(1) Let f(z) be the principal branch of z 4i. (a) Find f(i). Solution. f(i) = exp(4i Log(i)) = exp(4i(π/2)) = e 2π. (b) Show that

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1 Let fz be the principal branch of z 4i. a Find fi. Solution. fi = exp4i Logi = exp4iπ/2 = e 2π. b Show that fz fz 2 fz z 2 fz fz 2 = λfz z 2 for all z, z 2 0, where λ =, e 8π or e 8π. Proof. We have = exp4ilog z + Log z 2 Logz z 2 = exp4ilog z + Log z 2 Logz z 2 = exp 4Arg z + Arg z 2 Argz z 2 = exp 8nπ for some integer n. And since π < Arg z π, π < Arg z 2 π and π < Argz z 2 π, 3π < Arg z + Arg z 2 Argz z 2 < 3π and hence 3π < 2nπ < 3π. So n =, 0 or. Correspondingly, fz fz 2 = λfz z 2 with λ = e 8nπ = e 8π, or e 8π. 2 Let fz be an entire function satisfying fz z + for all z. Show that fz az + b for some constants a, b C satisfying a and b. Proof. Since fz is entire, fz = for all z. By Cauchy Integral Formula, f n 0 = n! 2πi f n 0 z n n! z =R fz dz zn+ for all n and R > 0. Since fz z +, fz z n+ z + z = R + n+ R n+

2 2 for z = R. Therefore, fz dz R + 2πi zn+ 2πR = R + 2π R n+ R. n z =R And since R R + /R n = 0 for n 2, fz dz = 0 R 2πi zn+ z =R and hence f n 0 = 0 for all n 2. And since R R + /R n = for n =, fz R 2πi z dz 2 z =R and hence f 0. Therefore, fz = f0 + f 0z = az + b with a = f 0. Finally, by az +b z +, we obtain that b by setting z = 0. 3 Let z 3 fz = z 2 5z + 4 Find the Laurent series of fz in each of the following domains: a < z < 4 b < z 2 < 2 Solution. We write fz as a sum of partial fractions: z 3 z 2 5z + 4 = z + 5 3z z 4. For < z < 4, z 3 z 2 5z + 4 = z z /z 3 = z + 5 3z z 6 n 3 = 3z n 3 3 z = n= n=2 3z z n n 3 4 n 2 n=2 z/4 z n 4 n z n 3 4 n 2

3 z 3 For < z 2 < 2, z 2 5z + 4 = z z z 2 2 = z z 2 + /z z 2/2 n = z z 2 z 2 n n+ 3 2 n 5 = n= 4 Do the following: π a Find tan 4 + i. n 3z 2 n z 2 Solution. π tan 4 + i = i eiπ/4+i e iπ/4+i e iπ/4+i + e iπ/4+i n=2 z 2 n 3 2 n 5 = i e cosπ/4 + i sinπ/4 ecosπ/4 i sinπ/4 e cosπ/4 + i sinπ/4 ecosπ/4 i sinπ/4 = e + e + ie e e + e ie e = 2e2 e e4 e 4 + i b Find the first three terms i.e. up to z 2 of the Taylor series of tan z at z = 0 and also its radius of convergence. Solution. Let fz = tan z. Then f n 0 fz = z n n! for z < R, where fz is analytic in z < R. Since tan z is odd, f n 0 = 0 for n even. And since tan z = sec 2 z, f 0 =. Therefore, the first three terms are 0 + z + 0z 2 = z The radius of convergence of the Taylor series is the largest R such that tan z is analytic in z < R. Since tan z is analytic in {z nπ + π/2 : n integers}, the largest R with this property is π/2. So the radius of convergence is π/2. 3

4 4 c Show that and tanz 2 + tanz 2 + cos 2 x sinh 2 y for all z C, where x = Rez and y = Imz. Here we take the right hand sides to be if cos x = 0 or sinh y = 0. Proof. We know that tan 2 z = sec 2 z and cosz 2 = cosx + yi 2 = cosx cosyi sinx sinyi 2 = cosx coshy i sinx sinhy 2 = cos 2 x cosh 2 y + sin 2 x sinh 2 y = cos 2 x + sinh 2 y + sin 2 x sinh 2 y = cos 2 x + cos 2 x + sin 2 x sinh 2 y = cos x 2 + sinh y 2. Therefore, tanz 2 = cos 2 z + cosz = + 2 cos x 2 + sinh y. 2 And since cos 2 x 0 and sinh 2 y 0, and tanz 2 + tanz 2 + cos 2 x sinh 2 y. d Let C N be the boundary of the square { x Nπ, y Nπ}, where N is a positive integer. Show that tanz dz = 0. N z 2 C N

