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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 X n =0 1 n 2 + z 2 converges if and only if X n =1 1 n 2 converges. Since the latter series is known to converge, the former must as well. That is, X 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 X 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 X n =1 1 n 2 + z 2 1 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 X 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 \ { } is contained in some A , this proves that the series gives us a function that is analytic on all of z C \ { } . As to the integral, since the series converges uniformly on A 1 / 2 and the unit circle is contained in this set, we have Z X n =1 1 n ! z n dz = X n =1 1 n !...
View Full Document

This note was uploaded on 02/29/2012 for the course MATH 4364 taught by Professor Ryandaileda during the Fall '11 term at Trinity University.

Page1 / 7

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online