Using Cauchy's Theorem on Contour Integral

Question

I need to solve $\int_\gamma (1-e^z)^{-1}$ if $\gamma (t) = 2i + e^{it}$. I would assume Cauchy's Integral theorem applies here, where $\gamma$ is a closed path on a convex open set. I'm having trouble with $f(z) = (1-e^z)^{-1}$, however: I can't figure out an antiderivative.

Answer 1

Note that $f(z)=1-\mathrm{e}^z$ vanishes if and only if $z=2k\pi i$, for some $k\in\mathbb Z$. No such roots of $f$ exists in the interior of the disk $D(2i,1)$ where the integration takes places. In particular, $$ g(z)=\frac{1}{1-\mathrm{e}^z}, $$ is holomorphic in an even larger disk: $D(2i,2)$, and thus its integral $$ \int_{\lvert z-2i\rvert=1}\frac{dz}{1-\mathrm{e}^z} $$ vanishes.