Let $C$ be the unit circle in the complex plane with positive orientation. Calculate $$\int_C \frac{dz}{sin^3z}$$
Solution: $\int_{|z|=1}\frac{dz}{sin^3z}=\bigg(sinz=0\Rightarrow z=k\pi,k\in\mathbb{Z}\Rightarrow z_1=0\in|z|=1\bigg)=\frac{2\pi i}{(3-1)!}(1)''\bigg|_{z_1=0}=0$
I am beginner in complex analysis, can you, please, check the solution
No, it is not correct. That integral is equal to$$2\pi i\operatorname{res}_{z=0}\left(\frac1{\sin^3z}\right)=\pi i.$$To see why, note that, since $\sin^3(z)=z^3-\frac12z^5+\cdots$, if$$\frac1{\sin^3z}=\frac a{z^3}+\frac bz+\cdots,$$we have\begin{align}1&=\left(z^3-\frac12z^5+\cdots\right)\left(\frac a{z^3}+\frac bz+\cdots\right)\\&=\left(1-\frac12z^2+\cdots\right)(a+bz^2+\cdots)\\&=a+\left(a-\frac12b\right)z^2+\cdots\end{align}So, $a=1$ and, since $a-\frac12b=0$, $b=\frac12$. And $b=\operatorname{res}_{z=0}\left(\frac1{\sin^3z}\right)$.