Find the residue of $\frac{e^{1/z}}{(z-1)^2}$ at $z=0$. Find the residue of $\frac{e^{1/z}}{(z-1)^2}$ at $z=0$.
How to obtain Laurent Series?
 A: Let $f(z)=\frac{e^{1/z}}{(z-1)^2}$.  
We can expand $e^{1/z}$ and $\frac{1}{(z-1)^2}$ in the series
$$e^{1/z}=\sum_{n=0}^\infty \frac{1}{n!z^n}$$
and
$$\frac{1}{(z-1)^2}=\sum_{m=0}^\infty (m+1)z^m$$
Then, we can write
$$f(z)=\sum_{m=0}^\infty\sum_{n=0}^\infty \frac{(m+1)z^{m-n}}{n!}\tag 1$$
The residue of $f(z)$ at $z=0$ is the coefficient of the series in $(1)$ when $m-n=-1$.  This reveals
$$\bbox[5px,border:2px solid #C0A000]{\text{Res}\left(f(z), z=0\right)=\sum_{m=0}^\infty \frac{1}{m!}=e}$$

NOTE:
We can express $(1)$ as a Laurent series by writing
$$\begin{align}
f(z)&=\sum_{m=0}^\infty\sum_{n=0}^\infty \frac{(m+1)z^{m-n}}{n!}\\\\
&=\sum_{m=0}^\infty\sum_{p=-\infty}^m \frac{(m+1)z^{p}}{(m-p)!}\\\\
&\sum_{p=-\infty}^\infty \left(\sum_{m=p}^\infty \frac{m+1}{(m-p)!}\right)\,z^p\tag 2
\end{align}$$
From $(2)$ we can see that the coefficient on the term $z^{-1}$ is given by
$$\begin{align}
\text{Res}\left(f(z), z=0\right)&=\sum_{m=-1}^\infty \frac{m+1}{(m+1)!}\\\\
&=\sum_{m=0}^\infty \frac{m+1}{(m+1)!}\\\\
&=\sum_{m=0}^\infty \frac{1}{m!}\\\\
&=e
\end{align}$$
as expected!
