Show that $\int_{|z|=1} \frac{e^z}{z^k} dz = \frac{2\pi i}{(k-1)!}$

Question

I'm supposed to show that $$\int_{|z|=1} \frac{e^z}{z^k} dz = \frac{2\pi i}{(k-1)!}$$ where $|z|=1$ is traversed counterclockwise and $k>0$.

We can parametrize this path as $\gamma(t)=e^{it}$ for $t\in [0,2\pi]$. Now I already know that if $n\neq -1$, we have $\int_{\gamma} z^n dz = 0$ by Cauchy's Integral Theorem (and $= 2\pi i$ if $n=-1$).

I also know that $e^z$ converges uniformly on any closed ball $\overline{D_r(0)}$. Lastly, I'm told that $$ \frac{e^z}{z^k} = z^{-k} + z^{1-k}+\frac{z^{2-k}}{2!} + \cdots$$ converges uniformly on any annulus $\{z\in\mathbb{C} \mid r\leq|z|\leq R\}$, where $0<r<R$.

I'm supposed to use all of these elements to calculate the integral, but I'm having a hard time putting all the pieces together.

Answer 1

By Cauchy Integral Formulas:

$$\oint\limits_{|z|=1}\frac{e^z}{z^k}=\left.\frac{2\pi i}{(k-1)!}\frac{d^{k-1}(e^z)}{dz^{k-1}}\right|_{z=0}=\frac{2\pi i}{(k-1)!}$$

Answer 2

There is one singularity to $\dfrac{e^z}{z^k}$ at $z=0$. The residue is the coefficient of the $\dfrac1z$ term of the Laurent expansion: $$ \begin{align} \frac{e^z}{z^k} &=\frac1{z^k}\sum_{j=0}^\infty\frac{z^j}{j!}\\ &=\sum_{j=0}^\infty\frac{z^{j-k}}{j!} \end{align} $$ $j-k=-1$ when $j=k-1$, so the coefficient of the $\dfrac1z$ term is $\dfrac1{(k-1)!}$ .

The integral along a contour circling a singularity counterclockwise gets a contribution of $2\pi i$ times the residue from that singularity. That is, $$ \oint\frac{e^z}{z^k}\mathrm{d}z=2\pi i\frac1{(k-1)!} $$