Contour integral: $ \oint_{C}\frac{e^{\frac{1}{z}}}{z-1}dz $ where $C: |z-1|=\frac{3}{2}$.

Question

I have some problems with finding contour integral: $$ \oint_{C}\frac{e^{\frac{1}{z}}}{z-1}dz $$ where $C: |z-1|=\frac{3}{2}$.

The function $e^\frac{1}{z}$ is not analytic on this circle; that's because it's impossible to use Cauchy integral formula. Also, I'm trying to find Laurent series in $z=1$ and then use residue. If I use residue theorem I have 2 residues in the this contour? Right? Any suggestions on how to solve it? Thanks in advance.

Answer 1

Substitute $z-1=\frac{1}{w}\implies dz=\frac{-1}{w^2}dw$ and the new contour is $|w|=\frac{2}{3}$ oriented clockwise, so that

$I=\int_{|w|=\frac{2}{3}}\frac{-e^{\frac{w}{w+1}}}{w}dw$

$w=0$ is simple pole inside $|w|=2/3$ and Res{$f(w),w=0$}$=-1$. Hence

$I=-2πi.(-1)=2πi$