How to find a Laurent Series for this function

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 $$f(z)=\frac{1}{2z}+\frac{1}{z-1}+\frac{1}{2(z-2)}$$

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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
$$f(z)=\frac{1}{2z}-\frac{1}{z}\frac{1}{1-\frac{1}{z}}+\frac{1}{2z}\frac{1}{1-\frac{2}{z}}$$

First, it is always useful to write $f$ in the form of partial fractions (while $\frac1z$ is ok anyway, since we need powers of $z$):
$$f(z)=\frac1{1(z-1)(z-2)}=\frac1z\left(\frac1{z-2}-\frac1{z-1}\right)=\frac1z\frac1{z-2}-\frac1z\frac1{z-1}$$
Finding the series in each domain is similar. For example, let’s find the series in $1<|z|<2$: we will always use the expansion $\frac1{1-x}=\sum_{n=0}^\infty x^n$ for $|x|<1$.
In our domain, $1<|Z|$, hence $\left|\frac1z\right|<1$, so in this domain $$\frac1z\frac1{z-1}=\frac1{z^2}\frac1{1-\frac1z}=\frac1{z^2}\sum_{n=0}^\infty\frac1{z^n}=\sum_{n=2}^\infty\frac1{z^n}$$
Do the same to $\frac1z\frac1{z-2}$, while remembering that $|z|<2$. Can you continue?

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.