Help evaluating an integral complex analysis

Question

Question is: Evaluate $\int_\gamma z^n e^{1/z} \, dz$ where $\gamma$ is the circle of radius $1$ centered at $0$ and traveled once in the counterclockwise direction.

I know that $\int_\gamma f(z) \, dz=b_1 \cdot2\pi i$

and that

$\int_\gamma e^{1/z} \, dz = 2\pi i$ for any circle $\gamma$ around $0$ (from the fact that

$e^{1/z} = 1 + \frac 1z+\frac 1{2!z^2}+\cdots$ where $b_1=1$ and $e^{1/z}$ has an essential singularity at $z=0$).

But what do I do with $z^n$?

Answer 1

Recall that if $f(z) = \displaystyle \sum_{k=-\infty}^{\infty} a_k z^k$, we then have $$\oint_{\Gamma} f(z) dz = 2 \pi i a_{-1}$$ if $\Gamma$ encloses the origin. In your case, you have $$z^n e^{1/z} = z^n \sum_{k=0}^{\infty} \dfrac1{k! \cdot z^k} = \sum_{k=0}^{\infty} \dfrac{z^{n-k}}{k!} = \sum_{k=0}^{n} \dfrac{z^{n-k}}{k!} + \dfrac1{(n+1)! \cdot z} + \sum_{k=n+2}^{\infty} \dfrac{z^{n-k}}{k!}$$Can you now finish this off?

Answer 2

Put $w=\dfrac{1}{z}.$ Then $$\int\limits_\gamma z^n e^{\tfrac{1}{z}}dz=-\int\limits_{\gamma^{-}} w^{-n} e^{w}\dfrac{dw}{w^2}=\int\limits_{\gamma^{+}} \dfrac{e^{w}}{w^{n+2}}dw=2\pi i \underset{w=0}{\operatorname{res}}{\dfrac{e^{w}}{w^{n+2}}}.$$