math185f08-hw10sol

# math185f08-hw10sol - MATH 185 COMPLEX ANALYSIS FALL 2008/09...

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Unformatted text preview: MATH 185: COMPLEX ANALYSIS FALL 2008/09 PROBLEM SET 10 SOLUTIONS For α,β ∈ C , we will use the shorthand [ α,β ] to denote the line segment from α to β , parameterized by z : [0 , 1] → C , z ( t ) = (1- t ) α + tβ . We will use the shorthand | z- a | = r to denote the circle of radius r centered at a , parameterized by z : [0 , 2 π ] → C , z ( t ) = a + re it . For a ∈ C , r > 0, we write D * ( a,r ) := { z ∈ C | < | z- a | < r } . We write C × = C \{ } . 1. Derive the following expansions. (a) For all z ∈ C , e z = e + e ∞ X n =1 1 n ! ( z- 1) n . Solution. By the Taylor expansion of exp, e z- 1 = ∞ X n =0 1 n ! ( z- 1) n = 1 + ∞ X n =1 1 n ! ( z- 1) n . Since e z- 1 = e z e- 1 , multiplying both sides by e gives the required expansion. (b) For all z ∈ D (1 , 1), 1 z = ∞ X n =0 (- 1) n ( z- 1) n . Solution. For | z- 1 | < 1, we may use the formula for geometric series 1 z = 1 1 + ( z- 1) = ∞ X n =0 (- 1) n ( z- 1) n . (c) For all z ∈ D (- 1 , 1), 1 z 2 = 1 + ∞ X n =1 ( n + 1)( z + 1) n . Solution. For | z + 1 | < 1, we have- 1 z = 1 1- ( z + 1) = 1 + ∞ X n =1 ( z + 1) n . Since the convergence of the power series is uniform in D (- 1 , 1), we may differentiate the power series on the rhs term-by-term to get d dz- 1 z = ∞ X n =1 d dz ( z + 1) n and thus 1 z 2 = ∞ X n =1 n ( z + 1) n- 1 = 1 + ∞ X n =0 ( n + 1)( z + 1) n . Date : December 12, 2008 (Version 1.0). 1 2. Let n ∈ N be n ≥ 2. Determine all entire functions f that satisfy f ( z n ) = [ f ( z )] n for all z ∈ C . Solution. This is a generalization of Homework 6 , Problem 2 (b). By Theorem 4.3 , f has a power series expansion f ( z ) = ∞ X m =0 a m z m with infinite radius of convergence. By the given condition a = f (0) = f n (0) = a n and so either a = 0 or a = e 2 pπi n- 1 for some p ∈ { 1 ,...,n- 1 } . Case I. Suppose a = e 2 pπi n- 1 for some p ∈ { 1 ,...,n- 1 } and f is non-constant. Let k ∈ N be the smallest positive number such a k 6 = 0. Hence f ( z n ) = 1 + a k z kn + higher order terms and [ f ( z )] n = 1 + na k z k + higher order terms. Since f ( z n ) = [ f ( z )] n , comparing coefficients tells us that na k = 0 and so a k = 0 — a contra- diction. In other words, if a = e 2 pπi n- 1 for some p ∈ { 1 ,...,n- 1 } , then f must be a constant function. Hence f ( z ) = a = e 2 pπi n- 1 for all z ∈ C . Case II. Suppose a = 0 and f is non-constant. Again we let k be as above and observe that f ( z ) = z k [ a k + a k +1 z + a k +2 z 2 + ··· ] =: z k g ( z ) ....
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## This note was uploaded on 06/15/2009 for the course MATH 185 taught by Professor Lim during the Fall '07 term at Berkeley.

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math185f08-hw10sol - MATH 185 COMPLEX ANALYSIS FALL 2008/09...

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