Solution Set III for 18.075–Fall 2004

SOLUTION SET III FOR 18.075–FALL 2004 10. Functions of a Complex Variable 10.7. Taylor Series. . 48. Obtain each of the following series expansions by any convenient method: (1) � � z 2n sin z z2 z4 =1− + − ... = (−1)n z 3! 5! (2n + 1)! n=0 (2) (3) (4) � � cosh z − 1 1 z2 z4 z 2n = + + + ... = z2 2! 4! 6! (2n + 2)! n=0 8 5 ez = 1 + 2z + z 2 + z 3 + ... 1−z 2 3 a2 z+a (z + a)2 = 1 + 2 + 3 + ... z2 a a2 (|z | < ∼), (|z | < ∼), (|z | < 1), (|z + a| < |a|). Solution. (a) We repeat what we did in class. For |z | < ∼, sin z = z − z3 z5 z2 z4 + − ... = z(1 − + − ...) 3! 5! 3! 5! Since the series in parenthesis is absolutely convergent (by the same criterion used to prove the absolute convergence of the Taylor series of e iz and sin z) we can divide by z both terms of the above equality. Thus we get � � z 2n sin z z2 z4 =1− + − ... = (−1)n . z 3! 5! (2n + 1)! n=0 From the uniqueness of the Taylor series we have that the series on the right hand side is the Taylor series of sin z/z. (b) From the formulas cos z = (eiz + e−iz )/2 and cosh z = cos(iz), we get for |z| < ∼: cosh z − 1 = 1 z2 z2 z4 + − ... = z 2 ( + + ...) 2! 4! 2! 4! Date: September 25, 2002. 1 2 SOLUTION SET III FOR 18.075–FALL 2004 Since the series in parenthesis is absolutely convergent (by the same criterion used to prove the absolute convergence of the Taylor series of e z and cosh z) we can divide by z 2 both terms of the above equality. Thus we get � � z2 z4 z 2n cosh z − 1 1 = + + + ... = z2 2! 4! 6! (2n + 2)! n=0 From the uniqueness of the Taylor series we have that the series on the right hand side is the Taylor series of (cosh z − 1)/z 2 . (c)From the definition of ez we have that, for |z | < ∼, ez = 1 + z + z2 z3 + + ....., 2! 3! while, for |z | < 1, 1 = 1 + z + z 2 + .... 1−z Since the above series are absolutely convergent, we can multiply them term by term and we obtain a series which converges absolutely to the product e z /(1 − z) in the disk |z | < 1. Therefore, in |z | < 1, we have 1 5 ez = (1 + z + z 2 + ....) + (z + z 2 + z 3 + ...) + (z 2 + z 3 + ...) = 1 + 2z + z 2 + ... 1−z 2! 2 From the uniqueness of the Taylor series, the series on the right hand side is the Taylor series of ez /(1 − z). (d) Observe�that a2 /z 2 = 1/[1 − (z + a)/a]2 . Therefore, using the geometric series � n 2 1/(1 − w) = n=0 w�, |w | < 1 and differentiating in w term by term as 1/(1 − w) = � n (d/dw)[1/(1 − w)] = n=0 (n + 1) w , |w| < 1 for w = (z + a)/a, we get: � � � z+a n a2 = (n + 1) , |z + a| < |a|. z2 a n=0 Then z+a (z + a)2 a2 =1+2 +3 + ... 2 z a a2 (|z + a| < |a|). 10.8. Laurent Series. . 