What is the residue of this essential singularity? I have this function $$ f(z) = \frac{e^{1/z}}{ 1 - z^2} $$ which has two poles of order $1$ with residues $- e / 2$ and $ 1 / (2e)$. Now I need to compute the residue at the essential singularity $z = 0$ where the numerator becomes undefined. I expanded this in series: \begin{align*} f(z) = \frac{1}{- (z + 1) (z - 1)} \big( 1 + \frac{1}{z} + \frac{1}{2} \frac{1}{z^2} + \ldots \big). \end{align*} But I'm not sure how to find the residue from this. 
 A: You can easily compute the sum of all three residues as
$$\frac{1}{2\pi i} \int_{\lvert z\rvert = R} \frac{e^{1/z}}{1-z^2}\,dz$$
with $R > 1$. Then just subtract the residues at the poles from the sum to find the residue at $0$.
A: HINT:
We can write for $|z|<1$
$$e^{1/z}=\sum_{n=1}^\infty \frac{z^{-n}}{n!}$$
and 
$$\begin{align}
\frac{1}{1-z^2}&=\sum_{n=0}^\infty z^{2n}
\end{align}$$
Therefore, we have
$$\begin{align}
\frac{e^{1/z}}{1-z^2}&=\sum_{n=1}^\infty \sum_{m=0}^\infty \frac{z^{2m-n}}{n!} 
\end{align}$$
The coefficient on $z^{-1}$ comes from the terms for which $2m=n-1$ and therefore the only terms implicated are those for which $n$ is odd.  The residue is thus $\sum_{n=1}\frac{1}{(2n-1)!}=\sinh(1)$
A: The approach of Daniel Fischer is clever. However, maybe you want a direct approach. We can use the expansion
$$ \frac{e^{1/z}}{1-z^2} = \sum_{j=0}^\infty z^{2j} \sum_{k=0}^\infty \frac{z^{-k}}{k!}$$
valid for $|z|<1$. We need to obtain the residue as the coefficient in from of $z^{-1}$. For that we replace $k$ by the new summation variables $l=2j-k$ with the result
  $$ \frac{e^{1/z}}{1-z^2} =  \sum_{l\in\mathbb{Z}}\sum_{j=\max(0,l)}^\infty \frac{z^l}{(2j-l)!}  .$$
The residue is thus given by
$$\operatorname{Res}_{z=0} \frac{e^{1/z}}{1-z^2} = \sum_{j=0}^\infty \frac1{(2j+1)!}= \sinh(1).$$ 
