I am following the Coursera course on Introduction to Complex Analysis and I am stuck at the following problem:
Evaluate the following using the residue theorem: $\int_{|z|=2}\frac{e^z}{(z-1)}dz$
I have successfully evaluated the problem using the following approach:
$\int_{|z|=2}\frac{e^z}{(z-1)}dz = \int_{|z|=2}e\frac{e^{(z-1)}}{(z-1)}dz$
$e\frac{e^{(z-1)}}{(z-1)}= e(\frac{1}{z-1}+1+\frac{z-1}{2}+...)$ by Taylor Expansion of $e$
Therefore the $a_{-1}$ is $e$. From there, we can evaluate that the answer is $2\pi i e$ using the residue theorem.
However, the answer given has a much shorter method which involves observing that $\frac{e^z}{(z-1)}$ has a simple pole. I am wondering how do you tell that this function has a simple pole without doing Laurent expansion.
Your help is appreciated!