5 5 Proof. When x = ±Nπ, z Nπ and tan z 2 + cos 2 x = 2. When y = ±Nπ, z Nπ and tan z 2 + sinh 2 y = + sinh 2 Nπ < 2. Therefore, tanz 2 z 2 N 2 π 2 for z C N. Consequently, C N tanz z 2 dz 2 N 2 π 8Nπ = Nπ. And since N 8 2/Nπ = 0, N C N tanz z 2 dz = 0. 5 Compute the integral Solution. Obviously, 0 0 x sin x x dx. x sin x x 2 + dxdx = x sin x 2 2 x 2 + dx 2 = 2 Im ze iz z 2 + dz 2 = R 2 Im ze iz R z 2 + dz 2 R

6 6 We integrate along the closed contour going from R to R and then the semicircle C R = { z = R, Imz 0} counterclockwise. Then R ze iz R z 2 + dz + ze CR iz 2 z 2 + dz 2 ze iz ze iz = 2πi Res z=i z 2 + = 2πi 2 z + i 2 z=i e iz + ize iz = 2πi 2zeiz = z + i 2 z + i z=i πi 3 2e by Cauchy Integral Theorem. For z on C R, e iz = e y, z 2 + z 2 = R 2 and hence z ze iz z R R 2 2 for z C R. Therefore, And since we conclude that Thus, and z 2 + dz 2 CR ze iz R R πr 2 R 2 2 = 0, CR ze iz πr2 R 2 2. dz = 0. z R ze iz πi dz = R R z e 0 x sin x x = π 4e. 6 For each of the following complex functions, do the following: find all its singularities in C; write the principal part of the function at each singularity; for each singularity, determine whether it is a pole, a removable singularity, or an essential singularity; compute the residue of the function at each singularity.

7 a fz = cos z 2 Solution. The function has singularities at {cos z = 0} = {z = kπ + π/2 : k integers}. At z = kπ + π/2, we let w = z kπ π/2 and then 7 cos z = 2 sin w = 2 = w 2 n w 2n 2n +! n w 2n+ 2n +! = n+ w 2n w 2 2n +! n= = n+ w 2n w 2 2n +! m=0 n= = + a w 2 n w n. n=2 m So the principal part at kπ + π/2 is z kπ π/2 2, the function has a pole of order 2 at kπ + π/2 and Res z=kπ+π/2 fz = 0. b fz = z 3 exp z

8 8 Solution. The function has a singularity at 0 where z 3 exp = z 3 z n!z n = = + n!z n n!z n 3 n= n!z n n=4 n!z n 3 z3 + z 2 + z = z 3 z 2 z n!z n n + 3!z n n= n= = z 3 z 2 z n! z n. n + 3! So the principal part is n= n! n + 3! n= z n, the function has an essential singularity at 0 and Res z=0 fz =! 4! = c fz = sin z z 200 Solution. The function has a singularity at z = 0 where sin z z = 200 z 200 = 004 = n z 2n+ 2n +! n z 2n n +! n z 2n n +! So the principal part is n z 2n 2009, 2n +! n z 2n n +! the function has a pole of order 2009 at 0 and Res z=0 fz = ! = 2009!.

9 ez d fz = z 2 Solution. The function has two singularities at and. At z =, e z z = e z 2 z + z = e z a n z + n n= = 2ez + + a n z + n. n= So the principal part at is 2ez +, the function has a simple pole at and At z =, e z z 2 = z Res z= fz = 2e. = z e z e z a n z n n= e = 2z a n z n. n= So the principal part at is e 2z the function has a simple pole at and Res z= fz = e 2. 9

10 0 7 Compute the following contour integrals. a C expsin zdz, where C is the ellipse 2x 2 + y 2 = oriented counterclockwise. Solution. Since fz = e z and gz = sin z are entire, fgz = expsin z is entire. And since C is a closed curve, expsin zdz = 0 by Cauchy Integral Theorem. C b z 20 C z 20 + z z dz, where C is the circle z = 2 oriented counter-clockwise. Solution. We first show that all zeroes of z 20 + z z lie in z < 2. Otherwise, suppose that z 20 + z z = 0 for some z 2. Then + z + z 2 + z 20 = 0. But + z + z 2 + z 20 z z 2 z > 0 20 for z 2. Contradiction. So all zeroes of z 20 + z z lie in z < 2. Therefore, z 20 /z 20 + z z is analytic in z 2. It follows that z 20 z 20 + z z dz C z 20 = 2πi Res z= z 20 + z z z 20 = 2πi Res z=0 z 2 z 20 + z z = 2πi Res z=0 z 2 + z + z 2 + z 20 = 2πi = 2πi. + z + z 2 + z z=0 20