math185f09-hw2sol - MATH 185: COMPLEX ANALYSIS FALL 2009/10...

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Unformatted text preview: MATH 185: COMPLEX ANALYSIS FALL 2009/10 PROBLEM SET 2 Throughout the problem set, i = - 1; and whenever we write x + yi , it is implicit that x,y R . For z C , recall that the argument of z , denoted arg( z ), is any R such that z = | z | e i . We write C := C \{ } . 1. Let ( z n ) n =1 be a sequence of complex numbers. (a) Show that if lim n z n = z , then lim n | z n | = | z | but that the converse is not true in general. Solution. We will first prove the inequality || u | - | v || | u- v | for u,v C . Since u = ( u- v ) + v , we have | u | | u- v | + | v | and so | u | - | v | | u- v | . Since v = ( v- u ) + u , we have | v | | v- u | + | u | and so | v | - | u | | u- v | . Hence-| u- v | | u | - | v | | u- v | as required. Since lim n z n = z , we have that for every > 0, there exists N N such that | z n- z | < whenver n > N . Now using the inequality that we proved, we see that || z n | - | z || | z n- z | < whenever n > N . Hence lim n | z n | = | z | as required. The converse is not true. Let z n = (- 1) n . Then lim n | z n | = 1 but lim n z n does not exist. (b) Is it true that if lim n z n = z , then lim n arg( z n ) = arg( z )? Solution. No. Let z n =- 1 + (- 1) n /n . Then lim n =- 1 but since arg( z 2 n ) = - tan- 1 1 / 2 n and arg( z 2 n +1 ) =- + tan- 1 1 / (2 n + 1), lim n arg( z n ) does not exist. [ Note : I take arg( z ) to be the angle that z makes with the positive real axis; if you use some other conventions, you could construct a similar counter example]. (c) Show that if lim n | z n | = r and lim n arg( z n ) = , then lim n z n = re i . Solution. Let z n = x n + iy n . Then x n = Re z n = | z n | cosarg( z n ) , y n = Im z n = | z n | sinarg( z n ) . Since cos and sin are continuous functions, lim n x n = lim n | z n | lim n cosarg( z n ) = r cos , lim n y n = lim n | z n | lim n sinarg( z n ) = r sin . Hence lim n z n = r (cos + i sin ) = re i . Date : September 29, 2009 (Version 1.0). 1 2. Which of the following limit exists? Prove your answer. lim n 1 + i 1- i n , X n =1 i n log n n + 1 , lim z 1 1- z 1- z . Solution. Note that 1 + i 1- i n = 1 if n 0mod4 , i if n 1mod4 ,- 1 if n 2mod4 ,- i if n 3mod4 . Since limit of a sequence, if exists, must be unique (Why?), lim n [(1 + i ) / (1- i )] n doesnt exist. Note that the usual way of summing a geometric progression yields m X n =1 i n = 1- i m +1 1- i = 1 if m 0mod4 , 1 + i if m 1mod4 , i if m 2mod4 , if m 3mod4 , and so m X n =1 i n 2 for all m N ....
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math185f09-hw2sol - MATH 185: COMPLEX ANALYSIS FALL 2009/10...

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