Stuck on contour integral, can I use Cauchy's theorem instead? I am trying to calculate
$$\int_{|z-2| = 3} e^{1/z}$$
I parametric the circle of radius three centered at 2 by $\gamma(t) = 3e^{it} +2$ and so I can instead evaluate
$$\int_\gamma f(\gamma(t))\gamma'(t) = \int_0^{2\pi}\exp\left(\frac{1}{3\exp(i\theta)+2}\right)3i\exp(i\theta)$$ and I am stuck here because I do not know how to compute the antiderivate of this. This makes me think there must be an easier way to do this
I am aware that the integrand is not defined at $z = 0$ is this what is causing problems? Since $\gamma$ is a closed path can I just say that the integral is zero, or does the fact that the integrand is not defined at $0$ mean that it is not analytic and hence I can not use Cauchy's theorem?
 A: You can apply the residue theorem here. Since$$e^{\frac1z}=1+\frac1z+\frac1{2z^2}+\cdots,$$you know that $\operatorname{res}_{z=0}\left(e^{\frac1z}\right)=1$. Therefore, your integral is equal to $2\pi i\times1=2\pi i$.
A: I'm going to assume you aren't versed in the residue theorem yet. (Well, you implied that you were in a comment, so oh well. You can take this as an alternate answer, or this might be helpful to those in the future looking at this who aren't as well versed. As you will.)

Hint #1:
Since the integrand is not defined at $z=0$, and the contour encloses that singularity, you cannot use the Cauchy Integral Theorem, at least not immediately.
My recommendation would be to use the power series expansion of $e^z$ (plugging in $1/z$ for $z$). You can then express the integral by
$$\int_{|z-2|=3} e^{1/z}dz = \int_{|z-2|=3} \sum_{k=0}^\infty \frac{1}{k! \cdot z^k}= \sum_{k=0}^\infty \int_{|z-2|=3}  \frac{1}{k! \cdot z^k}$$
You should find some pleasant surprises in that summation that simplify the process a bit.

Hint #2:
For any contour of positive orientation which is a circle of radius $r$ centered at the complex number $z_\star$, you can show
$$\int_{|z-z_\star|=r} \frac{1}{z^n} dz = \left\{\begin{matrix}
0 & \forall n \neq 1 \\ 
2\pi i & n = 1
\end{matrix}\right.$$
