Finding and defining singularities of $f(z)=\frac{1}{e^{2\pi i/ z}-1}$

Question

I'm stuck at finding singularities of $f(z)=\frac{1}{e^{2\pi i/ z}-1}$

I've found the roots of $e^{2\pi i/z}=1$

$e^{2\pi i/z}=e^{2 \pi n i}, z=1/n$ where $n=1,2,3...$

so there will be infinitely number of roots as $e^{2\pi i},e^{4\pi i},e^{6\pi i},e^{8\pi i}...$

Also I have no idea about $z=0$ since function goes to $e^{\infty \pi i}$

How can I determine the type of these roots? (removable, essential or a pole etc...)

Answer 1

$e^{2\pi i /z}=1$ iff $2\pi i /z=2\pi in$ for some integer $n$. So the singulrities are the numbers $\frac 1 n: n \in \mathbb Z, n \neq 0$. Thehn are all simple poles: $\lim_{z \to \frac 1 n} (z-\frac 1 n) e^{2\pi i /z}=\frac 1 {2\pi in^{2}}$ by L'Hopital's Rule.