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?
On
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 $f(z)=1/2$.
$e^{z}-1$ : 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 $z=0$ : $$C = \lim_{z \to 0} \frac{1}z-\frac{1}{e^z-1} +\sum_{k=1}^\infty (\frac{1}{z-2i k \pi}+\frac{1}{z+2i k \pi})$$ $$ = \lim_{z \to 0} \frac{1}z-\frac{1}{e^z-1} +\sum_{k=1}^\infty \frac{2z}{z^2+4 k^2 \pi^2}=\lim_{z \to 0} \frac{1}z-\frac{1}{e^z-1} = \frac{1}{2}$$