# Singularity of a function $f(z)$

Which type of singularity the function
$f(z)= \frac{e^z}{z(1- e^{-z})}$ at $z=0$

I Know, we find the singularity of a function by expanding the given function by using Laurent series expansion.

I tried like this. $f(z)= \frac{e^z}{z(1- e^{-z})}$
$f(z)= \frac{1}{z}{e^z}{(1- e^{-z})^{-1}}$
$f(z)= \frac{1}{z}({e^z+1+e^{-z}+e^{-2z}+\dots)}$
from this i concluded that this function has essential singularity at $z=0$.
Is my conclusion correct??

$z=0$ is a zero both for $z$ and for $1-e^{-z}$, while the numerator is never zero. Hence it is a pole of order $2$. –  Federica Maggioni Apr 26 '13 at 6:34
$e^{-z}$ has infinite terms, how pole of order 2? can you explain.... @FedericaMaggioni –  prasad Apr 26 '13 at 6:40
1. Function $f(z)$ can be rewritten as $$f(z)= \dfrac{e^z}{z(1- e^{-z})}=\dfrac{e^{2z}}{z({e^{z}-1})}=\dfrac{ze^{2z}}{z^2({e^{z}-1})}=\dfrac{e^{2z}}{z^2}h(z)$$ 2. Function $h(z)=\dfrac{z}{e^{z}-1}$ has a removable singularity at $z=0.$
Thus $f(z)$ has a pole of order $2$ at $z=0.$ –  M. Strochyk Apr 26 '13 at 6:56