So, as the question states, I'm trying to find the Laurent series expansion of $\dfrac{z^2}{z^2-1}$ about $z_0=1$.
I've tried fiddling with geometric series stuff (the form of the rational function makes me want to), but I have to keep reminding myself that that won't work because we're trying to get the series expansion about $z_0=1$, not $z_0=0$.
I've also noticed that it can be put in the form
$$\left(\frac{z^2}{z+1}\right)\frac{1}{z-1}$$
And so in that form all I'd have to do is find the series expansion of $\dfrac{z^2}{z+1}$ about $z_0=1$, and that doesn't seem to be any easier.
And finally, I've also tried to use partial fractions on it
$$\frac{z^2}{z^2-1}=\frac{z^2}{2}\left(\frac{1}{z-1}-\frac{1}{z+1}\right)$$
But still, that doesn't seem to help much.
Could you please give me some direction here?