Why is the circle of convergence for complex power series a circle (and not e.g. a square)? Power-Series have an "circle of convergence".
With real numbers this is an interval.
Expanding this to complex numbers this becomes a circle.
There are lots of book stating this, but I did not find any which explained why this is a circle.
Thanks
Joe
 A: Well, I know it as the radius of convergence, but it stems from how it is calculated.
Given a power series $\sum_{n=k}^\infty a_nx^n$, it is convergent if $$\lim_{n\to\infty} \frac{|a_{n+1}x^{n+1}|}{|a_nx^n|}=\lim_{n\to\infty} \frac{|a_{n+1}||x^{n+1}|}{|a_n||x^n|}=\lim_{n\to\infty} \frac{|a_{n+1}|}{|a_n|}|x|<1$$
Or 
$$ |x|<\lim_{n\to\infty}\frac{|a_n|}{|a_{n+1}|} $$
Note that, if $x=a+bi$ for $a,b\in\Bbb R$, then $|x|=\sqrt{a^2+b^2}<L$, which has a locus of a disk.
A: Assume the power series $\sum_{k=0}^\infty a_kz^k$ converges for some $z$; say for $z:=1$. Then the "numerical" series $\sum_{k=0}^\infty a_k$ is convergent, which implies $\lim_{k\to \infty}|a_k|=0$, and in particular that there is an $M$ with $|a_k|\leq M$ for all $k\geq0$. It follows that the series $\sum_{k=0}^\infty a_kz^k$ is absolutely convergent in the disk $D_1:=\{z\in{\mathbb C}\>|\>|z|<1\}$.
From the above we can conclude that the domain $C$ of convergence is a union of concentric disks, one for each $z$, for which convergence has been established. This union is obviously again a (possibly infinite) open disk with radius
$$\rho:=\sup\bigl\{|z|\>\bigm|\>\sum_{k=0}^\infty a_kz^k\ {\rm is\ convergent}\bigr\}\ .$$
