# Find the residue theorem at $\frac{e^z-1}{z^2(z-1)}$

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?

• 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 $1.$ – Brevan Ellefsen Apr 10 at 4:05