51. Expand the function f (z) = 1/(1 − z) in each of the following series: (a) a Taylor series of powers of z for |z | < 1; (b) a Laurent series of powers of z for |z | > 1; (c) a Taylor series of powers of z +1 for |z +1| < 2, by first writing f (z) = [2−(z +1)] −1 = 1 −1 2 [1 − (z + 1)/2] ; (d) a Laurent series of powers of z + 1 for |z + 1| > 2, by first writing f (z) = −[1/(z + 1)]/[1 − 2/(z + 1)]. SOLUTION SET III FOR 18.075–FALL 2004 3 Solution. (a)In the disk |z | < 1, we have � � 1 = zn, 1−z n=0 which is the familiar geometric series. (b)In |z | > 1 we have, 1 1−z 1/z 1 1 =− = 1/z − 1 z (1 − 1/z) � � � � � 1 n+1 1� 1 n =− = − z n=0 z z n=0 = 1 1 1 = − − 2 − 3 − ..... z z z where we have used the geometric series expansion for 1/(1 − 1/z) in |1/z | < 1 or, equiva­ lently, |z | > 1. (c)Using the geometric series expansion for the function 1/[1−(z +1)/2] in |(z +1)/2| < 1 or, equivalently, |z + 1| < 2, we have: � 1 1 1 1� = = 1−z 2 [1 − (z + 1)/2] 2 n=0 z+1 2 �n = � � (z + 1)n n=0 2n+1 . (d)Using the geometric series expansion for the function 1/[1−2/(z +1)] in |2/(z +1)| < 1 or, equivalently, |z + 1| > 2, we get: � 1 1 1 1 � =− =− 1−z z + 1 [1 − 2/(z + 1)] z+1 n=0 2 z+1 �n =− � � n=0 2n . (z + 1)n+1 52. Expand the function f (z) = 1/[z(1 − z)] in a Laurent (or Taylor)series which converges in each of the following regions: (a) 0 < |z | < 1, (b) |z | > 1, (c) 0 < |z − 1| < 1, (d) |z − 1| > 1, (e) |z + 1| < 1, (f) 1 < |z + 1| < 2, (g) |z + 1| > 2. Solution.(a)Using the geometric series expansion for 1/(1 − z) in 0 < |z | < 1 we get: 1 z(1 − z) = � � � 1� n z = zn z n=0 = n=−1 1 + 1 + z + z 2 + .... z 4 SOLUTION SET III FOR 18.075–FALL 2004 (b)Using part (b) of exercise 51 we get, in |z | > 1, 1 z(1 − z) � � � 1� 1 1 = − =− n+2 z n=0 z n+1 z n=0 = − 1 1 1 − 3 − 4 − ... 2 z z z (c)Using the geometric series expansion for 1/[1 − (1 − z)] in 0 < |z − 1| < 1, we get: 1 z(1 − z) = � � � 1 1 1 � n = (1 − z) = (1 − z)n−1 (1 − (1 − z)) (1 − z) 1−z n=0 = n=0 1 + 1 + (1 − z) + (1 − z)2 + ... 1−z (d)Using the geometric series expansion for 1/[1 − 1/(1 − z)] in 1/|z − 1| < 1 or, equiva­ lently, |z − 1| > 1, we get: 1 z(1 − z) = − � � � � 1 1 1 1 1 = − = − 2 2 n (1 − z) (1 − z)n+2 [1 − 1/(1 − z)] (1 − z) (1 − z) n=0 n=0 1 1 1 + ... − − = − 2 3 (1 − z) (1 − z)4 (1 − z) (e)Using the geometric series expansion for 1/[1 − (z + 1)] in |z + 1| < 1 and the one for 1/[1 − (z + 1)/2] in |z + 1| < 2, we get in |z + 1| < 1 (overlap region of the two disks of convergence): � � � � 1 −1 1 1 1 � z+1 n n = + =− = (z + 1) + z(1 − z) 1 − (z + 1) 2 (1 − (z + 1)/2) 2 2 n=0 n=0 � � � � � 1 3 7 1 n+1 = − (z + 1)n = − − (z + 1) − (z + 1)2 + ... 1− 2 2 4 8 n=0 (f)Using the geometric series expansion for 1/(1 − 1/(z + 1)) in |z + 1| > 1 and the one for 1/(1 − (z + 1)/2) in |z + 1| < 2, we get in 1 < |z + 1| < 2 (intersection of the two disks of convergence): 1 z(1 − z) = = 1 1 1 1 + = (z + 1) (1 − 1/(z + 1)) 2 (1 − (z + 1)/2) � � � 1 1 � z +1 n 1 � + = z + 1 n=0 (z + 1)n 2 n=0 2 � � � 1 � z+1 n 1 = + ( ) (z + 1)n+1 2 n=0 2 n=0 (g)Using the geometric series expansion for 1/[1 − 1/(z + 1)] in |z + 1| > 1 and the one for 1/(1 − 2/(z + 1)) in |z + 1| > 2, we get in |z + 1| > 2 (intersection of the two disks of SOLUTION SET III FOR 18.075–FALL 2004 5 convergence): 1 z(1 − z) = = = 1 1 1 1 − = (z + 1) (1 − 1/(z + 1)) (z + 1) (1 − 2/(z + 1)) �n � � 1 1 � 2 1 � − = z + 1 n=0 (z + 1)n z + 1 n=0 z + 1 � � n=0 � � = − � � 1 2n − = n+1 (z + 1) (z + 1)n+1 n=0 n=0 (2n − 1) 1 1 1 =− −3 − ... (z + 1)n+1 (z + 1)2 (z + 1)3 10.9. Singularities of Analytic Functions. . 