Find the residue of $\frac{e^{1/z}}{(z-1)^2}$ at $z=0$.

Question

Find the residue of $\frac{e^{1/z}}{(z-1)^2}$ at $z=0$.

How to obtain Laurent Series?

Answer 1

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

$$\bbox[5px,border:2px solid #C0A000]{\text{Res}\left(f(z), z=0\right)=\sum_{m=0}^\infty \frac{1}{m!}=e}$$

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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!