Laurent series of $\frac{1}{z^2}$ centered at $z=-1$ in the annulus $1<|z+1|<2$

Question

I have managed to compute the Laurent series for $\frac{1}{z}$, which is $\sum_{n=-\infty}^{-1} (z+1)^n$. Now, I'm no expert at power series, so I wonder how to compute the Laurent series for $\frac{1}{z^2}$. Is it possible to do the following:

$\frac{1}{z^2} = (\sum_{n=-\infty}^{-1} (z+1)^n)^2 = \sum_{n=-\infty}^{-1} (z+1)^{2n}$?

Answer 1

You can't do that. Do you really belive that the square of a sum is the sum of the squares?

Since$$\lvert z+1\rvert>1\implies\frac1z=(z+1)^{-1}+(z+1)^{-2}+(z+1)^{-3}+\cdots,$$you have, when $\lvert z+1\rvert>1$,\begin{align}\frac1{z^2}&=-\left(\frac1z\right)'\\&=(z+1)^{-2}+2(z+1)^{-3}+3(z+1)^{-4}+\cdots\end{align}