61. Locate and classify the singularities of the following functions: 2 (a) z 2z+1 , (b) z 31+1 , (c) log(z 2 + 1), (d) (z 2 − 3z + 2) 3 , (e) tan z, (f) tan−1 (z − 1). Solution. (a)We have z 1 1 1 = ( + ). z2 + 1 2 z−i z+i z is analytic Then z 2z+1 has a simple pole at z + i = 0, i.e., z = −i, since (z + i) z 2z+1 = z−i z at z = −i with value ≥= 0. Similarly, z 2 +1 has a simple pole at z − i = 0, i.e., z = i, since 1 ≥ 0. z+i is analytic at z = i with value = (b) We have 1 1 1 1 � � . · = · 3 1+i 1−i 3 3 z +1 z+1 ) ) z−( z−( So 1 z 3 +1 has simple poles at z = −1, z = (c) We have � 1+i 3 2 , 2 and z = � 1−i 3 2 . 2 log (z 2 + 1) = log [(z + i)(z − i)] = log (z + i) + log (z − i), where we possibly add integral multiples of 2�i to the right-hand side. We see that log (z + i) in the right-hand side has a branch point at z = −i, and log (z − i) is ana­ lytic at z = −i. So log (z 2 + 1) has a branch point at z = −i. Similarly, since log (z − i) has a branch point at z = i and log (z + i) is analytic at z = i, log (z 2 + 1) has a branch point at z = i. (d) We have 2 (z 2 − 3z + 2) 3 = [(z − 2)(z − 1)]2/3 = w2/3 , 2 w = (z − 2)(z − 1). So (z 2 − 3z + 2) 3 has branch points at w = (z − 2)(z − 1) = 0, i.e., at z = 2 and z = 1. (e) We have sin z tan z = . cos z 6 SOLUTION SET III FOR 18.075–FALL 2004 So tan z has poles at cos z = 0. Hence, the singularities of tan z are z = z n = n� + �2 , where n:integer, and each of these singularities is a pole. Note that (z − z n ) tan z is analytic in a vicinity of z = zn . In particular, lim [z − (n� + z�n�+ �2 � )] tan z = −1. 2 It follows that all these poles are simple. (f) First, we find a formula for tan−1 z (check with p. 550 of textbook). Let w = tan −1 z. sin w e2iw −1 2iw we get e2iw = 1+iz or Then z = tan w = cos w = −i e2iw +1 . By solving with respect to e 1−iz � z−i 1 1 log = [log(z − i) − log(z + i)], w = tan−1 z = 2i z+i 2i with the possible addition of integral multiples of 2�i to the right-hand side. It follows that tan−1 z has branch points at z = ±i. Hence, tan −1 (z − 1) has branch points at z − 1 = ±i, or z = 1 ± i. 62. Show that the function (5) f (z) = has a simple pole at the origin. cosh z − 1 sinh z − z Proof. Clearly, and � � � � z 2n z 2n 2 =z , cosh z − 1 = (2n)! (2n + 2)! n=1 n=0 sinh z − z = Let � � � � z 2n z 2n+1 = z3 . (2n + 1)! (2n + 3)! n=1 n=0 g(z) = and h(z) = � � z 2n , (2n + 2)! n=0 � � z 2n . (2n + 3)! n=0 Then both g(z) and h(z) are analytic functions in the complex plane (set C) and have nonzero values at the origin, g(0) ≥= 0 ≥= h(0). It’s clear that cosh z − 1 = z 2 g(z), and g(z ) . Thus, f (z) has a simple pole at the origin. sinh z − z = z 3 h(z). So f (z) = 1z h(z) SOLUTION SET III FOR 18.075–FALL 2004 7 10.12. Residues. . 