ! 0: 2 (1 + ") " Integrate along 3. and let " ! 0+ .e. we obtain that 35 I +0+I +0=2 i i . and evaluate it. Thus R1 0 f (x) dx < 1 . as x ! 1 R1 R" " g (x) dx < 1 . a < n + 1. 0 < a < n + 1. a > 0. a n < 1 . 0 f (x) dx < 1 . Show that xi Im a begränsad.8. as x ! 0 Rg(x) " R" 0 g (x) dx < 1 . 1 a<1. 0 f (x) dx < 1 . The integral is generalized in 0 and 1 " 2 R. x = 0 1 g (x) = (1 x) 1 a f (x) n g(x) = xxa ! 1. positive or negative. x = 0 g (x) = xa1 n f (x) 1 = (1 x) 1 a ! 1. Determine for which complex values of the parameter a the integral Z 1 xn dx a 0 x (1 x)1 a converges. ( Obs fel i lösningen argumentet) x 2 R+ 1 x 2 R+ a = Re a + i Im a xa = xRe a+i Im a = xRe a xi Im a The integral is generalized in 0 and 1 " 2 R. Solution.7. set 2 [arg x] xi Im a = ei Im a(log x+i ) = ei Im a log x = 1 On principal arg 0 Determination for which complex values of the parameter a the integral con- verges 144 .3 1 2 3 P L K Fix an integer n. 8.3 γ1 γ2 γ4 γ3 Set Z 1 xn I= a dx 0 xa (1 x)1 and integrate zn zn f (z) = = z a (1 z)1 a jzja eai arg z j1 zj1 a e(1 a)i arg0 (1 z) along the contour in Figure VII. VII. 145 .3.8. VIII.8. This gives 146 .13 and Rule 1. where by result from Exercise VII.0 = lim = ei =2 = i w(1 + w2 ) w!0 w2 + 1 Integrate along 1.0 = w2 w w2 1 1 1 a wa 1 w p p 1jei arg(w 1)=2 2 w2 1 jw2 = Res . " # 1 1 1 1 wn Res [f (z) . 1] = Res f .3 arg z arg0 z y y π π π 0 0 0 π π π 0 0 0 -2 -1 1 2 x -2 -1 1 2 x −π −π −π 0 0 0 π π π 2π 2π 2π arg z arg0 (1 z) y y π π π 0 0 0 2π 2π 2π 2π π π -2 -1 1 2 x -2 -1 1 2 x −π −π −π 0 0 0 0 0 0 0 π π zn The argument for the function f (z) = jzja eai arg z j1 zj 1 a (1 a)i arg (1 z) e 0 y 2π − απ 2π − απ 2π − απ 2π − 2απ π − απ π − απ -2 -1 1 2 x −απ −απ −απ 0 π − απ π − απ Residue at a simple pole at z3 = 1 .0 = Res . and let R ! 1 and " ! 0+ .8. This gives Z "n f (z) dz 1 a 2 " 2 "n+1 a ! 0: 4 "a (1 ") Using the Residue Theorem and letting R ! 1 and " ! 0+ . and let " ! 0+ . and let " ! 0+ . Z f (z) dz = 1 Z 1 " Z 1 " zn zn = = dx = " jzja eai arg z j1 zj1 a e(1 a)i arg0 (1 z) " jzja eai 0 j1 zj1 a e(1 a)i 0 Z 1 " zn = dx ! " z a (1 z)1 a Z 1 zn ! dx = I: a 0 z (1 z)1 a Integrate along 2. we obtain that (a 1) (a n) I + 0 e2 ia I +0=2 i ( 1)n : n! 147 . This gives Z f (z) dz = Z 3 " Z " zn zn = = dx = 1 " jzja eai arg z j1 zj1 a e(1 a)i arg0 (1 z) a ai 0 1 " jzj e j1 zj1 a e(1 a)i 2 Z 1 " 2 ia zn = e dx ! " z a (1 z)1 a Z 1 2 ia zn ! e dx = e2 ia I: a 0 z (1 z)1 a Integrate along 4. and let R ! 1 and " ! 0+ . This gives Z (1 + ")4 2 f (z) dz 2 " ! 0: 4 (1 + ")a (1 (1 + "))1 a " a Integrate along 3. : : : 148 . 1 e2 ia e ia e ia sin ( a) n! i. 2.. 1 a dx = ( 1)n . 1. 0 xa (1 x) sin ( a) n! n = 0. Z 1 xn (a 1) (a n) 0 < a < 1. we obtain that ( 1)n 2 i (a 1) (a n) ( 1)n 2 i (a 1)n!(a n) (a 1) (a n) I= n! = = ( 1)n .Solve for I.e. VII.4 γ1 γ2 γ4 γ3 Set Z 1 p 1 x2 I= dx 1 1 + x2 and integrate p p p 1 z2 j1 + zjei arg (1+z)=2 j1 zjei arg0 (1 z)=2 f (z) = 2 = 2 z +1 z +1 along the contour in Figure VII. 149 .4 Show that Z p 1 1 x2 p dx = 2 1 : 1 1 + x2 Solution VII.8.8.8.4. 8.4 arg z arg0 z y y π π π 0 0 0 π π π 0 0 0 -2 -1 1 2 x -2 -1 1 2 x −π −π −π 0 0 0 π π π 2π 2π 2π arg (1 + z) arg0 (1 z) y y π π 0 0 0 0 2π 2π 2π 2π π π -2 -1 1 2 x -2 -1 1 2 x −π −π 0 0 0 0 0 0 0 0 π π 4 The argument for the function f (z) = p pz jzjei arg z=2 j1 zjei arg0 (1 z)=2 y 3π/2 3π/2 π π π/2 π/2 -2 -1 1 2 x −π/2 −π/2 0 0 π/2 π/2 Residue at a simple pole at z1 = i.i = (z i) (z + i) p p j1 + zjei arg(1+z)=2 j1 zjei arg(1 z)=2 = = 2z z=i p p p p j1 + ijei arg(1+i)=2 j1 ijei arg(1 i)=2 j1 + ije(i =4)=2 j1 ije(i7 =4)=2 = 2i pp pp 2i i p p 2 2e 2( 1) 2 i = = = =p 2i 2i 2i 2 150 . "p p # j1 + zjei arg(1+z)=2 j1 zjei arg(1 z)=2 Res . VIII. where by Rule 3. 1] = Res f .0 = Res 4 .13 and Rule 1.Residue at a simple pole at z2 = i. "p p # j1 + zjei arg(1+z)=2 j1 zjei arg(1 z)=2 Res .0 = lim = ei =2 = i w(1 + w2 ) w!0 w2 + 1 Integrate along 1. where by Rule 3. 05 = w2 w w2 1+ w12 p p 1jei arg(w 1)=2 2 w2 1 jw2 = Res . 2 q 3 1 1 1 1 1 w2 Res [f (z) . This gives 2. i = z2 + 1 p p j1 + zjei arg(1+z)=2 j1 zjei arg(1 z)=2 = = 2z z= i p i arg(1 i)=2 p i arg(1+i)=2 p ( i =4)=2 p j1 ije j1 + ije j1 ije j1 + ije(i =4)=2 = = = 2i pp pp 2i p 2 2 2 1 = = = p i: 2i 2i 2 Residue at a simple pole at z3 = 1 . where by result from Exercise VII. This gives Z f (z) dz = Z 1 " p p p p 1 Z 1 " j1 + zjei arg (1+z)=2 j1 zjei arg0 (1 z)=2 j1 + xjei 0=2 j1 xjei2 =2 = 2 dz = dx ! 1+" z +1 1+" 1 + x2 Z 1 " p 1 x2 ! dx = I: 1+" 1 + x2 Integrate along and let " ! 0+ . q Z 1 (1 ")2 f (z) dz 2 " 2 "3=2 ! 0: 2 1 + (1 ")2 151 .8. and let " ! 0+ . This gives Z f (z) dz = Z 1+" p p p p 3 Z 1 " j1 + zjei arg (1+z)=2 j1 zjei arg0 (1 z)=2 j1 + xjei 0=2 j1 xjei 0=2 = 2 dz = dx ! 1 " z +1 1+" x2 + 1 Z 1 " p 1 x2 ! dx = I: 1+" 1 + x2 Integrate along 4.. Z p 1 1 x2 p dx = 2 1 : 1 1 + x2 152 . and let " ! 0+ .Integrate along 3. and let " ! 0+ . we obtain that 1 1 I +0 I +0=2 i p i+ p i i 2 2 i.e. This gives q Z 1 (1 ")2 f (z) dz 2 " 2 "3=2 ! 0: 4 1 + (1 ")2 Using the Residue Theorem and letting R ! 1 and " ! 0+ . VII. 1] jzj=R H H P Since = . we get 2 i Res = 0.5 1 2 3 P L K Show that the sum of the residues of a rational function on the extended complex plane is equal to zero. jzj=R I f (z) dz = 2 i Res [f. 153 . Solution I X f (z) dz = 2 i Residues in …nite plane.8. we have that z2 1 a 1 = lim zf (z) = lim 2 = lim = 1: z!1 z!1 z + 1 z!1 1 + 1=z 2 hence z Res . i = = : z2 +1 2z z= i 2 For residue at 1. and check that the sum of the residues is zero. z z 1 Res .6 1 2 3 P L K Find the residue of z= (z 2 + 1) at each pole in the extended complex plane.8. z a 1 1 (Since z 2 +1 = z +O z2 as z ! 1). z z 1 Res . where by Rule 3.i = = : z2 +1 2z z=i 2 Residue at a simple pole at z1 = i. 154 . where by Rule 3. z2 +1 and the sum of the residues is zero in the extended complex plane.VII.1 = a 1 = 1. Solution Residue at a simple pole at z1 = i. a 1 1 f (z) = z +O z2 as z ! 1.7 1 2 3 P L K Find the sum of the residues of (3z 4 + 2z + 1) = (8z 5 + 5z 2 + 2) at its poles in the (…nite) complex plane. 1]. thus the sum of residues in …nite complex plane is 3=8.VII. 1] = : 8 We have that the sum of the residues in the complex …nite plane is Res [f (z) . so again 155 .8. Solution For residue at 1. we have that 3z 5 + 2z 2 + z 3 + 2=z 3 + 1=z 4 3 a 1 = lim zf (z) = lim 5 2 = lim 3 4 = . z!1 z!1 8z + 5z + z z!1 8 + 5=z + 1=z 8 hence 3 Res [f. e wj = e2 ij=n where 0 j n 1. Let wj be an nth root of unity. where by Rule 3 and that wjn = 1 zk z k+1 wjk+1 wjk+1 Res [f. then 1 X k+1 n 1 1 1 1 wn S= wj = 1 + w + w2 + + wn 1 = = 0. we have zk zk n f (z) = = 1 = zk n 1+z n +z 2n + : : : = zk n +z k 2n +z k 3n +: : : : zn 1 1 zn Then Res [f. and otherwise Res [f. Then f (z) has simple poles at w0 . Solution. Find the residue of z k = (z n 1) at each …nite pole. At 1. n j=0 n n 1 w unless w = 1. Residue at a simple pole at zj = wj . wj ] = = = = : nz n 1 z=wj nz n z=wj nwjn n Let S be the sum of residues. : : : . 156 . Find the residue of z k = (z n 1) at 1 by expand- ing 1= (z n 1) in a Laurent series. 1] = 0. i.VII. thus k = mn 1 for some integer m 1. e n (k+1) = 1 . thus k = mn 1 for some integer m 1.8 1 2 3 P L K Fix n 1 and k 0. then wjk+1 = wj = 1 for every j and w = 1 . 2 i If w = 1. and verify that the sum of all the residues are zero. wn 1 . Thus we have that the sum of the residues in the complex plane are zero. zk Let f (z) = zn 1 . k+1 n = m.8. 1] = 1 if k mn = 1. 1] : z!1 157 . then Res [f (z) . we have a 1 1 f (z) = a0 + +O : z z2 Then a 1 1 1 z (f (z) f (1)) = z a0 + +O a0 =a 1 +O !a 1 z z2 z as z ! 1.VII. From this follows that lim z (f (z) f (1)) = a 1 = Res [f (z) .9 1 2 3 P L K Show that f (z) is analytic at 1. 1] = lim z (f (z) f (1)) : z!1 Solution Since f (z) is analaytic at 1.8. If D is an exterior domaina. 1 = f (z) f (1) 2 i @D z z IIII f( ) f( ) Res .VII. Cauchy integral formula for exterior domain.1 = lim = f (1) : z !1 z Or a 1 a 2 f ( ) = a0 + + 2 + ::: f( ) a0 1 = +O 2 z z Z Z 1 f( ) 1 f( ) d = f (z) + d = 2 i @D z 2 ijzj=R z Z 1 a0 = f (z) d = f (z) a0 : 2 i jzj=R z 158 .8. f (z) is analytic in D and att 1. z 2 D: 2 i @D z Solution. Suppose that f (z) is analytic on D [ @D and at 1.10 1 2 3 P L K Let D be an exterior domain. then Z 1 f( ) f( ) d = f (z) + Res . Show that Z 1 f( ) d = f ( ) f (1) . 