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$z_{n} = r_{n}e^{i\theta_{n}}$ and $z = re^{i\theta}$. If $z_{n} \rightarrow z$ then $r_{n} \rightarrow r$ and $\theta_{n} \rightarrow \theta$.

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I was trying this problem but couldn't solve it. could someone help? Thanks. – Brian Sep 18 '12 at 13:48
$\theta_n \to \theta$ will hold only if $z \ne 0$, in this case, use that $z \mapsto \sqrt{(\Re z)^2 + (\Im z)^2}$ and $z \mapsto \arctan\frac{\Im z}{\Re z}$ are continuous – martini Sep 18 '12 at 13:53

If you take the principal value of the argument to lie in the interval $[\theta, \theta+ 2\pi)$, and if we look at the sequence $z_n = r \exp \left(i \left(\theta+ 2\pi - 1/n \right) \right)$, then $z_n \to z = r \exp(i \theta)$, but $\theta_n \to \theta + 2\ \pi$.

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here $z_{n}, z \in G = C - \{z:z \leq 0\}$ – Brian Sep 18 '12 at 14:04
$ - \pi < \theta, \theta_{n} < \pi $ – Brian Sep 18 '12 at 14:14

The question probably needs two further assumptions: that $0 \leq \theta_n <2\pi$ for every $n \in \mathbb{N}$ and that $z\neq 0$. Otherwise, you can easily exploit the $2\pi$-periodicity of the angle to construct a counter-example.

Since $z \mapsto |z|$ is continuous, then $r_n = |z_n| \to |z|$ whenever $z_n \to z$. Now, assume that (up to a subsequence) $\theta_n \to \theta'\neq \theta$. Then, by the continuity of sine and cosine, $e^{i\theta_n} = \cos \theta_n + i \sin \theta_n \to \cos \theta' + i \sin \theta'$. Since $\theta' \neq \theta$ and both lie between $0$ and $2 \pi$, then either $\cos \theta' \neq \cos \theta$ or $\sin \theta' \neq \sin \theta$. Hence $$ z_n = r_n e^{i \theta_n} \to r e^{i\theta'} \neq r e^{i \theta}, $$ against the assumption.

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$ - \pi < \theta, \theta_{n} < \pi $ – Brian Sep 18 '12 at 14:17
@Brian Of course it is the same. The point is that your angles must lie in a fixed interval of length $2 \pi$. – Siminore Sep 18 '12 at 14:19
@ Siminore here you proved that $\theta_{n} $ does not approach $\theta$ ? – Brian Sep 18 '12 at 14:40

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