Compute $\int_\Gamma \frac{e^\frac{1}{z}}{z-1}dz$, where $\Gamma$ is the circle $|z-1|\le\frac{3}{2}$, positively oriented.

Question

Compute $\int_\Gamma \frac{e^\frac{1}{z}}{z-1}dz$, where $\Gamma$ is the circle $|z-1|\le\frac{3}{2}$, positively oriented.

The numerator is not analytic in $\Gamma$ so we can't use Cauchy integral formula. I'm thinking maybe I shold use residue theorem. But then I have these two questions:

1. Should I look for the Laurent series of $e^\frac{1}{z}$ around $z=0$? What if I look for the series around some other points in $\Gamma$?
2. After I find the Laurent series of $e^\frac{1}{z}$, how do I find the Laurent series for the $\frac{e^\frac{1}{z}}{z-1}$?

Answer 1

The residue in the essential singularity can be computed trough Taylor's expansions: $$e^{\frac{1}{z}}=1+\frac{1}{z}+\frac{1}{2z^2}+\ldots,$$ $$\frac{1}{z-1}=-1-z-z^2-z^3-\ldots,$$ from which (just multiply the two series): $$\operatorname{Res}\left(\frac{e^{\frac{1}{z}}}{z-1},z=0\right)=-\sum_{k\geq 1}\frac{1}{k!}=\color{red}{1-e}.$$

Answer 2

There are two resides Z=1 and Z=0, when z=1, it's simple Res($e^{1\over z}\over {z-1}$,z=1)=e when z=0, since $${1\over {z-1}}=-{(1+z+z^{2}+...+z^{n}...)}$$ and $e^{1\over z}$=1+1/z+... so Res($e^{1\over z}\over {z-1}$,z=0)=-1, so the answer :2$\pi$i(e-1)

Answer 3

Let $f(z)=\frac{e^{1/z}}{z-1}$.

We can expand $e^{1/z}$ and $\frac{1}{z-1}$ in the series

$$e^{1/z}=\sum_{n=0}^\infty \frac{1}{n!z^n}$$

for $|z|>0$ and

$$\frac{1}{z-1}=-\sum_{m=0}^\infty z^m$$

for $|z|<1$. Then, we can write

$$f(z)=-\sum_{m=0}^\infty\sum_{n=0}^\infty \frac{z^{m-n}}{n!}\tag 1$$

for $0<|z|<1$.

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The residue of $f(z)$ at $z=0$ is the coefficient of the series in $(1)$ when $m-n=-1$. This reveals

$$\text{Res}\left(f(z), z=0\right)=-\sum_{m=0}^\infty \frac{1}{(m+1)!}=1-e$$

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The residue at $z=1$ is trivially seen to be $e$ and the sum of the residues is $1-e+e=1$.

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Putting it all together, we find that

$$\bbox[5px,border:2px solid #C0A000]{\oint_{|z-1|=3/2}\frac{e^{1/z}}{z-1}\,dz=2\pi i}$$

Answer 4

It can be evaluated by the residue at infnity trick...

Since the outside of the counter contains no singularity,

$$\begin{align} \int_{\Gamma}f(z)dz&=-2\pi i\left(Res_{z=\infty}f(z)\right)\\ &=-2\pi i\left(Res_{z=0}f(z)-\frac{1}{z^2}f(\frac{1}{z})\right)\\ &=-2\pi i\left(Res_{z=0}-\frac{1}{z^2}\frac{e^z}{z^{-1}-1}\right)\\ &=-2\pi i\left(Res_{z=0}\frac{e^z}{z(z-1)}\right)\\ &=-2\pi i\left(\left.\frac{e^z}{z-1}\right\rvert_{z=0}\right)\\ &=2\pi i \end{align}$$