# Res f z 1 lim z 1 z 1 f z lim z 1 1 z z 2 z 10 1 9

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, Res ( f ( z ) , 1) = lim z 1 ( z - 1) f ( z ) = lim z 1 1 z ( z - 2) · · · ( z - 10) = - 1 9! . Then the path integral can be evaluated by Z C 3 2 (0) d z z ( z - 1) · · · ( z - 10) = 2 πi [ Res ( f ( z ) , 0) + Res ( f ( z ) , 1)] = 2 πi 1 10! - 1 9! = - 18 πi 10! . C 3 2 H 0 L - 2 - 1 1 2 - 2 - 1 1 2 Therefore, the path integral Z C 3 2 (0) d z z ( z - 1)( z - 2) · · · ( z - 10) = - 18 πi 10! .
Section 5.1 Cauchy’s Residue Theorem 111 21. The function f ( z ) = e z 2 z 6 has a pole of order 6 at 0. To compute the residue at 0, we find the coefficient a 1 in the Laurent series expansion about 0. We have 1 z 6 e z 2 = 1 z 6 X n =0 ( z 2 ) n n ! . It is clear that this expansion has no terms with odd powers of z , positive or negative. Hence a - 1 = 0 and so Z C 1 (0) e z 2 z 6 dz = 2 πi Res (0) = 0 . 25. Same approach as in Exercise 21: sin z z 6 = 1 z 6 X k =0 ( - 1) k z 2 k +1 (2 k + 1)! = 1 z 6 z - z 3 3! + z 5 5! - · · · Coefficient of 1 z : a - 1 = 1 5! , so Z C 1 (0) sin z z 6 dz = 2 πi Res (0) = 2 πi 5! . 29. (a) The Order of a pole of csc( πz ) = 1 sin πz is the order of the zero of 1 csc( πz ) = sin πz. Since the zeros of sin πz occur at the integers and are all simple zeros (see Example 1, Section 4.6), it follows that csc πz has simple poles at the integers. (b) For an integer k , Res ( csc πz, k ) = lim z k ( z - k ) csc πz = lim z k z - k sin πz = lim z k 1 π cos πz (l’Hospital’s rule) = ( - 1) k π . (c) Suppose that f is analytic at an integer k . Apply Proposition 1(iii), then Res ( f ( z ) csc( πz ) , k ) = ( - 1) k π f ( k ) .
112 Chapter 5 Residue Theory 33. A Laurent series converges absolutely in its annulus of convergence. Thus to multiply two Laurent series, we can use Cauchy products and sum the terms in any order. Write f ( z ) g ( z ) = X n = -∞ a n ( z - z 0 ) n X n = -∞ b n ( z - z 0 ) n = X n = -∞ c n ( z - z 0 ) n , where c n s obtained by collecting all the terms in ( z - z 0 ) n , after expanding the product. Thus c n = X j = -∞ a j b n - j ; in particular c - 1 = X j = -∞ a j b - 1 - j , and hence Res ( f ( z ) g ( z ) , z 0 ) = X j = -∞ a j b - j - 1 . 37. (a) We have, Exercise 35(a), Section 4.5, J 0 ( z ) = 1 2 πi Z C 1 (0) e z 2 ( ζ - 1 ζ ) ) ζ . Thus Z 0 J 0 ( t ) e - st dt = 1 2 πi Z 0 Z C 1 (0) e - t ( s - 1 2 ( ζ - 1 ζ ) ) ζ dt (b) For ζ on C 1 (0), we have ζ - 1 ζ = ζ - ζ = 2 i Im ( ζ ) , which is 0 if Im ( ζ ) = 0 (i.e., ζ = ± 1) or is purely imaginary. In any case, for all ζ C 1 (0), and all real s > 0 and t , we have e - t ( s - 1 2 ( ζ - 1 ζ ) ) = e - ts e t 2 ( ζ - 1 ζ ) = e - ts . So, by the inequality on integrals (Th.2, Sec. 3.2), 1 2 πi Z C 1 (0) e - t ( s - 1 2 ( ζ - 1 ζ ) ) ζ 2 π 2 π max ζ C 1 (0) e - t ( s - 1 2 ( ζ - 1 ζ ) ) 1 ζ = max ζ C 1 (0) e - t ( s - 1 2 ( ζ - 1 ζ ) ) ( | ζ | = 1) e - ts
Section 5.1 Cauchy’s Residue Theorem 113 Thus the iterated integral in (a) is absolutely convergent because 1 2 πi Z 0 Z C 1 (0) e - t ( s - 1 2 ( ζ - 1 ζ ) ) ζ dt Z 0 Z C 1 (0) e - t ( s - 1 2 ( ζ - 1 ζ ) ) ζ dt Z 0 e - ts dt = 1 s < (c) Interchange the order of integration, and evaluate the integral in t , and get Z 0 J 0 ( t ) e - st dt = 1 2 πi Z C 1 (0) Z 0 e - t ( s - 1 2 ( ζ - 1 ζ ) ) dt ζ = 1 2 πi Z C 1 (0) 1 ( s - 1 2 ( ζ - 1 ζ ) ) e - t ( s - 1 2 ( ζ - 1 ζ ) ) 0 ζ = 1 πi Z C 1 (0) 1 - ζ 2 + 2 + 1 because, as t → ∞ , e - t ( s - 1 2 ( ζ - 1 ζ ) ) = e - ts 0 .