I have to solve if $C=C(1,\frac{1}{2})$ $$I=\int_{C} ze^{\frac{1}{z-1}}$$

I know that $I=2\pi i \operatorname{Res}(f(z), 1)$, but I do not know how could I calculate that residue.

What I did:

$$f(z)=ze^{\frac{1}{z-1}}=z\sum_{n=0}^{\infty}\frac{1}{n!}\left( \frac{1}{z-1}\right)^n=\sum_{n=0}^{\infty}\frac{1}{n!} \frac{1}{(z-1)^{n-1}}+\sum_{n=0}^{\infty}\frac{1}{n!} \frac{1}{(z-1)^{n}}$$

Ok, someone could help me to calculate the residues $\operatorname{Res}(f(z),1)$


The residue is the coefficient of the $\frac{1}{z^1}$ term in the series.

Here you have two contributions: From $n=2$ in the first sum, and from $n=1$ in the second. So the residue is $\frac12 + 1 = \frac32$


The residue is simply the coefficient of $(z-1)^{-1}$, or in this case, $1/2! + 1/1! = 3/2$.


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