Calculate the following integral by transposing to a curve integral and then using the residue theorem: $\displaystyle \int_{0}^{2\pi}{\frac{e^{int}}{C-e^{it}}dt}, \qquad |C|\ne1, n\in \mathbb N$.
My idea: I use the curve $\gamma:[0;2\pi]\to\mathbb{C}, \gamma(t)=e^{it}.$ Then:
$\displaystyle\int_{\gamma}\frac{z^{n-1}}{C-z}dz=\int_{0}^{2\pi}{\frac{(e^{it})^{n-1}}{C-e^{it}}\cdot ie^{it} dt}=i\cdot\int_{0}^{2\pi}{\frac{e^{itn}}{C-e^{it}}dt}$.
Let $\displaystyle f(z):=\frac{z^{n-1}}{C-z}$. $f$ has a single pole at $C$ (for $C\ne 0$) and if $C=0$ it has a removable singularity there.
And then by applying the residue theorem:
$\displaystyle\int_{\gamma}\frac{z^{n-1}}{C-z}dz=2\pi i \cdot Res_{C}(f)=-2\pi i C^{n-1}.$
So $\displaystyle \int_{0}^{2\pi}{\frac{e^{itn}}{C-e^{it}}dt}=-2\pi C^{n-1}.$
But if I compare results to the results calculated by wolfram alpha they differ, so I assume my solution is wrong.