what is the laurent expansion of this function

Question

expandd $f(z)=\frac{z^2}{(z+1)(z^2-1)}$ in a laurent series valid for $\{z\in C : 0<|z-1|<2\}$

I am practicing Laurent series. and I have a question. is my idea below correct?
$f(z)=\frac{1}{z-1}.\frac{z^2}{(z+1)^2}$ , therefore it is enough to find the expansion of $ \frac{z^2}{(z+1)^2}$, since $\frac{1}{z-1}$, does already have term $z-1$.
then I m going to take the derivative of the expansion of $\frac{1}{(z-1)+1}$ to find the expansion of $\frac{1}{(z+1)^2}$. but then I am stuck then with $z^2$. I am not really confident on how to fully elaborate a laurent expansion. I would appreciate any help.

Answer 1

$\frac{z}{z+1}=1-\frac{1}{z+1}=1-\frac{1}{(z-1)+2}=1-\frac{1/2}{1+\frac{z-1}{2}}$. Squaring, we find $\frac{z^2}{(z+1)^2}=1-\frac{1}{1+\frac{z-1}{2}}+\frac{1/4}{\left(1+\frac{z-1}{2}\right)^2}$. The second term is an alternating geometric series. Finally, $$ \left[-\frac{1}{1+\frac{z-1}{2}}\right]'=\frac{1/2}{\left(1+\frac{z-1}{2}\right)^2}. $$ Notice that the third term above is just this quantity multiplied by $1/2$. At the end, just multiply through by $\frac{1}{z-1}$.