I am trying to show that
$\frac{1}{2\pi i}\int_{\gamma(0,3)}\frac{e^{(zk)}}{z^2+1}\mathrm dz = \sin k$ for all $k \in \mathbb{C}$. And $\gamma(z_0,R)$ is the circular contour $z_0+Re^{it}$, where $0\le t\le 2\pi.$ I know I need to use contour integration, but I cant seem to solve this problem.
Do I possibly have to use $z=e^{it}$ and $\sin t=\frac{1}{2i}(z-\frac{1}{z})?$ Its just a thought, I most probably might be wrong on this..