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First: Sorry that I wrote "normal" complex coordiantes. No idea how they are properly called =(

So I have this example given, it should show me (i think) about problems occurding when switchting to polar coordinates:

When

$z_n=-2 + i\frac{(-1)^n}{n^2}, \quad n=1,2,...$

we can write

$\lim_{n\to\infty} z_n = \lim_{n\to\infty} (-2) + i \lim_{n\to\infty}\frac{(-1)^n}{n^2}=-2 + i\cdot 0 = -2$

If, using polar coordinates, we write

$r_n = |z_n|$ and $\Phi=Arg z_n, \quad n=1,2,...$

where $Arg z_n$ denotes principal arguments $(-\pi < \Phi \leq \pi)$ of $z_n$, we find that

$\lim_{n\to\infty} r_n = \lim_{n\to\infty} \sqrt{4+\frac{1}{n^4}} = 2$

but that

$\lim_{n\to\infty} \Phi_{2n}=\pi$ and $\lim_{n\to\infty} \Phi_{2n-1}=-\pi$

Evidently, then, the limit of $\Phi$ does not exists as $n$ tends to infinity.

Question: I was wondering what I should take away from this. Does it mean, that the existence of a limit of a complex sequence depends on the chosen coordinate system? If not, how would I properly calculate it in polar coordiantes?

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The problem lies solely in the fact that you chose to work with the principal argument in a situation in which the limit is a negative real number. That's preceisely the line in which there are problems with this approach. If you had decided to work with the same sequence but used instead the argument that belongs to, say, $(0,2\pi)$, there would have been no problem.

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  • $\begingroup$ Thanks but that doesn't really solve my problem. I don't see the proble mat all. I mean, $\Phi_n \in (-\pi, \pi]$ no? Shouldn't that sovle the problem anyway? $\endgroup$ – xotix Jun 7 '18 at 10:14
  • $\begingroup$ That introduces a discontinuity along the negative reals. $\endgroup$ – José Carlos Santos Jun 7 '18 at 10:15

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