Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
I was trying this problem but couldn't solve it. could someone help? Thanks. –  Brian Sep 18 '12 at 13:48
1  
$\theta_n \to \theta$ 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

2 Answers 2

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 $\theta_n \to \theta + 2\ \pi$.

share|improve this answer
    
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.

share|improve this answer
    
$ - \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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.