# limit of a complex number [duplicate]

Possible Duplicate:
Limit of argz and r

$z_{n} = r_{n}e^{i\theta_{n}}$ and $z = re^{i\theta}$. If $z_{n} \rightarrow z$ then $r_{n} \rightarrow r$ and $\theta_{n} \rightarrow \theta$. $z_{n},z \in G = C - \{z:z \leq 0\}$ and $- \pi < \theta, \theta_{n} < \pi$.

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It is not clear what you are asking. If you browse a bit the site, you will immediately notice that people asking questions usually write up a little bit, explaining what the problem is. –  Mariano Suárez-Alvarez Sep 18 '12 at 14:31
There doesn't seem to be a question here. Are you asking how to prove that if $z_n\to z$ and have the properties stated then $r_n\to r$ and $\theta_n\to\theta$? Also, it would be helpful for you to state your background so that people know at what level they should answer the question. –  J. Loreaux Sep 18 '12 at 14:48
Also, you just asked this question here. Please refrain from doing this. –  J. Loreaux Sep 18 '12 at 14:52

## marked as duplicate by Norbert, Andres Caicedo, Did, William, Chris EagleSep 24 '12 at 9:22

1. the fact that $r_n \to r$ is an immediate consequence of the continuity of $z \mapsto |z|$ (in $G$, in $\mathbb{C}$, in whatever you like).
2. The convergence of the phase is a bit more complicated. Either you look for a closed formula in terms of $\Re z$ and $\Im z$, or you proceed as follows. First of all, $\theta$ is determined, since $z \neq 0$. The sequence $\{\theta_n\}_n$ is bounded (between $-\pi$ and $\pi$), hence $\liminf_n \theta_n$ and $\limsup_n \theta_n$ are both finite numbers. Assume that they are different. Therefore, you have two subsequences $\theta'_n$ and $\theta''_n$ such that $\theta'_n \to m=\liminf_n \theta_n$ and $\theta''_n \to M=\limsup_n \theta_n$. Recalling that $e^{ix}=\cos x + i \sin x$ for any $x$, the continuity of sine and cosine imply that $\cos \theta'_n \to \cos m$, $\sin \theta'_n \to \sin m$ and similarly for $\theta''_n$. But $(\cos m,\sin m ) = (\cos M, \sin M)$ means $m=M$, since $m$ and $M$ lie between $-\pi$ and $\pi$. Hence $e^{i\theta'_n}$ and $e^{i \theta''_m}$ converge to different limits. This contradicts the fact that $e^{i\theta_n} \to \frac{z}{|z|}$.
For the convergence of $|z|$ , notice that if the sequence ${z_{n}}$ converges, the sequence ${|z_{n}|}$ also converges. if $$lim_{n->\infty}z_{n} = lim_{n->\infty}(x+iy)= w = (a+bi)$$ it means: $lim_{n->\infty}x_{n}=a$ and $lim_{n->\infty}y_{n}=b$. Then $$lim_{n->\infty}|z_{n}|=lim_{n->\infty}\sqrt{x_{n}^{2}+y_{n}^{2}}=\sqrt{a^{2}+b^{2}}$$ The sequence of $arg(z_{n})$ need not to converge. To notice this look at this example: $$z_{n}=-1+(-1)^{n}\frac{i}{n}$$ However it is possible to find a convergent sequence of $Arg(z_{n})$