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 expansion of $f(z)$ to be $$f(z)=\frac{1}{2z}+\frac{1}{z-1}+\frac{1}{2(z-2)}$$
On
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.
On
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?