1 x a : i . Im a < 0 . Im a > 0. This gives 159 . a2 < 0. we de…ne the principal value Z 1 Z R PV f (x) dx = lim PV f (x) dx: 1 R!1 R Show that 8 Z 1 < i . Integrate and let R ! 1. This gives Z R Z R dx dx R a1 a2 i = = [log (x a1 a2 i)]RR = log = R x a R x a1 a2 i R a1 a2 i R a1 a2 i = log + i arg (R a1 a2 i) i arg ( R a1 a2 i) ! R a1 a2 i !0+i 0 i = i : Case 2. This gives Z R Z R Z a1 " Z R dx dx dx dx = = + = [log jx a1 j]a1R " +[log jx a1 j]R a1 +" = R x a R x a1 R x a 1 a1 +" x a 1 R a1 = log " log jR + a1 j + log jR a1 j log " = log ! 0: R + a1 Case 3. Im a = 0. Im a = 0 . where a1 and a2 is …xed.11 1 2 3 P L K If f (z) is not integrable at 1.8. Im a < 0: Solution Set a = a1 + a2 i. 1 PV dx = 0. Integrate and let R ! 1 and " ! 0+ . Case 1. We compute the principal value for the following 3 cases. Integrate and let R ! 1.VII. a2 = 0. a2 > 0. Im a > 0 . Im a < 0: 160 . Z R Z R dx dx R a1 a2 i = = [log (x a1 a2 i)]RR = log = R x a R x a1 a2 i R a1 a2 i R a1 a2 i = log + i arg (R a1 a2 i) i arg ( R a1 a2 i) ! R a1 a2 i !0+i 2 i =i : We have that 8 Z 1 < i . 1 x a : i . Im a = 0. 1 PV dx = 0. Im a > 0. in initial case Res Q(x) . Use the preceding exercise. formula holds for By Exercise b. By limiticity. 215. 1 = 0. Show that Z 1 X X P (x) P (x) P (x) PV dx = 2 i Res .7 Solution.VII. and set x0 = 1. it hods form sums. 1 By VII. 1 Q (x) Q (x) Q (x) summed over the poles zj of P (z) =Q (z) in the open upper half- plane and summed over the xk : s including 1.) it holds if h z a i P (x) deg Q deg P + 2. xm of Q (z) on the real axis are all simple. See also Exercise 6. : : : .12 1 2 3 P L K Suppose that P (z) and Q (z) are polynomials such that the degree of Q (z) is strictly greater than the degree of P (z).8.7 (p. 161 . Hint. zk . Suppose that the zeros x1 .8.11. hence whenever deg Q deg P + 1. zj + i Res . thus 1 1 Res f . 1] = a1 = Res f .0 = a 1 w w2 We have that 1 1 Res [f (z) . Now we take the residue for f (1=w) dw=w2 at w = 0. 1] = a1 if f (z) = 11 aj z j .13 1 2 3 P L K Show that the analytic di¤erential f (z) dz transforms the change of variable w = 1=z to f (1=w) dw=w2 . then dz = 1 w2 dw. we have that 1 X aj X 1 1 1 1 j 2 f = = ( aj ) w = w w2 w2 1 wj 1 = a 4 w2 a 3w a 2 a 1w 1 a0 w 2 . and f (z) dz = f w1 w12 dw. We transform f (z) dz by the change of variable z = w1 . Solution P We have that Res [f (z) .VII. Show that the residue of f (z) at z = 1 coincides with that of f (1=w) =w2 at w = 0.0 : w w2 162 .8. jzj > R. 1 Show using residue theory that Z 1 p 1 2 3 p 3 dx = : 0 x2 x3 3 Solution Set Z 1 1 I= p 3 dx 0 x2 x3 and integrate Z 1 1 1 f (z) = p 3 dx = q p 0 z2 z3 3 jzj2 e2i arg z=3 3 j1 zjei arg0 (1 z)=3 along the contour in Figure VII.1.9. 163 .9.VII. 2 3 1 1 1 1 Res [f (z) .1 arg z arg0 z y y π π π 0 0 0 π π π 0 0 0 -2 -1 1 2 x -2 -1 1 2 x −π −π −π 0 0 0 π π π 2π 2π 2π arg z arg0 (1 z) y y π π π 0 0 0 2π 2π 2π 2π π π -2 -1 1 2 x -2 -1 1 2 x −π −π −π 0 0 0 0 0 0 0 π π The argument for the function f (z) = p 3 p1 jzj2 e2i arg z=3 3 j1 zjei arg0 (1 z)=3 y 4π/3 4π/3 4π/3 2π/3 π/3 π/3 -2 -1 1 2 x −2π/3 −2π/3 −2π/3 0 π/3 π/3 Residue at a simple pole at z1 = 1 . where by result from Exercise VII. VIII. This gives 164 . and let " ! 0+ .8.0 = lim p = e w w 1 3 w!0 3 jw 1jei arg(w 1)=3 Integrate along 1.9. 05 = w2 w w2 3 1 1 1 w2 w 1 1 1 i=3 = Res p . 1] = Res f .0 = Res 4 q .13 and Rule 1. and let " ! 0+ . This gives Z f (z) dz = 3 Z " 1 = q p dz = 1 " 3 jzj2 e2i arg z=3 3 j1 zjei arg0 (1 z)=3 Z " Z 1 " 1 1 = q p dx = p3 dx ! x 2 x 3 1 " 3 jxj2 e2i 0=3 3 j1 xjei 0=3 " Z 1 1 ! p3 dx = I: 0 x2 x3 Integrate along 4. This gives Z "4 f (z) dz p 2 " 2 "9=2 ! 0: 4 " (1 ") Integrate along 3. and let " ! 0+ . Z f (z) dz = 1 Z 1 " 1 = q p dz = " 3 jzj2 e2i arg z=3 3 j1 zjei arg0 (1 z)=3 Z 1 " Z 1 " 1 2 i=3 1 = q p dx = e p 3 dx ! " 3 jxj2 e2i 0=3 3 j1 xjei 2 =3 " x2 x3 Z 1 2 i=3 1 2 i=3 !e p 3 dx = e I: 0 x2 x3 Integrate along 2. and let " ! 0+ . This gives Z "4 f (z) dz p 2 " 2 "9=2 ! 0: 4 " (1 ") Using the Residue Theorem and letting R ! 1 and " ! 0+ . we obtain that 165 . . Z 1 p 1 2 3 p 3 dx = : 0 x2 x3 3 166 . 1 e e e sin ( =3) 3 i.e. 2 i=3 i=3 e I +0 I +0=2 i e : Solve for I. we obtain that i=3 p 2 ie 2 i 2 3 I= 2 i=3 = i=3 i=3 = = . VII.9.2 γ2 γ3 γ4 γ1 -R γ5 γ8 γ7 R γ6 Set Z 1 1 I= p dx 0 x x2 1 and integrate 1 1 f (z) = p = p p z z2 1 i z jz + 1je 0arg (z+1)=2 jz 1jei arg (z 1)=2 along the contour in Figure VII.2 Show using residue theory that Z 1 1 p dx = : 0 x x2 1 2 Solution VII.9.2. 167 .9. where by Rule 3. " # 1 Res p p . and let R ! 1 and " ! 0+ .0 = z jz + 1jei arg(z+1)=2 jz 1jei arg(z 1)=2 1 1 p p = i=2 = i jz + 1jei arg(z+1)=2 jz 1jei arg(z 1)=2 e z=0 Integrate along 1. This gives 168 . VIII.2 arg0 z arg z y y π π π 0 0 0 π π π 0 0 0 -2 -1 1 2 x -2 -1 1 2 x π π π 2π 2π 2π −π −π −π 0 0 0 arg0 (z + 1) arg0 (z 1) y y π π 0 0 0 0 π π π π 0 0 -2 -1 1 2 x -2 -1 1 2 x π π 2π 2π 2π 2π −π −π −π −π 0 0 1p The argument for the function f (z) = p z jz+1jei arg0 (z+1)=2 jz 1jei arg (z 1)=2 y π π π/2 π/2 0 0 -2 -1 1 2 x 0 0 π/2 π/2 π π Residue at a simple pole at z1 = i.9. Z f (z) dz = 1 Z R 1 = p p dz = 1+" z jz + 1jei arg0 (z+1)=2 jz 1jei arg (z 1)=2 Z R 1 = p p dx = 1+" x jx + 1jei 0=2 jx 1jei 0=2 Z R 1 = p dx ! 1+" x x2 1 Z 1 1 ! p dx = I: 1 x x2 1 Integrate along 2. This gives Z "4 f (z) dz p 2 " 2 "9=2 ! 0: 4 " (1 ") Integrate along 5. This gives Z 1 f (z) dz p R ! 0: 2 R R2 1 R Integrate along 3. and let R ! 1 and " ! 0+ . This gives 169 . This gives Z f (z) dz = 3 Z 1 " 1 = p p dz = i z jz + 1je 0arg (z+1)=2 jz 1jei arg (z 1)=2 R Z 1 " 1 = p p dx = x jx + 1je i =2 jx 1jei =2 R Z R Z R 1 x= t 1 p dx = = p dt = I: 1 " x x 2 1 dx = dt 1 t t 2 1 Integrate along 4. and let " ! 0+ . and let " ! 0+ . and let R ! 1 and " ! 0+ . Z f (z) dz = 5 Z R 1 = p p dz = 1 " z jz + 1jei arg0 (z+1)=2 jz 1jei arg (z 1)=2 Z R 1 = p p dx = 1 " x jx + 1jei =2 jx 1jei ( )=2 Z R Z R 1 x= t 1 p dx = = p dt = I: 1 " x x2 1 dx = dt 1 t t2 1 Integrate along 6. and let " ! 0+ . This gives Z 1 f (z) dz p R ! 0: 6 R R2 1 R Integrate along 7. and let R ! 1 and " ! 0+ . and let " ! 0+ . we obtain that 170 . This gives Z f (z) dz = 7 Z 1+" 1 = p p dz = R z jz + 1jei arg0 (z+1)=2 jz 1jei arg (z 1)=2 Z 1+" 1 = p p dx = R x jx + 1jei 2 =2 jx 1jei 0=2 Z R 1 = p dx ! 2 1+" x x 1 Z 1 1 ! p dx = I: 1 x x2 1 Integrate along 8. This gives Z "4 f (z) dz p 2 " 2 "9=2 ! 0: 4 " (1 ") Using the Residue Theorem and letting R ! 1 and " ! 0+ . Z 1 1 p dx = : 0 x x2 1 2 171 . i.e. I + 0 + I + 0 + I + 0 + I + 0 = 2 i ( i) .. 3 Show using residue theory that Z 1 1 p dx = p 2 x2 1 (x + 1) 1 2 Solution VII.9.VII.3 γ8 γ2 γ7 γ1 -R γ5 γ3 R γ4 γ6 Set Z 1 1 I= p dx 1 (x2 + 1) 1 x2 and integrate 1 1 f (z) = p = p p (z 2 + 1) 1 x2 (z i) (z + i) j1 + zjei arg (1+z)=2 j1 zjei arg0 (1 z)=2 along the contour in Figure VII.3. 172 .9.9. " # 1 Res p p . where by Rule 3.3 arg z arg0 z y y π π π 0 0 0 π π π 0 0 0 -2 -1 1 2 x -2 -1 1 2 x −π −π −π 0 0 0 π π π 2π 2π 2π arg (1 + z) arg0 (1 z) y y π π 0 0 0 0 2π 2π 2π 2π π π -2 -1 1 2 x -2 -1 1 2 x −π −π 0 0 0 0 0 0 0 0 π π p 1 p The argument for the function f (z) = (z i)(z+i) j1+zjei arg (1+z)=2 j1 zjei arg0 (1 z)=2 y 3π/2 3π/2 π π π/2 π/2 -2 -1 1 2 x −π/2 −π/2 0 0 π/2 π/2 Residue at a simple pole at z1 = i.i = (z i) (z + i) j1 + zjei arg(1+z)=2 j1 zjei arg(1 z)=2 1 p p = j1 + zjei arg(1+z)=2 j1 zjei arg(1 z)=2 z=i (z + i) 1 1 p p = p p (i + i) j1 + zjei arg(1+i)=2 j1 zje i arg(1 i)=2 2i j1 + zje(i =4)=2 j1 zje(i7 =4)=2 1 1 1 1 = pp pp = p = p = p i 2i 2 2e i 2i 2( 1) 2 2i 2 2 173 . VIII.9. Residue at a simple pole at z2 = i. and let " ! 0+ . i = (z i) j1 + zjei arg(1+z)=2 j1 zjei arg(1 z)=2 1 = p p = (z i) j1 + zjei arg(1+z)=2 j1 zjei arg(1 z)=2 z= i 1 1 p p = p p = ( i i) j1 + zjei arg(1 i)=2 j1 zjei arg(1+i)=2 2i j1 + zje( i =4)=2 j1 zje(i =4)=2 1 1 1 1 = pp pp = p = p = p i 2i 2 2e0i 2i 2 2 2i 2 2 Integrate along 1. where by Rule 3. This gives 174 . and let " ! 