Evaluating complex integral $\int _{\left|z\right|=2}^{ }\:\frac{e^z}{z-1}dz$ using the residue theorem

Question

I am trying to solve $\int _{\left|z\right|=2}^{ }\:\frac{e^z}{z-1}dz$ using the residue theorem and keep getting $-4\pi i$, which is evidently wrong. I would greatly appreciate any helpful hints.

Answer 1

Let $f\left(z\right)=\frac{e^z}{z-1}$. Then f has has a simple pole at $z0=1$, $Res\left(f,1\right)=\lim _{z\to 1}\left(z-1\right)\frac{e^z}{z-1}=\lim _{z\to 1}e^z=e$. Hence, $\int _{\left|z\right|=1}^{ }\:f\left(z\right)dz=2\pi iRes\left(f,1\right)=2\pi ie$. By the way, thank you for the advice, I just solved the integral by Cauchy's theorem, but this is the answer provided under residue solution, so I just decided to post it here for reference.