= Solution
Inside the unit circle, use the <geometric series> for $1/(1-z)$:
$$
\boxed{\frac1{z^2(z-1)}=-\sum_{n=0}^{\infty}z^{n-2}=-\frac1{z^2}-\frac1z-1-z-z^2-\cdots,\qquad0<|z|<1.}
$$
Outside the unit circle, expand in powers of $1/z$ instead:
$$
\boxed{\frac1{z^2(z-1)}=\frac1{z^3}\frac1{1-z^{-1}}=\sum_{n=0}^{\infty}z^{-n-3},\qquad |z|>1.}
$$
The different <Laurent series> reflect different annuli separated by the <pole> at one. Both follow from absolutely convergent <geometric series> in their stated regions.
Back to article page
