How do I show that $\frac 1{e^z-1}$ has essential singularities (instead of say, poles) at $z=2n\pi i(n\in \mathbb Z)$?
I can't figure out how to show that the function does not go to infinity near $0$, or that it assumes every possible value near $0$. Exhibiting the laurent series around $0$ isn't general enough to show that essential singularities occur at all the stated points.
Best Answer
You don't!
The function in question has simple poles. The easiest way to see this is to note that $\exp(2n\pi i)=1$, so by continuity $e^z-1$ tends to zero as $z \rightarrow 2n\pi i$ and thus the reciprocal tends to infinity.
To show that the poles are simple, note that $e^z-1$ has a zero of degree $1$ at $2n\pi i$ so the reciprocal has a simple pole and so the residue is:
$$\frac{1}{\frac{d}{dz}(e^z-1)}=\frac{1}{e^{2n\pi i}}=1$$