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We have $z_n=r_ne^{i\theta_n}$ and $z=re^{i\theta}$, where $z_n,z\in\mathbb{C}\setminus{\{x\in\mathbb{R}:x\le 0\}}$ and $-\pi<\theta,\theta_n<\pi$. I wish to prove that if $z_n\to z$ then $r_n\to r$ and $\theta_n\to\theta$.

Well, if this is the polar form then $r_n=\lvert z_n \rvert$ and $r=\lvert z\rvert$ and we need only to use that $\big\lvert \lvert z_n\rvert -\lvert z\rvert \big\rvert \leqslant \lvert z_n-z \rvert$ to show that if $z_n\to z$ then $\lvert z_n \rvert \to \lvert z\rvert$.

I can't prove $\theta_n\to\theta$ yet. Could anyone give me a hint?

Thanks.

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Hint: expanding out the polar form of $\lvert z_n-z \rvert^2$ gives $$ r_n^2 +r^2-2rr_n \cos{(\theta_n-\theta)}, $$ if you group the terms correctly. Now find necessary conditions for this to tend to zero.

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  • $\begingroup$ I see. As both $r_n^2+r^2$ and $2rr_n$ converge to the same number we can see that $\cos (\theta_n-\theta)\to 1$ and then $\theta_n-\theta\to 0$. When did we use that $z_n,z\in\mathbb{C}\setminus\{x\in\mathbb{R}:x\le 0\}$? I think this works for non-zero complex numbers. $\endgroup$ – Tanius Oct 1 '15 at 0:04
  • $\begingroup$ You have to choose a branch for the argument, since it is not a continuous function on all of $\mathbb{C}$. Note that if $\theta_n-\theta=2\pi$, you still get zero. $\endgroup$ – Chappers Oct 1 '15 at 0:07
  • $\begingroup$ I see. Thank you very much. $\endgroup$ – Tanius Oct 1 '15 at 0:23

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