78. Calculate the residues of the following functions at each of the poles in the finite part of the plane: z 1 sin z sin z 1+z 2 1 eaz ez−1 −1 (a) z 2e+a2 , (b) z 4 −a , 4 , (c) z 2 , (d) z 3 , (e) z(z−1)2 , (f) (z 2 +a2 )2 , (g) 2z 2 −5z+2 , (h) 1−z 2 (i) 1−cos az , z9 (j) sinh z cosh z−1 , (k) z , sin2 z (l) (1−cos z)2 . z7 Solution. In the following, we use the notation Res(a) = Resz=a [f (z)] � Res[f (z), a]. (a) We have the formula So ez z 2 +a2 ez ez = . z 2 + a2 (z + ai)(z − ai) has simple poles at z = ±ai. Res[ Res[ (b) We have z4 So 1 z 4 −a4 ez ez eai | , ai] = = , z=ai z 2 + a2 z + ai 2ai ez ez e−ai , −ai] = | = − . z=−ai 2ai z 2 + a2 z − ai 1 1 = . 4 −a (z + a)(z − a)(z + ai)(z − ai) has simple poles at ±a, ±ai. Res[ 1 1 1 |z=a = , a] = , z 4 − a4 (z + a)(z + ai)(z − ai) 4a3 1 1 1 , −a] = |z=−a = − 3 , z 4 − a4 (z − a)(z + ai)(z − ai) 4a 1 1 1 |z=ai = − 3 , Res[ 4 , ai] = 4 z −a (z + a)(z − a)(z + ai) 4ia 1 1 1 , −ai] = . Res[ 4 |z=−ai = 4 z −a (z + a)(z − a)(z − ai) 4ia3 Res[ (c) We have So sin z z2 sin z 1 sin z = · . 2 z z z has a simple pole at the origin, and Res[ sin z sin z , 0] = |z=0 = 1. z2 z (d) We have 1 sin z sin z = 2 · . 3 z z z 8 So SOLUTION SET III FOR 18.075–FALL 2004 sin z z3 has a pole of order 2 (i.e., double pole) at the origin, and � � 2n−2 2n+1 � sin z 1 � n z n z = (−1) = (−1) . z3 z3 (2n + 1)! (2n + 1)! n=0 Note that the right-hand side has a n=0 z −2 term but no z −1 term. Thus, sin z Res[ 3 , 0] = 0. z (e) We have So 1+z 2 z(z−1)2 1 + z2 1 1 = (z 2 + 1) · · . 2 z(z − 1) z (z − 1)2 has a simple pole at the origin and pole of order 2 at z = 1. Res[ 1 + z2 1 , 1] = ((z 2 + 1) · )|z=0 = 1, 2 z(z − 1) (z − 1)2 2 2 1+z d((z − 1)2 z(z−1) 2) d( 1+z 1 1 + z2 z ) , 0] = Res[ |z=1 = |z=1 z(z − 1)2 (2 − 1)! dz dz �� 1 + z 2 �� = 2− = 0. � z2 z=1 (f) We have 1 1 1 = · . 2 2 2 2 (z − ai)2 (z + a ) (z + ai) So (z 2 +1a2 )2 has two poles of order 2 (double poles). One pole is at z = ai and the other one is at z = −ai. d(z + ai)−2 1 1 , ai] = , Res[ 2 |z=ai = 2 2 (z + a ) dz 4a3 i Res[ d(z − ai)−2 1 1 |z=−ai = − 3 . , −ai] = 2 2 2 (z + a ) dz 4a i (g) We have Thus, eaz 2z 2 −5z+2 eaz eaz . = 2z 2 − 5z + 2 2(z − 2)(z − 21 ) has simple poles at z = 2 and z = 12 . Res[ eaz eaz e2a | , 2] = = , z=2 2z 2 − 5z + 2 3 2(z − 21 ) a (h) We have 1 eaz e2 eaz , ] = Res[ 2 |z= 1 = − . 2 3 2z − 5z + 2 2 2(z − 2) � � (z−1)n � −1 + � ez−1 − 1 (z − 1)n 1 n=0 n! · = = ]. [ 1 − z2 (1 − z)(1 + z) 1+z (n + 1)! n=0 SOLUTION SET III FOR 18.075–FALL 2004 9 � (z−1)n Note that � n=0 (n+1)! is analytic for every (finite) z. Thus, the given function is analytic at z = 1, with residue equal to 0 at z = 1. Since � � (z − 1)n n=0 it’s clear that ez−1 −1 1−z 2 (n + 1)! |z=−1 = ez−1 − 1 e−2 − 1 ≥= 0, |z=−1 = 1−z 2 has a simple pole at z = −1, and Res[ e−2 − 1 ez−1 − 1 , −1] = . 