What is the residue of this function, where both functions at denominator vanish?

$$\frac{1}{z^2*\sin(z)}, |z| < 1$$

I am having a hard time to find the residue of this function. Particularly, it is obvious that the singularity occurs at $z=0$, but what is the order of the pole? Since sin also is zero at $z=0$, i have no idea how to find the residue..

Any tips? Certainly to expand $cossec$ will help here, but since i don't know the expansion, i ask myself if there is other way to solve it. Maybe a faster way.

$$f(z)=\frac{1}{z^3\left(1-\frac{1}{6}z^2+\frac{1}{120}z^4-\cdots\right)}=\frac{1}{z^3}\left(1+\frac{1}{6}z^2+\cdots\right)\;,$$

so the residue of $f(z)$ at $z=0$ is $\frac{1}{6}\;.$

Also, this expansion is valid for $0\lt|z|\lt π$ as $z=\pm π$ are the nearest other singularities.

Why does $z^3=0$ have 3 complex roots and not simply just $0$?

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