Let $\displaystyle C$ be the unit circle with the positive sense.

Evaluate $\displaystyle \int _C \frac{dz}{\sin z}$.

My attempt :

Ok so first I notice that $\displaystyle \sin (z)=0$ only when $\displaystyle z=0$ (at least in the circle, if I'm not wrong).

So there is one singularity in the domain of integration.

My idea was to determine the order of the pole (I'm all confused with poles and singularities) of $\displaystyle \frac{1}{\sin z}$ at $\displaystyle z=0$.

I found the Laurent series of $\displaystyle \frac{1}{f(z)}$ to be worth $\displaystyle 1-\frac{z^3}{3!}+\frac{z^5}{5!}-...=1-z^3 \left ( \frac{1}{3!}-\frac{z^2}{5!}+... \right )$ thus the zero of $\displaystyle \frac{1}{f(z)}$ is of order 3, hence the order of the pole of $\displaystyle f$ is of order $\displaystyle 3$.

This implies that the integral is worth $\displaystyle 3\cdot 2 \pi i= 6\pi i$.

Am I right?

(sorry for all these questions, tomorrow I'm going to ask help at university since I've a lecture. I might be asking here at night though

)