= Solution
The <Laurent theorem> says that if $f$ is analytic on an annulus
$$
r<|z-a|<R,
$$
then it has a unique <Laurent series>
$$
f(z)=\sum_{n=-\infty}^{\infty}c_n(z-a)^n
$$
converging locally uniformly on that annulus, where
$$
c_n=\frac1{2\pi i}\oint_C\frac{f(\zeta)}{(\zeta-a)^{n+1}}\,d\zeta
$$
for any positively oriented circle $C$ in the annulus around $a$.
An <isolated singularity> at $a$ is a point at which $f$ is not analytic although it is analytic on some punctured neighbourhood. It is removable when every $c_n$ with $n<0$ vanishes; it is a pole of order $m$ when $c_{-m}\ne0$ and $c_n=0$ for $n<-m$; and it is essential when infinitely many negative-index coefficients are nonzero.
For $0<|z|<1$,
$$
\frac1{z(z-1)}
=-\frac1z\frac1{1-z}
=-\sum_{n=0}^{\infty}z^{n-1}.
$$
For $|z|>1$,
$$
\frac1{z(z-1)}
=\frac1{z^2}\frac1{1-z^{-1}}
=\sum_{n=0}^{\infty}z^{-n-2}.
$$
The coefficients are unique after the annulus is fixed; these expansions differ because they represent the function on different annuli. At zero the first expansion has principal part $-z^{-1}$, so zero is a simple pole with residue $-1$.
Back to article page