0+ . " # 1 Res p p . This gives Z "4 f (z) dz p 2 " 2 "9=2 ! 0: 4 " (1 ") Integrate along 3. and let " ! 0+ . This gives Z f (z) dz = 1 Z 1 " 1 = p p dz = 1+" j1 + zjei arg (1+z)=2 j1 zjei arg0 (1 z)=2 (z 2 + 1) Z 1 " Z 1 " 1 1 = p p dx = p dx ! (z 2 + 1) j1 + xjei0=2 j1 xjei2 =2 2 x2 1+" 1+" (x + 1) 1 Z 1 1 p dx = I: 2 1 x2 1 (x + 1) Integrate along 2. and let " ! 0+ . This gives Z f (z) dz = 5 Z R 1 = p p dz = 1 " (z 2 + 1) j1 + zjei arg(1+z)=2 j1 zjei arg(1 z)=2 Z R Z R 1 1 = p p dx = p dx ! (z 2 + 1) j1 + xjei( )=2 j1 xjei0=2 2 1 x2 ( i) 1 " 1 " (x + 1) Z 1 1 ! p dx = J: 1 (x + 1) 1 x2 ( i) 2 Integrate along 6. This gives 175 . Z f (z) dz = 1 Z 1+" 1 = p p dz = 1 " (z 2 + 1) j1 + zjei arg (1+z)=2 j1 zjei arg0 (1 z)=2 Z 1+" Z 1 " 1 1 = p p dx = p dx ! 1 " (z 2 + 1) j1 + xjei 0=2 j1 xjei 0=2 1+" (x2 + 1) 1 x2 Z 1 1 ! p dx = I: 1 (x2 + 1) 1 x2 Integrate along 4. and let R ! 1. and let R ! 1 and " ! 0+ . This gives Z f (z) dz ! 0: 4 Integrate along 5. This gives Z 1 2 f (z) dz p 2 R ! 0: 6 (R2 1) R2 1 R3 Integrate along 7. and let R ! 1 and " ! 0+ . This gives Z "4 f (z) dz p 2 " 2 "9=2 ! 0: 4 " (1 ") Using the Residue Theorem and letting R ! 1 and " ! 0+ . Z f (z) dz = 7 Z 1 " 1 = p p dz = R (z 2 + 1) j1 + zjei arg(1+z)=2 j1 zjei arg(1 z)=2 Z 1 " Z R 1 1 = p p dx = p dx ! R (z 2 + 1) j1 + xjei =2 j1 xjei2 =2 1 " (x2 + 1) 1 x2 ( i) Z 1 1 ! p dx = J: 1 (x2 + 1) 1 x2 ( i) Integrate along 8.. Z 1 1 p dx = p : 1 (x2 + 1) 1 x2 2 176 .e. we obtain that 1 1 I +0 I J +0+J +0=2 i p i+ p i 2 2 2 2 i. and let " ! 0+ . VII.4.9. 177 .4 Show using residue theory that Z 1 1 2 q dx = p : 1 3 (1 x) (1 + x)2 3 Solution Set Z 1 1 I= q dx 1 3 2 (1 x) (1 + x) and integrate 1 1 f (z) = q = p q 3 (1 z) (1 + z)2 3 j1 zjei arg0 (1 z)=3 j1 + zj2 e2i arg (1+z)=3 along the contour in Figure VII.9. where by result from Exercise VII. VIII.9.8. 178 .13 and Rule 1.4 arg0 z arg z y y π π π 0 0 0 π π π 0 0 0 x -2 -1 1 2 x π -2 π -1 π 2π 1 2π 2 2π −π −π −π 0 0 0 arg0 (1 z) arg (1 + z) y y 2π 2π 2π 2π π π π π 0 0 0 0 -2 -1 1 2 x -2 -1 1 2 x 0 0 0 0 π π −π −π 0 0 0 0 1 p The argument for the function f (z) = p 3 j1 zjei arg0 (1 z)=3 j1+zj2 e2i arg (1+z)=3 y 4π/3 4π/3 2π/3 2π/3 π/3 π/3 -2 -1 1 2 x −2π/3 −2π/3 0 0 π/3 π/3 Residue at a simple pole at z1 = 1 . and let " ! 0+ . 2 3 1 1 1 1 Res [f (z) . 0 = Res 4 2 q . This gives Z f (z) dz = 1 Z 1 " 1 = p q dz = 1+" 3 j1 zjei arg0 (1 j1 + zj2 e2i arg (1+z)=3 z)=3 Z 1 " Z 1 " 1 2 i=3 1 = p q dx = e q dx ! 1+" 3 j1 xjei2 =3 j1 + xj2 e2i 0=3 1+" 3 (1 x) (1 + x)2 Z 1 2 i=3 1 !e q dx = e 2 i=3 I: 1 3 2 (1 x) (1 + x) Integrate along 2. 05 = w 3 2 (w 1) (w + 1) 2 3 1 lim 4 p q 5= e i=3 w!0 3 jw 1jei arg(w 1)=3 jw + 1j2 e2i arg(w+1)=3 Integrate along 1. This gives Z "4 f (z) dz p 2 " 2 "9=2 ! 0: 4 " (1 ") Integrate along 3. 05 = w w w 3 1 1 2 1 w 1+ w 2 3 1 1 = Res 4 q . and let " ! 0+ . and let " ! 0+ . 1] = Res 2 f . This gives 179 . we obtain that i=3 2 ie 2 i 2 I= 2 i=3 = i=3 i=3 = =p . we obtain that 2 i=3 i=3 e I +0 I +0=2 i e : Solve for I. Z 1 1 2 q dx = p : 1 3 (1 x) (1 + x)2 3 180 . Z f (z) dz = 3 Z 1+" 1 = p q dz = 1 " j1 zjei arg0 (1 z)=3 j1 + zj2 e2i arg (1+z)=3 3 Z 1+" Z 1 " 1 1 = p q dx = q dx ! 1 " 3 i 0=3 2 2i 0=3 1+" 3 2 j1 xje j1 + xj e (1 x) (1 + x) Z 1 1 ! q dx = I: 1 3 (1 2 x) (1 + x) Integrate along 4. This gives Z "4 f (z) dz p 2 " 2 "9=2 ! 0: 4 " (1 ") Using the Residue Theorem and letting R ! 1 and " ! 0+ .e. and let " ! 0+ . e 1 e e sin ( =3) 3 i.. 5 Show using residue theory that Z 1 1 q dx = p 4 : 0 (x + 1) 4 x (1 x) 3 2 Solution Set Z 1 1 I= q dx 3 0 (x + 1) 4 x (1 x) and integrate 1 1 f (z) = q = p q (z + 1) 4 z (1 z)3 (z + 1) 4 jzjei arg z=4 4 j1 zj3 e3i arg0 (1 z)=4 along the contour in Figure VII.VII.9.9. 181 .5. VIII. where by Rule 3.9.5 arg z arg0 z y y π π π 0 0 0 π π π 0 0 0 -2 -1 1 2 x -2 -1 1 2 x −π −π −π 0 0 0 π π π 2π 2π 2π arg z arg0 (1 z) y y π π π 0 0 0 2π 2π 2π 2π π π -2 -1 1 2 x -2 -1 1 2 x −π −π −π 0 0 0 0 0 0 0 π π 1 p The argument for the function f (z) = p z=4 4 (z+1) 4 jzjei arg j1 zj3 e3i arg0 (1 z)=4 y 7π/4 7π/4 7π/4 3π/2 3π/4 3π/4 -2 -1 1 2 x −π/4 −π/4 −π/4 3π/4 3π/4 3π/4 Residue at a simple pole at z1 = 1. 182 . where by result from Exercise VII.0 = Res 4 q . 05 = 0 w!0 4 3 (1 + w) (w 1) Residue at a simple pole at z2 = i. where by Rule 3. 05 = w2 w w2 1 4 1 1 3 w +1 w 1 w 2 3 1 = lim 4 q . 1] = Res f . 2 3 1 1 1 1 Res [f (z) . and let R ! 1 and " ! 0+ . 2 3 1 Res 4 p q .8. Integrate along 1 .13 and Rule 1. This gives Z f (z) dz = 1 Z 1 " 1 = p q dz = " (z + 1) 4 jzjei arg z=4 4 j1 zj3 e3i arg0 (1 z)=4 Z 1 " Z 1 " 1 1 = p q dx = i q dx ! " (x + 1) 4 jxjei 0=4 4 j1 xj3 e3i 2 =4 " (x + 1) 4 x (1 x)3 Z 1 1 !i q dx = iI: 3 0 (x + 1) 4 x (1 x) 183 . 15 = (z + 1) 4 jzjei arg(z)=4 4 j1 zj3 e3i arg(1 z)=4 1 p q = 4 jzjei arg(z)=4 4 j1 zj3 e3i arg(1 z)=4 z= 1 1 = p q = 4 jzjei arg(z)=4 4 j1 zj3 e3i arg(1 z)=4 1 1 1 1 1 1 = p 4 = p 4 p + ip = p 4 +i p 8e i=4 8 2 2 2 2 242 Residue at a simple pole at z2 = 1 . e. 2 2 242 i. This gives Z "4 f (z) dz p 2 " 2 "9=2 ! 0: 4 " (1 ") Using the Residue Theorem and letting R ! 1 and " ! 0+ . Z 1 1 q dx = p 4 : 0 (x + 1) 4 x (1 x) 3 2 184 .Integrate along 2. This gives Z f (z) dz = 3 Z " 1 = p q dz = 1 " (z + 1) 4 jzjei arg z=4 4 j1 zj3 e3i arg0 (1 z)=4 Z " Z 1 " 1 1 = p q dx = q dx ! 1 " 4 (x + 1) jxjei 0=4 4 j1 xj3 e3i 0=4 " (x + 1) 4 x (1 3 x) Z 1 1 ! q dx = I: 3 0 (x + 1) 4 x (1 x) Integrate along 4. This gives Z "4 f (z) dz p 2 " 2 "9=2 ! 0: 4 " (1 ") Integrate along 3. and let " ! 0+ . and let " ! 0+ . and let R ! 1 and " ! 0+ .. we obtain that 1 1 iI + 0 I +0=2 i p 4 +i p +0 . VIII 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 2 3 4 5 6 7 8 1 . First of all p (z) have no real zeros. Since the variation in argument is determined by the dominating term R4 e4it for R large the change of argument on this arc is 4 times the variation in argument for the arc. Solution VIII. 4 4it 2 2it it and our function p (z) becomes p (t) = R e + 2R e Re + 1.1. and thus 4 arg p (z) 4 2 = 2 . The positive imaginary axis 3 from iR to 0 we parametizize as 3 : z = it.1. Since p (x) > 0 for t 0. and compute 4 arg p (z) along the three segments in the sector path in …gure VIII. thus 4 arg p (z) = 0.1 1 2 3 P L K LLL Show that z 4 + 2z 2 z + 1 has exactly one root in each quadrant. and our function p (z) becomes f (t) = (it)4 + 2 (it)2 it + 1 2 = (t2 1) it .1 Set p (z) = z 4 +2z 2 z +1. 0 t R. We make the following table and do the sketches. We …nd that the real part have a zero at t = 1. The arc 2 from R to iR we parametizize as 2 : z = Reit . 2 and our function p (z) becomes p (t) = t4 + 2t2 t + 1 = (t2 + 1) t > 0.VIII. 0 t =2. their is no change of argument on this line segment on the real axis. and the imaginary part have a zero for t = 0. 0 t R.1. The positive real axis 1 from 0 to R we parametizize as x : z = t.1. 2 . 1 (R = 4) VIII. 3 . the argument for p (z) remain in the 4-th quadrant except for touching the imaginary axis at i and terminating at 1.1. and their is no change of argument on this line segment on the imaginary axis. 2 Becauce that p (x) = (x2 + 1) x > 0 it is clear that p (z) have no real roots and p (z) must have on has one zero in each quadrant. VIII. so we have exactly one zero in the …rst quadrat. Now we have that the total change of argument for p (z) is 2 .1.1 (R = 4) t Re z Im z arg z 4 1 + 0 3 1 0 2 2 0 1 0 0 1 -2 -1 1 2 3 4 -1 -2 Thus when we move from iR to 0. thus 4 arg p (z) 0. Because that the roots come in complex conjugate pairs it is plain that p (z) have a root in the fourth quadrant too. their is no change of argument on this line segment on the real axis. thus 4 arg p (z) = 0. and 4 4it 3 3it 2 2it it our function p (z) becomes p (t) = R e + R e + 4R e + 3Re + 2. First of all p (z) have no real zeros. and our function p (z) becomes p (t) = t4 +t3 +4t2 +3t+2 > 0. 