1 − z2 2 1−cos az z9 � � (i) Evidently, the only pole of is at z = 0. We have � � (−1)n (az)2n (−1)n+1 a2n z 2n−9 1 − cos az −9 = z (1 − ) = . z9 (2n)! (2n)! n=0 n=1 Specially, the coefficient of z −1 is −a8 8! . So −a8 1 − cos az , 0] = . z9 8! (j) It’s easy to check that cosh z − 1 = 0 for z = i2n� with n: integer. Thus, the only possible singularities are poles at z = z n = i2n�. In order to examine what sort of poles these are, let t = z − zn . Then, �� �� t2n+1 t2n sinh z sinh t n=0 (2n+1)! n=0 (2n+1)! −1 = t �� , = = � t2n t2n cosh z − 1 cosh t − 1 −1 + � n=0 n=0 Res[ (2n)! and t2n n=0 (2n+1)! |t=0 �� t2n n=0 (2n+2)! �� So t = 0, i.e., z = zn = i2n�, is a simple pole of (2n+2)! = 2 = ≥ 0. sinh z cosh z−1 , and sinh z , 0] = 2. cosh z − 1 (k) sin z has zeros at z = k�, where k: integer, and each of these zeros is simple. Accordingly, sinz2 z has a simple pole at z = 0 and poles of order 2 at z = k�, where k ≥= 0. Res[ Res[ z z lim z 2 = 1, 2 , 0] = z�0 sin z sin z and, for k ≥= 0 (k: integer), z d z ((z − k�)2 2 ) = 1. lim 2 , k�] = z�k� dz sin z sin z Alternatively, by setting t = z − k�, Res[ z t + k� t + k� k� 1 � 2 + . = � 2 2 2 t t t sin z sin t 10 SOLUTION SET III FOR 18.075–FALL 2004 For k = 0, the t−2 term vanishes and, hence, z = 0 is a simple pole. For k = ≥ 0, the t −2 term is nonzero and, hence, z = k�, k = ≥ 0, is a double pole. In each case, the coefficient of the t−1 term is 1. Thus, the residue is 1. z)2 is z = 0. Let (l) The only possible pole of (1−cos z7 g(z) = cos z − 1 . z2 Then g(z) is analytic, g(0) = − 12 ≥= 0, g → (0) = 0 , and g →→ (0) = 1 12 . We have (1 − cos z)2 = z −3 (g(z))2 . z7 So (1−cos z)2 z7 has a pole of order 3 at z = 0, and 1 d2 (g(z))2 |z=0 (3 − 1)! dz 2 1 d(2g(z)g → (z)) |z=0 = 2 dz = ((g → (0))2 + g(0)g →→ (0)) −1 −1 1 . = = 2 12 24 Alternatively, we expand this function in Laurent series as follows: � (1 − cos z)2 1 cos z − 1 2 = 3 z7 z z2 �2 � 1 1 2 1 1 2 = z −3 − + − z + ... z + ... = z −3 2 24 4 24 Res[ (1 − cos z)2 , 0] = z7 1 1 1 1 − + ... 4 z 3 24 z Clearly, the residue is −1/24. I personally find this alternative way faster! Note: In the above, we use the symbol � to mean “approximately equal to” in cases where we neglect the other terms in Laurent series. = 79. If f (z) has a pole of order m at z = a, prove that � M −1 � 1 d M {(z − a) f (z)} Res(a) = (M − 1)! dz M −1 z=a for any positive integer M such that M ≡ m. Solution. By definition of the point z = a as a pole of order m, f (z) admits the Laurent expansion c−1 c−m+1 c−m + + ... + + c0 + c1 (z − z0 ) + . . . , f (z) = (z − z0 )m (z − z0 )m−1 z − z0 SOLUTION SET III FOR 18.075–FALL 2004 11 where c−1 is the residue. It follows that (z−z0 )M f (z) = c−m (z−z0 )M −m +c−m+1 (z−z0 )M −m+1 + . . . +c−1 (z−z0 )M −1 +c0 (z−z0 )M + . . . We see that, for M ≡ m, this last expansion is a Taylor series. In particular, the coefficient c−1 multiplies (z − z0 )M −1 and thus must be equal to the derivative of order M − 1 of (z − z0 )M f (z) at z = a divided