I am attempting to evaluate the following contour integral: \begin{equation*} \oint_{C(i,2)} \frac{e^z}{(z-1)^n} \,\text{d}z \end{equation*} where $C(i,2)$ is the circle of radius 2 around $i$, $\forall\; n \in \mathbb{N}$. I attempted to use Cauchy's Integral formula, letting \begin{equation*} f_{n}(z) = \frac{e^z}{(z-1)^{n-1}} \end{equation*} $\forall\; n \in \mathbb{N}$, so then \begin{align*} f_n(z) = \frac{1}{2 \pi i} &\oint_{C(i,2)} \frac{f_n(z)}{z - 1} \,\text{d}z \\ \frac{2 \pi i e}{(1-1)^{n-1}} = &\oint_{C(i,2)} \frac{e^z}{(z-1)^n} \,\text{d}z \end{align*} But this is clearly undefined. Is there some key piece of understanding I'm missing with regards to this method or is there an entirely different tactic I am meant to take (or better yet, a result that simplifies this entirely)?
Any help is appreciated. Thanks in advance.
You can't apply this formula in this case because $f_n$ is not holomorphic in the contour. But you can use this generalised Cauchy formula : $$f^{(n)}(a) = \frac{n!}{2\pi i} \oint_\gamma \frac{f(z)}{(z-a)^{n+1}}\, dz$$ $$\oint_{C(i,2)} \frac{e^z}{(z-1)^n} \,\text{d}z=\frac{2\pi i}{(n-1)!}\frac{de^z}{dz^{n-1}}(1)=\frac{2\pi i}{(n-1)!}e^1= \frac{2\pi e i}{(n-1)!}$$