0 t =2.1. The positive real axis 1 from 0 to R we parametizize as x : z = t. The arc 2 from R to iR we parametizize as 2 : z = Reit . We …nd that the real part have a zero 4 . Since p (x) > 0 for t 0.2 1 2 3 P L K Find the number of zeros of the polynomial p (z) = z 4 +z 3 +4z 2 +3z+2 in each quadrant. Solution VIII. and compute 4 arg p (z) along the three segments in the sector path in …gure VIII. Since the variation in argument is determined by the dominating term R4 e4it for R large the change of argument on this arc is 4 times the variation in argument for the arc. and our function p (z) becomes f (t) = (it)4 + (it)3 + 4 (it)2 + 3it + 2 = t4 4t2 + 2 + i ( t3 + 3t) .1. 0 t R. and thus 4 arg p (z) 4 2 = 2 . The positive imaginary axis 3 from iR to 0 we parametizize as 3 : z = it. 0 t R.2.VIII.2 Set p (z) = z 4 + z 3 + 4z 2 + 3z + 2.1. p p p p at t = 2 p 2 and t = 2 + 2, and the imaginary part have a zero for t = 0 and t = 3. We make the following table and do the sketches, t Re z Im z arg z VIII.1.2 (R = 10) VIII.1.2 (R = 10) 1 p + 0 p 4 p 2 + 2 0 2 2 p3 p 0 3 2 2 0 + 2 -4 -2 2 4 0 + 0 2 -2 -4 Thus when we move from iR to 0, the argument for p (z) on this line segment on the imaginary axis changes so that, 4 arg p (z) 2 . Now we have that the total change of argument for p (z) is 0, so we have no zero in the 1st quadrant. Since p (z) has real coe¢ cients, the roots come in conjugate pairs (since there are nor real roots), therefore there are no roots in the 4th quadras as well. By symmetry, there are 2 roots each in the 2nd and 3rd quadrants. 5 VIII.1.3 1 2 3 P L K Find the number of zeros of the polynomial p (z) = z 6 + 4z 4 + z 3 + 2z 2 + z + 5 in the …rst quadrant fRe z > 0; Im z > 0g. Solution VIII.1.3 Set p (z) = z 6 + 4z 4 + z 3 + 2z 2 + z + 5, and compute 4 arg p (z) along the three segments in the sector path in …gure VIII.1.3. First of all p (z) have no real zeros. The positive real axis 1 from 0 to R we parametizize as x : z = t; 0 t R, and our function p (z) becomes p (t) = t6 + 4t4 + t3 + 2t2 + t + 5 > 0. Since p (t) > 0 for t 0, their is no change of argument on this line segment on the real axis, thus 4 arg p (z) = 0. The arc 2 from R to iR we parametizize as 2 : z = Reit , 0 t =2, and our function p (z) becomes p (t) = R6 e6it +4R4 e4it +3R3 e3it +2R2 e2it +Reit +5. Since the variation in argument is determined by the dominating term R6 e6it for R large the change of argument on this arc is 6 times the variation in argument for the arc, and thus 4 arg p (z) 6 2 = 3 . The positive imaginary axis 3 from iR to 0 we parametizize as 3 : z = it; 0 t R, and our function p (z) becomes f (t) = (it)6 + 4 (it)4 + (it)3 + 2 (it)2 + it + 5 = t6 + 4t4 2t2 + 5 + i ( t3 + t) . We …nd that the real part have a zero at t 1; 95, and the imaginary part have a zero for t = 0 and t = 1. We make the following table and do the sketches, 6 VIII.1.3 (R = 3) VIII.1.3 (R = 3) t Re z Im z arg z 1 1000 1; 95 0 2 -2 2 4 6 8 10 500 1 + 0 0 -2 0 + 0 0 -4 -1000 -500 500 1000 -500 -6 Thus when we move from iR to 0, the argument for p (z) on this line segment on the imaginary axis changes so that, 4 arg p (z) . Now we have that the total change of argument for p (z) is 4 , so we have that p (z) has exactly two zeros in the …rst quadrat. (None on real or imaginary axis) 7 VIII.1.4 1 2 3 P L K LLL Find the number of zeros of the polynomial p (z) = z 9 +2z 5 2z 4 +z+3 in the right half-plane. Solution VIII.1.4 Set p (z) = z 9 + 2z 5 2z 4 + z + 3, and compute 4 arg p (z) along the three segments in the sector path in …gure VIII.1.4. First of all p (z) have no real zeros. The positive real axis 1 from 0 to R we parametizize as x : z = t; 0 t R, and our function p (z) becomes p (t) = t9 + 2t5 2t4 + t + 3. Since p (x) > 0 for t 0, their is no change of argument on this line segment on the real axis, thus 4 arg p (z) = 0. The arc 2 from R to iR we parametizize as 2 : z = Reit , 0 t =2, and 9 9it 5 5it 4 4it it our function p (z) becomes p (t) = R e + 2R e 2R e + Re + 3. Since the variation in argument is determined by the dominating term R9 e9it for R large the change of argument on this arc is 9 times the variation in argument for the arc, and thus 4 arg p (z) 9 2 = 92 . The positive imaginary axis 3 from iR to 0 we parametizize as 3 : z = it; 0 t R, and our function p (z) becomes f (t) = (it)9 + 2 (it)5 2 (it)4 + 2 3 = 2t4 + 3 + iy (y 4 + 1) . We …nd that the real part have a zero at it + q t = 4 32 , and the imaginary part have only a zero for t = 0. We make the following table and do the sketches, 8 t Re z Im z arg z VIII.1.4 (R = 2) VIII.1.4 (R = 2) 1 + 600 q 2 400 30 4 3 2 0 + 2 200 20 0 + 0 0 -600-400-200 200 400 600 10 -200 -400 -600 -20 -10 10 20 Thus when we move from iR to 0, the argument is changed from =2 to 0, so the charge of argument on this on the imaginary axis is , thus 4 arg p (z) = =2. Now we have that the total change of argument for p (z) is 4 , so we have exactly two zeros in the …rst quadrat. Because that the roots come in complex conjugate pairs it is plain that p (z) has four zeros in the right half-plane. 9 VIII.1.5 1 2 3 P L K For a …xed real number , …nd the number of zeros z 4 +z 3 +4z 2 + z+3 satisfying Re z < 0. (Your answer depends on .) Solution VIII.1.5 Set p (z) = z 4 + z 3 + 4z 2 + z + 3, and compute 4 arg p (z) along the three segments in the contuour in …gure VIII.1.5. The positive real axis 1 from 0 to R we parametizize as x : z = t; 0 t R, 2 and our function p (z) becomes p (t) = t4 + t3 + 4t2 + at + 3 = (t2 + 1) t > 0. Since p (x) > 0 for t 0, their is no change of argument on this line segment on the real axis, thus 4 arg p (z) = 0. The arc 2 from R to iR we parametizize as 2 : z = Reit , 0 t =2, and 4 4it 3 3it 2 2it it our function p (z) becomes p (t) = R e + R e + 4R e + aRe + 3. Since the variation in argument is determined by the dominating term R4 e4it for R large (independent of the value of a) the change of argument on this arc is 4 times the variation in argument for the arc, and thus 4 arg p (z) 4 2 = 2 . The positive imaginary axis 3 from iR to 0 we parametizize as 3 : z = it; 0 t R, and our function p (z) becomes f (t) = (it)4 + (it)3 + 4 (it)2 + ait + 3 = (t2 1) (t2 p3) + i ( t3 + at). We …nd that the real part have a zero at t = 1 and t = p3 and the imaginary part have a zero for t = 0 and and a second root t = a if a 0. We make the following table and do the sketches, 10 a<0 VIII.1.5 (a = -4, R = 10) VIII.1.5 (a = -4, R = 10) t Re z Im z arg z 1 p + 0 3 0 2 1 0 2 0 + 0 0 a=0 VIII.1.5 (a = 0, R = 10) VIII.1.5 (a = 0, R = 10) t Re z Im z arg z 1 p + 0 3 0 2 1 0 2 0 + 0 0 0<a<1 VIII.1.5 (a = 1/2, R = 10) VIII.1.5 (a = 1/2, R = 10) t Re z Im z arg z 1 p + 0 3 0 2 1 0 p 2 a + 0 0 0 + 0 0 11 a=1 VIII.1.5 (a = 1, R = 10) VIII.1.5 (a = 1, R = 10) t Re z Im z arg z 1 p + 0 3 0 2 1 0 0 0 + 0 0 1<a<3 VIII.1.5 (a = 2, R = 10) VIII.1.5 (a = 2, R = 10) t Re z Im z arg z 1 p + 0 3 0 p 2 a 0 3 1 0 + 2 0 + 0 2 a=3 VIII.1.5 (a = 3, R = 10) VIII.1.5 (a = 3, R = 10) t Re z Im z arg z 1 p + 0 3 0 0 1 0 + 2 0 + 0 0 12 a>3 VIII.1.5 (a = 9, R = 10) VIII.1.5 (a = 9, R = 10) t Re z Im z arg z 1 + 0 p pa + 0 0 3 0 + 2 1 0 + 2 0 + 0 0 Thus when we move from iR to 0, the argument for p (z) only change if 1 < a < 3 and we have that 4 arg p (z) 2 , for all other value on a we have that 4 arg p (z) 0. Remark that we can se that for a = 1 and a = 3 we have zeros on the imaginary axis. Now we have that the total change of argument for p (z) is 2 , so we have exactly one zero in the …rst quadrat. Because that the roots come in complex conjugate pairs it is plain that p (z) has one zero in each quadrant. If 1 3, the total change of argument for p (z) is 0 and no zeros on the imaginary axis, thus we must have four zeros in the open left half-plane. If < 1 and > 3, the total charge of argument for p (z) is 2 and no zeros of the imaginary axis, thus we must have two zeros in the open left half-plane. If = 1 and = 3, the total charge of argument for p (z) is 2 and we have two zeros on the imaginary axis, thus we must have remaining two zeros in the open left half-plane. 