# Evaluate integral using residues at $\infty$

Find an integral $\int_{|z|=3} \frac{e^{\frac1{1-z}}}{z}$ using residues.

In the book it is proposed to calculate the residue in $z=\infty$. But I'm not quite understand, why we do it here? $Res_{\infty}=0$ here and then an integral is equal to $0$ .

When you do the change of variables, you need to consider the $dz$ too. Setting $w=1/w$, you have $dz=d(1/w)=-dw/w^2$, by the chain rule. So $$\oint_{|z|=3} dz \frac{\exp(\frac{1}{1-z})}{z} = -\oint_{|w|=1/3}\left(\frac{-dw}{w^2}\right)w\exp\left( \frac{w}{w-1}\right).$$ [The extra $-1$ at the front on the RHS comes from the fact that an anticlockwise rotation in $z$ corresponds to a clockwise rotation in $w$.]
Now, you should be able to read off the residue at $w = 0$...
$\int_{|z|=3} \frac{e^{\frac1{1-z}}}{z}=2\pi i(\text{residue at}~ 0+\text{residue at }~1)=-2\pi i(\text{residue at} \infty)$.