I need to find the residue of $f(z) = z^2 \sin\left(\frac{1}{z^2}\right)$ at $z = 0$
I did the following:
$\sin (w)=\sum_{n=0}^{+\infty}\frac{(-1)^n}{(2n+1)!}w^{2n+1}$ $\implies$ $\sin\left(\frac{1}{z^2}\right)=\sum_{n=0}^{+\infty}\frac{(-1)^n}{(2n+1)!}\frac1{z^{4n+2}}$
$\implies f(z) = z^2\sin\left(\frac{1}{z^2}\right)=\sum_{n=0}^{+\infty}\frac{(-1)^n}{(2n+1)!}\frac1{z^{4n}} = 1-\frac{1}{6}z^{-4}+\frac{1}{120}z^{-8}-+...$
Now what do I do?
Let $g(z) = \sin z/z$. It has removable singularity at $0$, so it has Laurent series $$g(z) = a_0 + a_1z + a_2z^2+\ldots$$ on the punctured plane $\mathbb C\setminus\{0\}$. Then $$f(z) = g(1/z^2) = a_0 + a_1z^{-2} + \ldots$$ on $\mathbb C\setminus\{0\}$ and thus the residue is $0$.