Laurent expansion of function with singularities

Question

I'm trying to get the power series expansion of the following function about $ z=0$:

$$f(z)=\frac{z^3}{1-z^4}$$

I'm having some troubles with this. Is it possible to do the following:

$$\frac{1}{1-z^4} = \sum_{n=0}^\infty z^{4n}$$

Because if it is, then I know how to proceed.

Many thanks !

Answer 1

There's nothing shady here. It's just the geometric series $$\dfrac 1{1-z}=\sum_{n\ge0}z^n$$, after an obvious substitution.