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Suppose that $$z_n,z\in G:=\mathbb{C}-\{z\,:\,z\leq 0\}$$ and $$z_n=a_n e^{i\theta_n},z=ae^{i\theta}$$ where $-\pi<\theta,\theta_n<\pi$. Prove that if $z_n\to z$ then $\theta_n\to\theta$ and $a_n\to a$.

My attempt:

Since $z_n\to z$. For each $\epsilon>0\,\exists\,m\in\mathbb{Z}^+$ s.t. $$|z_n-z|=|a_ne^{i\theta_n}-ae^{i\theta}|<\epsilon\quad\forall n\geq m$$

And hints for how to proceed?

lemon
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johny
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1 Answers1

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$a_n\to a$ follows from $$|a_n-a|=||z_n|-|z||\leq|z_n-z|<\epsilon$$

and $\theta_n\to\theta$ can be proved from $$\lim_{n\to\infty}\frac{z_n}{z}=1$$

lemon
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  • How would the continuity of the function $f(z)=f(ae^i\theta)=log a+i\theta$ for $\theta$ between $-\pi$ and $\pi$ follow from this ? – johny Aug 25 '14 at 10:53
  • @johny Hint: $|f(z_n)-f(z)|=|\log\frac{a_n}{a}+i(\theta_n-\theta)|\leq |\log\frac{a_n}{a}|+|\theta_n-\theta|$. – lemon Aug 25 '14 at 12:43
  • So using this i will get that $f(z_n)$ converges uniformly to $f(z)$ and since the uniform limit of a continuous function is continuous in complex numbers it would follow that $f(z)$ is continuous. – johny Aug 25 '14 at 15:02
  • http://math.stackexchange.com/questions/117717/how-to-show-f-is-continuous-at-x-iff-for-any-sequence-x-n-in-x-converg – lemon Aug 25 '14 at 15:24