13 VIII.1.6 1 2 3 P L K For a …xed real number , …nd the number of solutions of z 5 +2z 3 z 2 +z = satisfying Re z > 0. Solution VIII.1.6 Set p (z) = z 5 + 2z 3 z 2 + z , and compute 4 arg p (z) along the two segments in the contuour in …gure VIII.1.6. The arc 1 from iR to iR we parametizize as 2 : z = Reit , =2 t =2, and our function p (z) becomes p (t) = R5 e5it +2R3 e3it R2 e2it +Reit . Since the variation in argument is determined by the dominating term R5 e5it for R large (independent of the value of ) the change of argument on this arc is 5 times the variation in argument for the arc, and thus 4 arg p (z) 5 2 = 52 . The positive imaginary axis 2 from iR to iR we parametizize as 2 : z = it; R t R, and our function p (z) becomes f (t) = (it)5 + 2 (it)3 2 (it)2 +pit = t2 p + it (t2 1) . We …nd that the real part have a zero at t= and t = and the imaginary part have a zero for t = 1, t = 0 and t = 1. We make the following table and do the sketch, 14 VIII.1.6 (a = 4, R = 4) VIII.1.6 (a = 4, R = 4) t Re z Im z arg z 1 VIII.1.5 (a = 2, R = 10) VIII.1.5 (a = 2, R = 10) t Re Im arg 1 p + 0 0 3 0 p 2 a 0 3 1 0 + 2 0 3 0 2 VIII.1.5 (a = 0, R = 10) VIII.1.5 (a = 0, R = 10) t Re z Im z arg z 1 15 Thus when we move from iR to 0, the argument for p (z) remain in the 4-th quadrant except for touching the imaginary axis at i and terminating at 1, and their is no change of argument on this line segment on the imaginary axis, thus 4 arg p (z) = 0. Now we have that the total change of argument for p (z) is 2 , so we have exactly one zero in the …rst quadrat. Because that the roots come in complex conjugate pairs it is plain that p (z) has one zero in each quadrant. 16 VIII.1.7 1 2 3 P L K For a …xed complex number , show that if m and n are large integers, then the equation ez = z + has exactly m + n solutions in the horizontal strip f 2 im < Im z < 2 ing. Solution VIII.1.7 Set p (z) = z 4 +2z 2 z +1, and compute 4 arg p (z) along the three segments in the contuour in …gure VIII.1.7. The positive real axis 1 from 0 to R we parametizize as x : z = t; 0 t R, 2 and our function p (z) becomes p (t) = t4 + 2t2 t + 1 = (t2 + 1) t > 0. Since p (x) > 0 for t 0, their is no change of argument on this line segment on the real axis, thus 4 arg p (z) = 0. The arc 2 from R to iR we parametizize as 2 : z = Reit , 0 t =2, and our function p (z) becomes p (t) = R4 e4it + 2R2 e2it Reit + 1. Since the variation in argument is determined by the dominating term R4 e4it for R large the change of argument on this arc is 4 times the variation in argument for the arc, and thus 4 arg p (z) 4 2 = 2 . The positive imaginary axis 3 from iR to 0 we parametizize as 3 : z = it; 0 t R, and our function p (z) becomes f (t) = (it)4 + 2 (it)2 it + 1 2 = (t2 1) it . We …nd that the real part have a zero at t = 1, and the imaginary part have a zero for t = 0 on the positive imaginary axis. We make the following table and do the sketch, 17 VIII.1.7 t Re Im arg 20 1 + 0 10 1 0 2 0 + 0 0 -20 -10 10 20 -10 -20 Thus when we move from iR to 0, the argument for p (z) remain in the 4-th quadrant except for touching the imaginary axis at i and terminating at 1, and their is no change of argument on this line segment on the imaginary axis, thus 4 arg p (z) = 0. Now we have that the total change of argument for p (z) is 2 , so we have exactly one zero in the …rst quadrat. Because that the roots come in complex conjugate pairs it is plain that p (z) has one zero in each quadrant. 18 0 t R. Since the variation in argument is determined by the dominating term R4 e4it for R large the change of argument on this arc is 4 times the variation in argument for the arc. 19 . and the imaginary part have a zero for t = 0 on the positive imaginary axis.8 Set p (z) = z 4 +2z 2 z +1. their is no change of argument on this line segment on the real axis. The arc 2 from R to iR we parametizize as 2 : z = Reit . The positive imaginary axis 3 from iR to 0 we parametizize as 3 : z = it. 0 t =2. Since p (x) > 0 for t 0. and our function p (z) becomes f (t) = (it)4 + 2 (it)2 it + 1 2 = (t2 1) it . The positive real axis 1 from 0 to R we parametizize as x : z = t.1. then the equation ez = z + has exactly one solution in the left half-plane.1.1.VIII. 2 and our function p (z) becomes p (t) = t4 + 2t2 t + 1 = (t2 + 1) t > 0.8 1 2 3 P L K LLL Show that if Re > 1. and thus 4 arg p (z) 4 2 = 2 . We make the following table and do the sketch. Solution VIII. and our function p (z) becomes p (t) = R4 e4it + 2R2 e2it Reit + 1. 0 t R. thus 4 arg p (z) = 0. and compute 4 arg p (z) along the three segments in the contuour in …gure VIII. We …nd that the real part have a zero at t = 1.8. 4 (R = 2) 1 q + 2 100 30 4 3 2 0 + 2 50 20 0 + + 0 -100 -50 50 100 10 -50 -100 -20 -10 10 20 20 . f (iy) remains in the left half-plane and the change in arg f is . thus the change in arg f is again . and get f (iy) = cos y Re + i (sin y y) i Im . When we move from f (iR) to f ( iR) along f Rei . and their is no change of argument on this line segment on the imaginary axis.) t Re Im arg VIII. Thus when we move from f ( iR) to f (iR). thus 4 arg p (z) = 0. 2 < < 32 .1. Because that the roots come in complex conjugate pairs it is plain that p (z) has one zero in each quadrant. Set f (z) = ez z .8 t Re Im arg 20 1 + 0 10 1 0 2 0 + 0 0 -20 -10 10 20 -10 -20 Thus when we move from iR to 0. Re > 1.4 (R = 2) VIII. so we have exactly one zero in the …rst quadrat.1. Now we have that the total change of argument for p (z) is 2 . we cross the real axis from the 4-th to the 1 st quadrant.1. VIII. (May also use Rouche’s Theorem. the argument for p (z) remain in the 4-th quadrant except for touching the imaginary axis at i and terminating at 1. 1. t Re Im arg VIII.4 (R = 2) 1 q + 2 400 30 4 3 2 0 + 2 200 20 0 + + 0 -400 -200 200 400 10 -200 -400 -20 -10 10 20 (it)4 (it)3 + 13 (it)2 it +36 = t4 + it3 13t2 it + 36 21 .4 (R = 2) VIII.1. Solution 22 .VIII. 0] then the increase in the argument of f (z) around is zero.1. and if is a closed curve in D such that the values of f (z) on lie in the slit plane Cn ( 1.9 1 2 3 P L K Show that if f (z) is analytic in a domain D. 2. where f2 (z) = 6z and h2 (z) = 2z 5 1. jf2 (z)j = j6zj = 6 and jh2 (z)j = j2z 5 1j 3. thus this root must be in the interval 0 < x < 1.VIII. the root in the disk jzj = 1 must be real. 23 . jf1 (z)j = j2z 5 j = 2 25 = 64 and jh1 (z)j = j6z 1j 13. then jf1 (z)j > jh1 (z)j on jzj = 2 so by Rouche’s Theorem p (z) has all 5 roots in the disk jzj = 2. we have that p (0) = 1 and p (1) = 7. Set now p (z) = f2 (z) + h2 (z). Solution Set …rst p (z) = f1 (z) + h1 (z). On the circle jzj = 1 we have.1 1 2 3 P L K LLL Show that 2z 5 + 6z 1 has one root in the interval 0 < x < 1 and four roots in the annulus f1 < jzj < 2g. then jf2 (z)j > jh2 (z)j on jzj = 1 so by Rouche’s Theorem p (z) has 1 roots in the disk jzj = 1. On the circle jzj = 2 we have. where f1 (z) = 2z 5 and h1 (z) = 6z 1. It follows that p (z) = 2z 5 + 6z 1 has 5 1 = 4 roots in the annulus 1 < jzj < 2. As any complex roots come in conjugate pairs that have the same magnitude. On the circle jzj = 2 we have. then jf2 (z)j > jh2 (z)j on jzj = 2 so by Rouche’s Theorem p (z) has 3 roots in the disk jzj = 1. Set now p (z) = f2 (z) + h2 (z). then jf1 (z)j > jh1 (z)j on jzj = 2 so by Rouche’s Theorem p (z) has 9 roots in the disk jzj = 2. where f2 (z) = z 9 + z 5 + 2z + 1 and h2 (z) = 8z 3 .VIII. Thus it follows that p (z) = z 9 + z 5 8z 3 + 2z + 1 has 9 3 = 7 roots in the annulus 1 < jzj < 2.2. where f1 (z) = z 9 and h1 (z) = z 5 8z 3 +2z+1. jf1 (z)j = jz 9 j = jzj9 = 29 = 512 and jh1 (z)j = jz 5 8z 3 + 2z + 1j 101. jf2 (z)j = j 8z 3 j = 8 and jh2 (z)j = jz 9 + z 5 + 2z + 1j 5.2 1 2 3 P L K LLL How many roots does z 9 + z 5 8z 3 + 2z + 1 have between the circles fjzj = 1g and fjzj = 2g? Solution Set …rst p (z) = f1 (z)+h1 (z). On the circle jzj = 1 we have. 24 . Solution 2 m Set p (z) = f (z) + h (z). 2 m jf (z)j = j3z n j = 3 jzjn = 3 1n = 3 and jh (z)j = 1 + z + z2! + + zm! P1 1 k=0 k! = e. where f (z) = 3z n and h (z) = 1 + z + z2! + + zm! .3 1 2 3 P L K LLL Show that if m and n are positive integers. the the polynomial z2 zm p (z) = 1 + z + + + + 3z n 2! m! has exactly n zeros in the unit disk. On the circle jzj = 1 we have.VIII. 25 .2. then jf (z)j > jh (z)j on jzj = 1 so by Rouche’s Theorem 2 m p (z) = 1 + z + z2! + + zm! + 3z n has exactly n roots in the unit disk jzj = 1. 