partial fraction expansion of $\frac{1}{e^z-1}$ how can you determine the partial fraction expansion of the following meromorphic function :$$f(z)=\frac{1}{e^z-1}$$
Is there a general way (or at least ways that work out for most of the common exercises concerning that topic) to determine partial fraction expansions of meromorphic functions?
 A: $e^{z}-1$ : its zeros are of order $1$ at $2i k \pi$ where its derivative $ = 1$. Let
$$g(z) = \lim_{K \to \infty}\sum_{k=-K}^K \frac{1}{z-2i k \pi}=\frac{1}{z}+ \sum_{k=1}^\infty (\frac{1}{z-2i k \pi}+\frac{1}{z+2i k \pi})$$
It is meromorphic with poles at $2i \pi k$ of order $1$ and residue $1$
Thus $$\frac{1}{e^{z}-1}-g(z)$$ is holomorphic on $\mathbb{C}$ (it is entire). 
A little work shows $g(z)$ and hence $\frac{1}{e^{z}-1}-g(z)$ is $1$-periodic and bounded as $z \to \pm i \infty$.
Thus $\frac{1}{e^{z}-1}-g(z)$  is a bounded entire function, and by the Liouville theorem it is constant, ie.
$$\frac{1}{e^{z}-1} = C+\frac{1}{z}+ \sum_{k=1}^\infty (\frac{1}{z-2i k \pi}+\frac{1}{z+2i k \pi})$$
For finding $C$ you can look at $z=0$ : 
$$C = \lim_{z \to 0} \frac{1}z-\frac{1}{e^z-1} +\sum_{k=1}^\infty (\frac{1}{z-2i k \pi}+\frac{1}{z+2i k \pi})$$
$$ = \lim_{z \to 0} \frac{1}z-\frac{1}{e^z-1} +\sum_{k=1}^\infty \frac{2z}{z^2+4 k^2 \pi^2}=\lim_{z \to 0} \frac{1}z-\frac{1}{e^z-1}  = \frac{1}{2}$$
A: The function is periodic with period $2\pi i$. Therefore we only have to look at the poles in a strip of width $2\pi i$, and in particular, the one at $z=0$, which is simple and has residue $1$. Therefore
$$ \frac{1}{e^z-1} = f(z) + \frac{1}{z} + \sum_{n\neq 0} \frac{1}{z-2\pi i n}+ \frac{1}{2\pi i n}, $$
where $f(z)$ is periodic and analytic. In fact, the latter function is $\frac{1}{2}\coth{\frac{1}{2}z}$ (there are various ways to show this: Fourier series, a certain integral, or the contour integral trick you've probably seen). We can subtract and conclude that $f(z)=1/2$.
