A question about complex integral: $\int_{|z| = 1}\frac{1}{z^{2}\sin z}\,dz.$

Question

I am stuck computing the following complex integral $$\int_{|z| = 1}\frac{1}{z^{2}\sin z}\,dz.$$ Any help will be appreciated.

Answer 1

Hint. Use the Residue Theorem. The integrand is a meromorphic function which has a pole of order $3$ at $0$ inside the closed simple path $|z|=1$: $$\int_{|z| = 1}\frac{1}{z^{2}\sin z}\,dz=2\pi i\cdot\mbox{Res}\left(\frac{1}{z^{2}\sin z},0\right).$$ In order to evaluate the residue, note that $\sin z=z-z^3/6+o(z^3)$ and therefore $$\frac{1}{z^{2}\sin z}=\frac{1}{z^{2}(z-z^3/6+o(z^3))}= \frac{1}{z^{3}(1-z^2/6+o(z^2))}=\frac{1+z^2/6+o(z^2)}{z^{3}}.$$