2. hur funkar den med nollställen *********** 26 . For n 1. show that n z (z 1) e has n zeros satisfying jz 1j < 1 and no other zeros in the right half-plane. Since f (1) = . On the i circle jz 1j = 1 which we parametrize by z = 1 + e . 0 2 we have.VIII. n z i n 1+i 1+cos jf (z)j = j(z 1) e j = 1 + e 1 e =e 1. where the only zeros in the right half-plane are at z = 1. the zeros of f (z) must be simple unless = 0. thus the zeros of of f 0 (z) are at z = 1 and at z = 1 n. in which case we ge a zero of order n at z = 1. *********** Här är lite konstigt den derivarande satsen. We have that f 0 (z) = n (z 1)n 1 ez + (z 1)n ez = (n + z 1) (z 1)n 1 ez . then jf (z)j > jh (z)j on jz 1j = 1 so by Rouche’s Theorem p (z) = (z 1)n ez has n roots in the disk jz 1j = 1. where f (z) = (z 1)n ez and h (z) = .4 1 2 3 P L K LLL Fix a complex number such that j j < 1. then 0 2 and jh (z)j = j j = j j < 1 from the text. Solution Set p (z) = f (z) + h (z). Determine the multiplicity of the zeros. VIII. which are all simple if 6= 0. Solution 27 . show that (z 1)n ez + (z + 1)n has n zeros in the right half-plane.5 1 2 3 P L K For a …xed satisfying j j < 1.2. Solution 28 . (b) Determine which quadrants contain the two zeros of p (z) that lie outside the unit circle. (c) Show that the two zeros of p (z) that lie outside the unit circle satisfy fjz 3ij < 1=10g.2.VIII. (a) Determine which quadrants contain the four zeros of p (z) that lie inside the unit circle.6 1 2 3 P L K Let p (z) = z 6 + 9z 4 + z 3 + 2z + 4 be the polynomial treated in the example in this section. This is a variant of Rouche’s theorem. Rouche’s theorem is obtained by setting h (z) = f (z) g (z). Remark.VIII. counting multiplicity. Solution 29 . For the solution of the exercise. see Exercise 9 in the preceding section.7 1 2 3 P L K LLL Let f (z) and g (z) be analytic functions on the bounded domain D that extend continuously to @D and satisfy jf (z) + g (z)j < jf (z)j + jg (z)j on @D. Show that f (z) and g (z) have the same number of zeros in D.2. in which the hypotheses are symmetric in f (z) and g (z). 2.VIII. What can be said about f (z) and f (z) + g (z)? Prove your assertion. Show by example that f (z) and f (z) + h (z) can have di¤erent numbers of zeros on D.8 1 2 3 P L K Let D be a bounded domain. and let f (z) and h (z) be meromorphic functions on D that extend to be analytic on @D. Suppose that jh (z)j < jf (z)j on @D. Solution 30 . Solution 31 .VIII. Show that f (z) is analytic.9 1 2 3 P L K Let f (z) be a continuously di¤erentiable function on a domain D. Suppose that for all complex constants a and b. the increase in the argument of f (z)+ az +b around any small circle in D on which f (z)+az +b 6= 0 is nonnegative.2. Solution 32 . and suppose that f (z) has a zero of order N at z0 2 D.1 1 2 3 P L K LLL Let ffk (z)g be a sequence of analytic functions on D that converges normally to f (z). Use Rouche’s theorem to show that there exists > 0 such that for k large. fk (z) has exactly N zeros counting multiplicity on the disk fjz z0 j < g.3.VIII. but more di¢ cult to prove. that is. Show that S is closed under normal convergence. It is also true. if a sequence in S converges normally to f (z).VIII. that S is a compact family of analytic functions. Solution 33 . every sequence in S has a normally convergent subsequence.3. that is. Remark. then f 2 S.2 1 2 3 P L K LLL Let S be the family of univalent functions f (z) de…ned on the open unit disk fjzj < 1g that satisfy f (0) = 0 and f 0 (0) = 1. 2 i @D f (z) j=1 where z1 . Solution 34 . Suppose that both f (z) and g (z) extend analytically across the boundary of D. Show that I Xn 1 f 0 (z) g (z) dz = mj g (zj ) .4.VIII. and mj is the order of f (z) at zj . Let f (z) be meromorphic and g (z) analytic on D. : : : .1 1 2 3 P L K LLL Suppose D is a bounded domain with piecewise smooth boundary. zn are the zeros and poles of f (z). and that f (z) 6= 0 on @D. Hint.2 1 2 3 P L K Let f (z) be a meromorphic function on the complex plane that is doubly periodic.4. Show that mj zj is a period of f (z). Solution 35 . : : : . and suppose no zj is a translateP by a period of another zk .5.VIII. Integrate zf 0 (z) =f (z) around the boundary of the fundamental parallelogram P constructed in Section VI. Let mj be the order of f (z) at zj . Suppose that the zeros and the poles of f (z) are at the points z1 . zn and at their translates by periods of f (z). Show that either f (z) is constant.4.VIII. or f (z) attains each value w at most m times in D. Solution 36 . Suppose that fk (z) attains each value w at most m times (counting multiplicity) in D.3 1 2 3 P L K LLL Let ffk (z)g be a sequence of analytic functions on a domain D that converges normally to f (z). Suppose there is an annulus U = fr < jzj < 1g such that the restriction of f (z) to U is one-to-one.VIII.4. Show that f (z) is one-to-one on D. Solution 37 .4 1 2 3 P L K Let f (z) be an analytic function on the open unit disk D = fjzj < 1g. VIII. Denote the degree of f (z) by d. (a) Show that each value w 2 C.4. Solution 38 . where p (z) and q (z) are polynomials that are relatively prime (no common zeros). w 6= f (1). is assumed d times by f (z) on C. We de…ne the degree of f (z) to be the larger of the degrees of p (z) and q (z).5 1 2 3 P L K Let f (z) = p (z) =q (z) be a rational function. (b) Show that f (z) attains each value w 2 C d times on C (as always. counting multiplicity). 4. 39 .6 1 2 3 P L K Let f (z) be a meromorphic function on the complex plane. and suppose there is an integer m such that f 1 (w) has at most m points for all w 2 C. Show that f (z) is a rational function.VIII. (b) Show that Z 1 F1 ( . w) with respect to w. w) : (d) Derive the inverse function theorem given in this section. w) be a continuous function of z and w that depends analytically on z for each …xed w.VIII. (a) Show that there exists > 0 such that if jw w0 j < . Show that g (w) is ana- lytic. w) = 0.7 1 2 3 P L K Let F (z. Note that a speci…c formula is given for the function g (w) de…ned implicitely by F (g (w) .4. w) denote the derivative of F (z. Suppose F (z0 . w) is analytic in w for each …xed z. there is a unique z = g (w) satisfying jz z0 j < and F (z. w) =F1 (g (w) . and F1 (z0 . this is the implicit function theorem for analytic functions. w0 ) 6= 0 for 0 < jz z0 j . w0 ) 6= 0. w) with respect to z. Remark. (b). jw w0 j < : 2 i j z0 j= F ( . w) (c) Suppose further that F (z. and let F1 (z. and g 0 (w) = F2 (g (w) . together with the formula for the derivative of the inverse function. w0 ) = 0. w) = 0. as a corollary of (a). 40 . w) denote the derivative of F (z. and (c). Choose such that F (x. and let F2 (z. w) g (w) = d . Show that @ (f (D)) f (@D). that is.VIII.4. and let f (z) be a continuous function on D[@D that is analytic on D. Solution 41 . the boundary of the open set f (D) is contained in the image under f (z) of the boundary of D.8 1 2 3 P L K Let D be a bounded domain. 1 1 2 3 P L K Find the critical points and critical values of f (z) = z + 1=z. 42 . Solution. Sketch the curves where f (z) is real. Sketch the regions where Im f (z) > 0 and where Im f (z) < 0.VIII.5. Sketch and label the curves passing through z = 0 where Re g (z) = 2 and Im g (z) = 0.VIII.2 1 2 3 P L K Suppose g (z) is analytic at z = 0. with power series g (z) = 2 + iz 4 + O (z 5 ).5. Solution 43 . 3 1 2 3 P L K Find the critical points and critical values of f (z) = z 2 + 1.VIII. and locate the critical points of f (z) on the sketch. Sketch the set of points z such that jf (z)j 1.5. Solution 44 . 