Is there another way to compute this Laurent series? $\frac{1}{(1+z)^2}$ Consider $g(z) = \frac{1}{(1+z)^2}$.  To compute a Laurent series of $g$ at $0$ on the region $|z| > 1$, I let $w = \frac{1}{z}$, so $$g(z) = \frac{w^2}{(1+z)^2w^2} = w^2 \frac{1}{(1+w)^2} $$ Since $|w| < 1$, we have $\frac{1}{1+w} = 1 - w + w^2 - \cdots$, so $$\frac{1}{(1+w)^2} = (1 - w + w^2 - w^3 + \cdots)(1 - w + w^2 - w^3 + \cdots)$$ $$ = 1 - 2w + 3w^2 - 4w^3 + \cdots$$ and therefore $$g(z) = w^2 \sum\limits_{n=0}^{\infty} (-1)^n (n+1)w^n =  \sum\limits_{n=0}^{\infty} (-1)^n (n+1)z^{-n-2}$$ My question is, is there a way to do this problem without having to resort to formal power series multiplication? 

HINT:
Note that $\frac{d}{dz}\left(\frac{1}{1+z}\right)=-\frac{1}{(1+z)^2}$.  Find the Laurent series for $\frac{1}{1+z}$ (this is pretty easy), differentiate term by term, and multiply by $-1$ should do it.
