math185f09-hw2sol

# 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 θ ∈ R such that z = | z | e iθ . 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 ε > 0, there exists N ∈ N such that | z n- z | < ε whenver n > N . Now using the inequality that we proved, we see that || z n | - | z || ≤ | z n- z | < ε 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 ) = π- tan- 1 1 / 2 n and arg( z 2 n +1 ) =- π + 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 ) = θ , then lim n →∞ z n = re iθ . 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 θ, lim n →∞ y n = lim n →∞ | z n | × lim n →∞ sinarg( z n ) = r sin θ. Hence lim n →∞ z n = r (cos θ + i sin θ ) = re iθ . 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 doesn’t 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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