5. Show also that if the set of z such that jf (z)j = jf (z0 )j consists of just one curve passing through z0 . Solution 45 . Show that if the set of z such that Re f (z) = Re f (z0 ) consists of just one curve passing through z0 . then f 0 (z0 ) 6= 0. then f 0 (z0 ) 6= 0.4 1 2 3 P L K Suppose that f (z) is analytic at z0 .VIII. VIII.5.5 1 2 3 P L K How many critical points. Solution 46 . does a polynomial of degree m have in the complex plane? Justify your answer. counting multiplicity. Use the chain rule.VIII.6 1 2 3 P L K Find and plot the critical points and critical values of f (z) = z 2 + 1 and of 2 its iterates f (f (z)) = (z 2 + 1) + 1 and f (f (f (z))). Suggestion.5. Solution 47 . Solution 48 .5.VIII.7 1 2 3 P L K Let f (z) be a polynomial of degree m. How many (…nite) critical points does the N fold iterate f f (N times) have? Describe them in terms of the critical points of f (z). w 6= w0 . Show that with this de…nition. Remark.5. We say that f (z) is a (k + 1) sheeted covering of f 1 (V n fw0 g) \ U over V n fw0 g. Solution 49 . a point z0 2 C is a critical point of order k for a meromorphic function f (z) if and only if there are open sets U containing z0 and V containing w0 = f (z0 ) such that each w 2 v.8 1 2 3 P L K We de…ne a pole of f (z) to be a critical point of f (z) of order k if z0 is a critical point of 1=f (z) of order k. We de…ne z = 1 to be a critical point of f (z) of order k if w = 0 is a critical point of g (w) = f (1=w) of order k. has exactly k + 1 preimages in U .VIII. 9 1 2 3 P L K Show that a polynomial of degree m.5. has a critical point of order m 1 at z0 = 1.VIII. Solution 50 . regarded as a meromorphic function on C . Determine the order of each critical point. Solution 51 .10 1 2 3 P L K Locate the critical points ana critical values in the extended complex plane of the polynomial f (z) = z 4 2z 2 .VIII. Sketch the set of points z such that Im z 0.5. VIII. What can be said about the critical values of g f ? What can be said about the critical points and critical values f g? Solution 52 .5. and if g is a fractional linear trans- formation.11 1 2 3 P L K Show that if f is a rational function. then f and g f have the some critical points in the extended complex plane C . so that p (z) and q (z) are reactively prime. Solution 53 .VIII. then the number of critical points of f (z) in the …nite plane C is deg (qp0 q 0 p) = deg p + deg q 1. while the order of the critical points at 1 is jdeg p deg qj 1. and d is the larger of the degrees of p (z) and q (z). counting multiplicity.) Show that f (z) has 2d 2 critical points.12 1 2 3 P L K Let f (z) = p (z) =q (z) be a rational function of degree d. (See Exercise 4. Hint if deg p 6= deg q.5.5. in the extended complex plane C . VIII. z) of the equation z 2 2 (cos w)+1 = 0 consists of the graphs of two entire functions z1 (w) and z2 (w) of w.5. The solutions set forms a reducible one-dimensional analytic variety in C2 . Solution 54 . Remark.13 1 2 3 P L K Show that the set of solution points (w. and determine where their graphs meet. Specify the entire functions. Show that Q (z. : : : . Consider the monic polynomial in z whose coe¢ cients are analytic functions in w. w) = 0 determine analytic functions z1 (w) . am 1 (w) be analytic in a neighborhood of w = 0 and vanish at w = 0. there are m distinct solutions of P (z. w) = (z z1 (w)) (z zm (w)) is irreducible. w) arise from subsets of the zj (w)’s in this way. w) is an irreducible factor of P (z. jwj < : Suppose that for each …xed w.7). 0 < jwj < . (c) Suppose that the indices are arranged so that for some …xed k. w) = (z z1 (w)) (z zk (w)) determines a polynomial in z whose coe¢ cients are analytic functions of w for jwj < .VIII. : : : . and that all irreducible factors of P (z. (a) Show that the m roots of the equation P (z. w) = 0. and z1 (w) . 0]g. (d) Show that if f (z) is an analytic function that has a zero of order m at z = 0. the continuation of zj (w) once around w = 0 is zj+1 (w) for 1 j k 1. P (z. zm (w) in the slit disk fjwj < n ( . while the continuation of zk (w) once around w = 0 is z1 (w).5. then the polynomial P (z. w). Use the implicit function theorem (Exercise 4. (b) Glue together branch cuts to form an m sheeted (possibly disconnected) surface over the punctured disk f0 < jwj < g on which the branches zj (w) determine a continuous function. Hint. 1 k n. : : : . w) = z m + am 1 (w) z m 1 + + a0 (w) .14 1 2 3 P L K Let a0 (w) . Show further that the polynomial Q (z. Solution 55 . zm (w) are solutions of the equation w = f (z). w) and P1 (z. w) and P1 (z. w) is less then the degree of P1 (z. and the factorization is unique up to the order of the factors and multiplication of a factor by a meromorphic function in w. w) = A2 (z. Using the division algorithm. (a) Show that D (z. w) = 1). w) and P2 (z. (e) 56 . w) is the greatest common divisor of P0 (z. w) : Let D (z. w). w). and consider the following algorithm. w) be the monic polynomials in z obtained by dividing Pl (z. Pl 1 (z. w) Pl (z. : : : . w)+ P2 (z. w) and Pj+1 (z. w) such that D = AP0 +BP1 . w) divides both P0 (z. …nd polynomials A2 (z. w) such that Pj 1 (z. (b) Show that there are polynomials A (z.5. w) be two such poly- nomials. w) = 0. that if P0 (z. w). w) P1 (z.15 1 2 3 P L K Consider monic polynomial in z of the form P (z. w) and B (z. w) = Aj+1 (z. (c) Show. w) and the degree of P2 (z. w) and P1 (z. w) also divides D (z. w) < deg Pj (z. and each polynomial that divides both P0 (z. w) = Al (z. 0 < jwj < ". w) have no common zeros. then there is " > 0 such that for each …xed w. w) are relatively prime (that is D (z. w) and P1 (z. w) and deg Pj+1 (z. the polynomials P0 (z. w) such that P0 (z. …nding polynomials Aj+1 (z.VIII. where the functions a0 (w) . am 1 (w) are de…ned and meromorphic in some disk centered at w = 0. w) = z m + am 1 (w) + + a0 (w) . until eventually we reach Pl (z. w). w) by the coe¢ cient of the highest power of z. Con- tinue in this fashion. Let P0 (z. w) and P1 (z. in the sense that D (z. w). (d) Show that any polynomial P (z. w) + Pj+1 (z. w) Pj (z. w) as above can be factored as a product of irreducible polynomials. w) and P1 (z. (g) Show that the results of Exercise 14(a)-(c) hold without the supposition that the solutions of P (z. then there is " > 0 such that for each …xed w. the roots of P (z. w) is irreducible. w) can be chosen so that their coe¢ cients are analytic at w = 0. then the irreducible factors of P (z. Solution 57 . w) are analytic at w = 0. (f) Show that if P (z. w) are distinct.Show that if the coe¢ cients of P (z. 0 < jwj < ". w) = 0 are distinct. VIII. and determine the winding number W ( .6. ) for each point not on the path.1 1 2 3 P L K Sketch the closed path (t) = eit sin (2t). 0 t 2t. Solution 58 . VIII.6. 0 t 2 . ) for each point not on the path Solution 59 . and determine the winding number W ( .2 1 2 3 P L K Sketch the closed path (t) = e 2it cos t. Solution 60 .VIII. and suppose f (z) 6= 0 on .6. Show that the increase in arg f (z) around is 2 W (f . 0).3 1 2 3 P L K Let f (z) be analytic on an open set containing a closed path . Solution 61 .VIII.6. (b) Show that (z z0 ) (z z1 ) has an analytic square in D. (a) Show that the increase in the argument of f (z) = (z z0 ) (z z1 ) around any closed curve in D is an even multiple of 2 .4 1 2 3 P L K Let D be a domain. (c) Show by example that (z z0 ) (z z1 ) does not unnecessarily have an ana- lytic cube in D. and suppose z0 and z1 lie in the same connected com- ponent of CnD. 5 1 2 3 P L K Show that if is a piecewise smooth closed curve in the complex plane. and if z0 62 .VIII.6. then Z 1 dz = 0. n 2: (z z0 )n Solution 62 . with trace . and use Exercise 5. ) = 0 for all 62 D.6 1 2 3 P L K Let be a closed path in a domain D such that W ( . Consider the Laurent decomposition at each zk . zk ) Res [f. Solution 63 . zm 2 Dn . zk ] : Hint. : : : . Suppose that f (z) is analytic on D except possibly at a …nite number of isolated singularities z1 .VIII.6. Show that Z X f (z) dz = 2 i W ( . Either use Exercise 6. 64 .VIII. or proceed directly with partial fractions.7 1 2 3 P L K Evaluate Z 1 dz . Hint. 2 i z (z 2 1) where is the closed path indicated in the …gure.6. VIII. (b) Show that if j (t) (t)j < j (t)j for a a t b. ) = W ( . Solution 65 . ) = W ( .6.8 1 2 3 P L K Let (t) and (t). ). be closed paths. then W ( . then W ( . ). (a) Show that if 2 C does not lie on the straight line segment between (t) and (t). for a t b. a t b. 9 1 2 3 P L K Let f (z) be a continuous complex-valued function on the complex plane such that f (z) is analytic for jzj < 1. f (z) 6= 0 for jzj 1. Show that f (z) 6= 0 for jzj < 1.6.VIII. and f (z) ! 