<s>[INST] Laurent Series of $(z^2-1)^{-2}$ I have problems to determine the Laurent series of the function $(z^2-1)^{-2}$ in the regions:
$$0<|z-1|<2$$
and
$$|z+1|>2$$.
My idea was as follows:
$$\frac{1}{z^2-1}=\frac{1}{4z(z-1)^2}-\frac{1}{4z(z+1)^2}$$
However, this procedure only makes it harder to make the series, and tried to do it by Taylor.
 [/INST] Make a partial fraction decomposition:
$$\begin{align}
\frac{1}{z^2-1} &= \frac{1}{2}\left(\frac{1}{z-1} - \frac{1}{z+1}\right)\\
\frac{1}{(z^2-1)^2} &= \frac{1}{4}\left(\frac{1}{(z-1)^2} - \frac{2}{z^2-1} + \frac{1}{(z+1)^2}\right)\\
&= \frac{1}{4}\left(\frac{1}{(z-1)^2} - \frac{1}{z-1} + \frac{1}{z+1} + \frac{1}{(z+1)^2}\right).
\end{align}$$
Each of these terms can be expanded easily in the regions as geometric series (possibly once differentiated), when they can't remain as is.
</s>