Application for using Cauchy's derivative formula in Complex integration

Exercise:

Show that $\oint_{\left| z \right|=2\pi }{\frac{{{z}^{2}}\sin z}{{{\left( z-\pi  \right)}^{3}}}dz}=-4\pi^2 i$ where $z,i\in \mathbb{C}$

Solution: It’s clear that the above integral is similar to Cauchy’s formula for k-th derivative .

That is ${{f}^{\left( k \right)}}\left( w \right)=\frac{k!}{2\pi i}\oint_{\gamma }{\frac{f\left( z \right)}{{{\left( z-w \right)}^{k+1}}}dz}$  $\forall z\in D,\,k\ge 0$

Hence $f\left( z \right)={{z}^{2}}\sin z\,,\,w=\pi \,\And \,k+1=3$ thus $k=2$

Thus we need to find $f''\left( \pi  \right)$

we have $f\left( z \right)={{z}^{2}}\sin z\Leftrightarrow f'\left( z \right)=2z\sin z+{{z}^{2}}\cos z$

$\Leftrightarrow f''\left( z \right)=2\sin z+2z\cos z+2z\cos z-{{z}^{2}}\sin z$hence $f''\left( \pi  \right)=-4\pi $

But $-4\pi =f''\left( \pi  \right)=\frac{2!}{2\pi i}\oint_{\left| z \right|=2\pi }{\frac{{{z}^{2}}\sin z}{{{\left( z-\pi  \right)}^{3}}}dz}$

Thus $\oint_{\left| z \right|=2\pi }{\frac{{{z}^{2}}\sin z}{{{\left( z-\pi  \right)}^{3}}}dz}=\frac{2\pi i}{2}f''\left( \pi  \right)=\frac{2\pi i}{2}\left( -4\pi  \right)=-4{{\pi }^{2}}i$