1 as z ! 1. Solution 66 . VIII. Solution 67 . Show that every value attained by f (z) on C = C [ f1g is attained by f (z) somewhere on K. that is f (C ) = f (K). and let f (z) be a continuous complex-valued function on the complex plane that is analytic on CnK and at 1.10 1 2 3 P L K Let K be a nonempty closed bounded subset of the complex plane.6. and suppose g ( ) is analytic for in the open upper hand lower half-planes and across the interval ( 1.7. Solution 68 .1 1 2 3 P L K Let f (z) be an entire function. What is the value of the integral when is in the lower half-plane? Justify your answer carefully.VIII. 1) on the real line. Suppose that Z 1 f (x) dx = g ( ) 1 z for in the upper half-plane. ) Reconcile this result with your solution to Exercise 1. Solution 69 .VIII.7.2 1 2 3 P L K Show that Z 1 dx 1 = Log . +1] : 1 x +1 (Note that we use the principal branch of the logarithm here. 2 Cn [ 1. VIII. positively oriented.3 1 2 3 P L K Find the Cauchy integrals of the following functions around the unit circle = fjzj =g.7. (a) z (b) z1 (c) x = Re z (d) y = Im (z) Solution 70 . and f1 ( ) = F ( ) for j j > r. Remark The formula f (z) = f0 (z) + f1 (z) re‡ects the jump theorem for the Cauchy integral of f (z) around circles jzj = r. Show that f0 ( ) = F ( ) for j j < r.7. Solution 71 . and let F ( ) be the Cauchy integral of f (z) around the circle jzj = r. and let f (z) = f0 (z)+ f1 (z) be the Laurent decomposition of f (z).) Fix r between and . Show further that f0 ( ) = F ( ) and f1 ( ) = F+ ( ).1.VIII.4 1 2 3 P L K Suppose f (z) is analytic on an annulus f < jzj < g. (See Section VI. 7.VIII. then Z ZZ @g g (z) f (z) dz = 2i F (z) dx dy: C @z Hint.1) of a continuous function f (z) on . Show that if g (z) is a smooth function on the complex plane that is zero o¤ some bounded set. Recall Pompeiu’s formula (Section IV. 72 . and let F ( ) be the Cauchy integral (7.8).5 1 2 3 P L K Let be a piecewise smooth curve. 7.VIII. Solution 73 . Explain.6 1 2 3 P L K Determine whether the point z lie inside or outside. ) for 2 Cn . 6= D0 . extend h (z) to a continuous function on C . Find at continuous determination h (z) of log (z z0 ) on . Hint. For z0 = (t0 ) 2 . and de…ne f (z) = z z0 on the component Cn containing z0 . Hint. (e) 74 . and component U of Cn . (c) Show that for any z0 = (t0 ) 2 . ) = W ( . and let be the simple closed curve in D0 obtained by the following the segment of in D0 from z0 to z1 and returning to z0 along the broken line segment. Consider the increase in the argument of f (z) around circles centered at z0 . You may use the Tietze extension theorem. where I is a small open parameter interval containing t0 . Show that W ( .7. (d) With notation as in (b). (a) Show that each component Cn has boundary . b] to the complex plane. that a continuous real-valued function on a closed subset of the complex plane can be extended to a continuous real- valued function on the entire complex plane.VIII. there are points z1 = (t1 ) and z2 = (t2 ) such that the image of the parameter segment between t1 and t2 is contained in D0 and such that z1 and z2 can be joined by a broken line segment in U \ D0 . Prove the Jordan curve theorem for a simple closed curve by …lling in the following proof outline. that is Cn is connected. and f (z) = eh(z) on the remainder of C n .7 1 2 3 P L K A simple arc in C is the image of a continuous one-to-one function (t) from a closed interval [a. ) = 0 and W ( . let be the simple curve obtained by replacing the segment of in D0 between z0 and z1 by the broken line segment in U \ D0 between them. any small disk D0 containing z0 . apply the preceding exercise to the simple arc n (I). Show that a simple arc in C has a connected complement. (b) Prove the Jordan curve theorem in the case where contains a straight line segment. Suppose z0 belongs to a bounded complement of Cn . Solution 75 . ) = 1 for in each bounded component of Cn . show that Cn has at least two components and that W ( . show that W ( . ) = 0 for in any other component of Cn . (f) By taking U in (c) to be a bounded component of Cn .Using (b) and (d). 1]. the upper half-plane with a vertical slit from i to 2i. Solution 76 . the complex plane with an interval deleted. (d) D = Cn [ 1. (c) D = Cn [0.1 1 2 3 P L K Which of the following domains in C are simply connected? Justify your answers. (a) D = fIm z > 0g n [0. the complex plane slit along the positive real axis. 2i]. the upper half-plane with a vertical slit from 0 to i. +1].VIII.8. (b) D = fIm z > 0g n [i. i]. VIII. Hint. Solution 77 .2 1 2 3 P L K Show that a domain D in the extended complex plane C = C [ f1g is simply connected if and only if its complement C nD is connected. If D = C . move a point in the complement of D to 1. …rst deform a given closed path to one that does not cover the sphere. then deform it to a point by pulling along arcs of great circles.8. If D 6= C . 0. Solution 78 .3 1 2 3 P L K Which of the following domains in C are simply connected? Justify your answers. (b) D = C n f 1.8.VIII. the extended plane with an interval deleted. 1g. (a) D = C n [ 1. 1]. the thrice-punctured sphere. 8.4 1 2 3 P L K Show that a domain D in the complex plane is simply connected if and only if any analytic function f (z) on D does not vanish at any point of D has an analytic logarithm on D.VIII. consider the function Z z 0 f (z) G (z) = dw: z0 f (w) Solution 79 . Hint if f (z) 6= 0 on D. Solution 80 .8.5 1 2 3 P L K Show that a domain D is simply connected if and only if any analytic function f (z) on D that does not vanish at any point of D has an analytic square root on D. Show that this occurs if and only if for any point z0 62 D the function z z0 has an analytic square root on D.VIII. 8.6 1 2 3 P L K Show that a domain D is simply connected if and only if each continuous function f (z) on D that does not vanish at any point D has continuous logarithm on D.VIII. Solution 81 . Show that each connected component C nE is simply connected.7 1 2 3 P L K Let E be a closed connected subset of the extended complex plane C . Solution 82 .8.VIII. 8 1 2 3 P L K Show that simple connectivity is a "topological property" that is.VIII. if U and V are domains. then U is simply connected if and only if V is simply connected.8. Solution 83 . and ' is a continuous map of U onto V such that ' 1 is also continuous. 8. Suppose also that f (D) is simply connected. and that there is a branch g (w) of f 1 that is analytic on w0 = f (z0 ) and that can be continued analytically along any path in f (D) starting at w0 .9 1 2 3 P L K Suppose that f (z) is analytic on a domain D. and f 0 (z) has no zeros on D. Solution 84 . Show that f (z) is one-to-one on D.VIII. : : : . . where 1 . Show that if h ( ) is a continuous integer-valued function on C nD such that h (1) = 0. P We de…ne the winding number of about to be W ( . ) = kj W j .VIII.10 1 2 3 P L K We P de…ne an integral 1 cycle in D to be an expression of the form = kj j . m are closed paths in D and k1 . ) = h ( ) for all 2 CnD. 2 CnD. : : : .8. then there is an integral 1-cycle on D such that W ( . Solution 85 . km are integers. m . such that P U together with its boundary is contained in D. Show that the 1-cycle @U = j is homologous to zero in D.11 1 2 3 P L K An integral 1-cycle is homologous to zero i D if W ( . : : : . oriented positively with respect to U . Solution 86 .VIII. ) = 0 whenever 62 D. Let U be a bounded domain whose boundary consists of a …nite number of piecewise smooth closed curves 1 .8. where the kj ’s are integers and 0 is homologous to zero in D. Solution 87 . m such thatPevery integral 1-cycle can be expressed uniquely in the form = 0 + kj j . Show that there are m piecewise smooth closed curves 1 .12 1 2 3 P L K Let D be a domain in C such that C nD consists of m + 1 disjoint closed con- nected sets. : : : . Remark. The j ’s form a homology basis for D.VIII.8.