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hw9_soln - Complex Analysis Fall 2007 Homework 9 Solutions...

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Complex Analysis Fall 2007 Homework 9: Solutions 3.1.4 (a) Let z C \ { ni : n Z } . Then lim n →∞ 1 / ( n 2 + z 2 ) 1 /n 2 = lim n →∞ n 2 n 2 + z 2 = 1 . According to the limit comparison test from calculus, the series n =0 1 n 2 + z 2 converges if and only if n =1 1 n 2 converges. Since the latter series is known to converge, the former must as well. That is, n =0 1 n 2 + z 2 converges absolutely for z C \ { ni : n Z } . (b) Let D be any bounded subset of C \ { ni : n Z } . Then there is an C > 0 so that | z | < C for all z D . For z D we then have | n 2 + z 2 | ≥ | n 2 | - | z 2 | = n 2 - | z | 2 > n 2 - C 2 . If n > C then the above implies that 1 n 2 + z 2 1 n 2 - C 2 = M n for all z D . Appealing to the limit comparison test as above, we conclude that n>C M n converges. Hence, the Weierstrass M -test implies that n>C 1 n 2 + z 2 converges uniformly and absolutely on D . Since neither of these modes of convergence is altered by adding finitely many terms (Why? Convince yourself of this.) we conclude that n =1 1 n 2 + z 2 1
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converges uniformly and absolutely on D . Since the functions 1 / ( n 2 z 2 ) are all analytic on D , we conclude that the series is an analytic function on D . Finally, since every point z C \ { ni : n Z } belongs to some bounded subset, we see that the series represents an analytic function on z C \ { ni : n Z } . 3.1.12 Let > 0 and A = { z C : | z | ≥ } . I will prove that n =1 1 n ! z n converges uniformly on A . Let z A . Then | z | ≥ so that 1 n ! z n 1 n ! n = M n . Since lim n →∞ 1 / ( n + 1)! n +1 1 /n ! n = lim n →∞ 1 n = 0 the ratio test implies that M n converges. Since 1 /n ! z n is analytic on A for each n , the Weierstrass M -test implies that the series in question converges uniformly and absolutely to an analytic function on A . Since every point z C \ { 0 } is contained in some A , this proves that the series gives us a function that is analytic on all of z C \ { 0 } . As to the integral, since the series converges uniformly on A 1 / 2 and the unit circle γ is contained in this set, we have γ n =1 1 n ! z n dz = n =1 1 n ! γ 1 z n dz = 2 πi since γ 1 z n dz is zero unless n = 1, in which case we know it’s 2 πi .
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