Compute $\int_{\gamma}\frac{dz}{z\sin(z)}$ Compute $\int_{\gamma}\frac{dz}{z\sin(z)}$ where $\gamma:[0,2\pi]\to \Bbb{C}$ is given by $\gamma(t)=e^{it}$
Im having a problem with this integral since, $\frac{1}{\sin(z)}$ is not holomorphic in the unit disc. I tried by doing series expansion but Im getting nowhere. Any hint?

The only singularity of $f(z)=1/(z \sin z)$ in the unit circle is at the origin. There, we have the Laurent-series
$$ f(z) = \frac{1}{z [z -z^3/6 + O(z^5)]} = \frac{1}{z^2} + \frac{1}{6} + O(z^2) .$$
The integral you try to find, can then be obtained either by residue theorem (of simply integrating the Laurent series. We have that $$\text{Res}_{z=0} f(z) =0$$ and thus your integral vanishes.
