Forgive me if this takes a while to post, attempting to get the LaTeX right.

Attempting to find a Laurent series expansion for the function:

$\displaystyle f (z)=\frac{4z^2}{1-z^2}$

Over 2 regions:

$\displaystyle 0<|z|<1$

$\displaystyle 1<|z|< {\infty} $

I have seen similar problems like:

$\displaystyle f(z)=\frac{z+2}{z-2}$ where you can simplify by taking

$\displaystyle f(z)=\frac{z+(-2+2)+2}{z-2}=\frac{z-2}{z-2}+\frac{4}{z-2}=1+\frac{2}{\frac{z}{2}-1}=1-\frac{2}{1-\frac{z}{2}}$

Which is solved here

I can't seem to find a way to simplify my equation in a similar manner.

I know that (obviously):

$\displaystyle f(z)=\frac{4z^2}{(1+z)(1-z)}$

Not sure where to go from there. If I knew a Maclaurin series for something like $\displaystyle \frac{z}{1+z}$ and $\displaystyle \frac{z}{1-z}$ or how to find one, that would make things much simpler as I could just multiply the series together...?

Thanks in advance to anyone who read all that.