by (M − 1)!: � M −1 � 1 d M {(z − a) f (z)} c−1 = . (M − 1)! dz M −1 z=a 80. (a) If f (z) is that branch of log z for which 0 � � P < 2�, determine the sum of the residues of f (z)/(z 2 + 1) at its poles. (b) Proceed as in part (a) when the restriction on � P is −� < �P � �. Solution. The denominator in f (z)/(z 2 + 1) vanishes at z = ±i. Thus, the possible poles (z) is analytic in a vicinity of z = ±i in = fz±i are z = ±i. Because the function (z ∞ i) zf2(z) +1 the respective branch of log z, and its value at z = ±i is nonzero, these poles are simple. The value of log z at z = rei� is log z = log r + i(� + 2k�) where k: integer and log r (r > 0) is the usual logarithm for real functions. The principal value is found by setting k = 0 and � = �P . (a) At z = i = ei�/2 , �P = �/2 while at z = −i = e−i�/2 , �P = −i�/2 + 2� = 3�/2. Hence, log i = i�/2 and log(−i) = i3�/2. It follows that f (z) f (z) log i � , z = i] = lim[(z − i) 2 ]= = , 2 z�i z +1 z +1 2i 4 3� f (z) log(−i) f (z) , z = −i] = lim[(z + i) 2 ]= =− . Res[ 2 z�i z +1 z +1 −2i 4 So, the desired sum is � Res[f (z), z = i] + Res[f (z), z = −i] = − . 2 (b) In this case, z = i has �P = �/2 and z = −i has �P = −�/2. Hence, log i = i�/2 and log(−i) = −i�/2. Accordingly, Res[ f (z) f (z) log i � , z = i] = lim[(z − i) 2 ]= = , 2 z�i z +1 z +1 2i 4 f (z) � f (z) log(−i) Res[ 2 , z = −i] = lim[(z + i) 2 ]= = . z�i z +1 z +1 −2i 4 So, the desired sum is Res[ Res[ f (z) f (z) � , z = i] + Res[ 2 , z = −i] = . z2 + 1 z +1 2 1/2 81. (a) If f (z) is that branch of the function e az for which z 1/2 = r 1/2 ei�P /2 with 0 � �P < 2�, determine the sum of the residues of f (z)/(z 2 + 1) at its poles. 12 SOLUTION SET III FOR 18.075–FALL 2004 (b) Proceed as in part (a) when −� < �P � �. Solution. Similarly to Prob. 80 above, the denominator in f (z)/(z 2 + 1) vanishes at (z) = fz±i z = ±i. Thus, the possible poles are z = ±i. Because (z ∞ i) zf2(z) is analytic in +1 1/2 the vicinity of z = ±i in the respective branch of e az and its value at z = ±i is nonzero, these poles are simple. (a) At z = i = ei�/2 , �P = �/2 while at z = −i = e−i�/2 , �P = −i�/2 + 2� = 3�/2. � i and (−i)1/2 = ei3�/4 = −1+i � . It follows that Hence, (i)1/2 = ei�/4 = 1+ 2 2 f (z) f (z) ea(1+i)/ Res[ 2 , z = i] = lim[(z − i) 2 ]= z�i z +1 z +1 2i Res[ So, the desired sum is � 2 , f (z) f (z) ea(−1+i)/ , z = −i] = lim[(z + i) ] = z�i z2 + 1 z2 + 1 −2i � 2 . � � � ea/ 2 − e−a/ 2 f (z) f (z) Res[ 2 , z = i] + Res[ 2 , z = −i] = −ieia/ 2 z +1 z +1 2 � ≤ ia/ 2 sinh(a/ 2). = −ie (b) In this case, z = i has �P = �/2 and z = −i has �P = −�/2. Hence, (i)1/2 = ei�/4 � . Accordingly, and (−i)1/2 = e−i�/4 = 1−i 2 f (z) f (z) ea(1+i)/ Res[ 2 , z = i] = lim[(z − i) 2 ]= z�i z +1 z +1 2i Res[ So, the desired sum is � , f (z) ea(1−i)/ f (z) , z = −i] = lim[(z + i) ] = z�i z2 + 1 z2 + 1 −2i � eia/ f (z) f (z) Res[ 2 , z = i] + Res[ 2 , z = −i] = ea/ 2 z +1 z +1 � ≤ = ea/ 2 sin(a/ 2). � 2 2 � 2 . − e−ia/ 2i � 2