Let $m \in \mathbb{Z}$, compute $\oint_{|z|=1} \frac{e^z}{z^m}dz$.
This exercise is supposed to be solve using Cauchy's Formula. The case when $m\leq 0$ could be solved by setting the auxiliar function $f(z)=e^z z^{-m+1}$ and the integral becomes $\oint_{|z|=1} \frac{f(z)}{z-0}dz$, but if $m>0$ $f(z)$ is not holomorphic in $z=0$.
Let $m \in \Bbb{Z}$ be such that $m \geq 1$. As the any derivative of $e^z$ is just $e^z$, one has, by Cauchy's integral formula
$$\int_{\vert z \vert = 1}\frac{e^z}{z^m}dz=\frac{2\pi i}{(m-1)!} \bigg(\frac{d^{m-1}}{dz^{m-1}} e^z \bigg\vert_{z=0}\bigg)=\frac{2\pi i}{(m-1)!} (e^0)=\frac{2\pi i}{(m-1)!}.$$
If $m \leq -1$ the answer is zero as the function is analytic everywhere by Cauchys Theorem. You tell me, what happens if its $m=0$.