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Use the residue theorem for:

$\int_{|z|=2} \frac{e^z-1}{z^2(z-1)}dz$

We have a a pole of order 2 at z=0, and a simple pole at z=1. Then we have:

$\operatorname{Res}[f(z),1]=\frac{(z-1)(e^z-1)}{z^2(z-1)} \rightarrow \lim_{z\rightarrow1}\frac{e-1}{1}=e-1$

and $\operatorname{Res}[f(z),0]= \frac{(z-0)e^z-1}{z^2(z-1)} \rightarrow$ as $\lim_{z \rightarrow0}$ Using L'Hospital I'm getting $\frac{(0)^2e^0-1}{3(0)^2-1}=1$. Am I heading the right direction?

edit: since $z^2$ has double pole order at z=0, that means that I should have $\frac{d}{dz}[(z-z_0)^2*f(z)]$ instead of $(z-0)$ That I have right?

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  • $\begingroup$ It would be a double pole except the numerator has a zero of order $1$ at the origin, so you end up with a pole of order $1.$ $\endgroup$ – Brevan Ellefsen Apr 10 at 4:05

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