I must find the residues of $z^2\sin(\frac{1}{z})$ at $z = 0$.
Since $z = 0$ seems to be an Essential Singularity, i'm not sure how I can continue to find the residue of the function. Usually I am able to apply the Taylor Series and then find the $z^{-1}$ coefficient, but in this case I do not get a $z^{-1}$ term.
Since $w\mapsto\sin w$ is an entire function, we can write it as its Taylor series around $0$: $$ \sin w=\sum_{n=0}^{+\infty}\frac{(-1)^n}{(2n+1)!}w^{2n+1} $$ thus writing $w=\frac1z$ we get $$ \sin\frac1z=\sum_{n=0}^{+\infty}\frac{(-1)^n}{(2n+1)!}\frac1{z^{2n+1}} $$ which is valid on the punctured plane $\Bbb C^{\times}$, hence $$ f(z):=z^2\sin\frac1z=\sum_{n=0}^{+\infty}\frac{(-1)^n}{(2n+1)!}\frac1{z^{2n-1}}=z-\frac{1/6}z+\dots\;\; $$
which allows us to conclude that $\operatorname{Res}(f,0)=-\frac16$.
