Closed form geometric series for complex numbers Say I want to express the following series of complex numbers using a closed expression:
$$S(x) = \sum_{n=0}^\infty\left(r\mathrm e^{2\pi\mathrm ix}\right)^n.$$
If we interpret the summands the same way as real numbers then we may derive such an expression:
$$\left(1-r\mathrm e^{2\pi\mathrm ix}\right)S(x)=1\iff S(x)=\frac1{1-r\mathrm e^{2\pi\mathrm ix}},$$
and I "know" this holds if
$$\left|r\mathrm e^{2\pi\mathrm ix}\right|=r<1.$$
But, I cannot really show why this in fact holds for complex numbers, in particular why the convergence condition $\left|r\mathrm e^{2\pi\mathrm ix}\right|<1$ makes sense (the requirement that the summands being less then one in case of real numbers is clear). I tried to write
$$S(x)=\sum_{n=0}^\infty r^n\cos(2n\pi x)+\mathrm i\sum_{n=0}^\infty r^n \sin(2n\pi x),$$
but didn't manage to make progress.
 A: If you remember how the proof of the convergence and sum for a real geometric series goes, that proof works directly for the complex case too.
The first part of the proof concludes that
$$ s_k = \sum_{n=0}^k c^n = \frac{1-c^{k+1}}{1-c} $$
and this part depends only on the field axioms, so it works just as well in $\mathbb C$ as it does in $\mathbb R$.
The next part takes the limit of this expression as $k\to\infty$, and what you need there is just that $c^{k+1}\to 0$ when $|c|<1$. That's a simple consequence of the fact that the complex absolute value is compatible with multiplication, so (by induction on $k$ if you insist on strict formality) $|c^{k+1}| = |c|^{k+1} \to 0$.

A different way to think about it is that the real and complex part of each term in $\sum_n (re^{2\pi ix})^n$ is absolutely bounded by $r^n$ -- so if $r<1$, then the real and complex parts of the sum converge separately by the comparison test.
This argument is more general and leads to the idea of a disc of convergence in general.
