Finding the types of singularities of $f(z)=\frac{1}{z\cdot (e^z -1 )}$

Question

I am getting trouble to find the types of singularities of $$f(z)=\frac{1}{z\cdot (e^z -1 )}$$
What I tried to do is:

1. $z=0$
2. $z=2\pi k i$
   for $z=2\pi k i$ its in order 1, but for the first one I dont know how to find the order
   Any suggestions?
   thanks.

Answer 1

$$ z(e^z-1)=z\Bigl(z+\frac{z^2}{2!}+\frac{z^3}{3!}+\dots\Bigr)=z^2\Bigl(1+\frac{z}{2!}+\frac{z^2}{3!}+\dots\Bigr) $$ You should consider also the point at $\infty$.