TITLE: How to find a Laurent Series for this function
QUESTION [5 upvotes]: How do I give a Laurent Series on various ranges of $|z|$? I need to find the Laurent series expansion for $$f(z)=\frac{1}{z(z-1)(z-2)}$$
for the following ranges of $|z|$:
$0<|z|<1$
$1<|z|<2$
$2<|z|$
I've calculated the partial fractions expasion of $f(z)$ to be $$f(z)=\frac{1}{2z}+\frac{1}{z-1}+\frac{1}{2(z-2)}$$

REPLY [8 votes]: Prove the equality
$$\frac{1}{z(z-1)(z-2)}=\frac{1}{2z}-\frac{1}{z-1}+\frac{1}{2(z-2)}$$
Then write each summand in a suitable way in order to be able to apply geometric series expansion
$$\frac{1}{1-w}=\displaystyle\sum_{n=0}^{+\infty}w^k$$
when $|w|<1$.
1) 0<|z|<1
Write 
$$f(z)=\frac{1}{2z}+\frac{1}{1-z}-\frac{1}{4}\frac{1}{1-\frac{z}{2}}$$
2) 1<|z|<2
write
$$f(z)=\frac{1}{2z}-\frac{1}{z}\frac{1}{1-\frac{1}{z}}-\frac{1}{4}\frac{1}{1-\frac{1}{z}}$$
3) |z|>2
write
$$f(z)=\frac{1}{2z}-\frac{1}{z}\frac{1}{1-\frac{1}{z}}+\frac{1}{2z}\frac{1}{1-\frac{2}{z}}$$

REPLY [5 votes]: Consider a term $$\frac{a}{z-b}.$$
If $|z| < |b|$ write this as
$$-\frac{a}{b} \frac{1}{1-\frac{z}{b}}$$
and note that $|z/b| < 1$.  If $|z| > |b|$ then write it as
$$
\frac{a}{z} \frac{1}{1 - \frac{b}{z}}
$$
and note that $|b/z| < 1$.  Use the geometric series in both cases to get the proper Laurent expansion.  Do this for each term in the partial fraction decomposition.