Calculate the residues of $f(z)=\frac{1}{z^2\sin z}$

Question

Calculate the residues of this complex function

$$\frac{1}{z^2\sin(z)}$$

I can notice that we have singularities at $z=n\pi$, where $n=0,1,2,3,\dots$ But, how to find the residues?

Answer 1

Separate by cases:

$$z=0\;\implies\;\frac1{z^2\sin z}=\frac1{z^2\left(z-\frac{z^3}6+\ldots\right)}=\frac1{z^3\left(1-\frac{z^2}6+ \ldots\right)}=$$

$$=\frac1{z^3}\left(1+\frac{z^2}6+\frac{z^4}{36}+\ldots\right)=...+\frac1{6z}+\ldots$$

Thus, $\;z=0\;$ is a pole of order three and the function's residue at it is $\;\frac16\;$ .

$$z=n\pi,\,n\in\Bbb Z\;\implies\;\text{the poles are the same as those of sine and thus are simple, so:}$$

$$\text{Res}_{z=n\pi}(fg)=\lim_{z\to n\pi}\frac{z-n\pi}{z^2\sin z}\stackrel{\text{l'H}}=\lim_{z\to n\pi}\frac1{2z\sin z+z^2\cos z}=\frac1{n^2\pi^2(-1)^n}=\frac{(-1)^n}{n^2\pi^2}$$