So I wanted to find the Laurent series expansion for $f(z) = \frac{z^3}{z^2-3z+2}$. So I guess more specifically, my question pertains how the expansion may change for different domains: $|z| < 1; 1 < |z| < 2; |z| > 2; 0 < |z − 1| < 1; |z − 1| > 1$.
I first did partial fraction decomposition, leading to:$$f(z) = \frac{z^3}{z^2-3z+2} = z+3+ \frac{20}{z-2} - \frac{13}{z-1}$$
From here, my question is, how do I know if I take out a $\frac{1}{z}$ from the fractions or not? I'm guessing it depends on the domain, but if anyone could outline this for me given my domain choices, that would help immensely. Would I consider taking out the $\frac{1}{z}$ if $|z| >1$?
$$z+3+ \frac{20}{z-2} - \frac{13}{z-1} = z+3+ \frac{20}{z}\frac{1}{1-(\frac{2}{z})} + 13\frac{1}{1-z} $$ $$= z+3+\frac{20}{z}\sum_{n=0}^\infty (\frac{2}{z})^n +13\sum_{n=0}^\infty z^n $$
(in writing this, I'm forming my own conclusions: I'm thinking that if I wanted $|z| >1$ then I would do exactly what I did in the above line, but reversing the power series in terms of taking of a $\frac{1}{z}$ and doing it instead for the $\frac{1}{1-z}$. Is this correct (i.e, what I did above would be for the case $|z|>2$? And in general, if $|z|>n$ we would take out a $\frac{1}{z}$ for that fraction?)
And in the case the domain is $0 < |z − 1| < 1$, would I then try taking out $z-1$ from the denominator of $\frac{1}{z-2} = -\frac{1}{1-(z-1)}$ and substitute $z-1$ for $z$ in the summation notation?
As always any help would be awesome!
If you want to find the Laurent expansion of $\frac{1}{z-b}$ in an annulus centred at $a$ with some given radii, you may rewrite the fraction as $$ \frac{1}{z-b}=\begin{cases} \frac{1}{a-b}\frac{1}{1-w} &\text{when $b\ne a$ and $|z-a|<|b-a|$, where $w=\frac{z-a}{b-a}$,}\\ \frac{1}{b-a}\frac{w}{1-w} &\text{when $b\ne a$ and $|z-a|>|b-a|$, where $w=\frac{b-a}{z-a}$,}\\ \frac{1}{z-b} &\text{when $b=a$.}\\ \end{cases} $$ In the first two cases, the whole point is to expand $(1-w)^{-1}$ as $1+w+w^2+\cdots$, where $w$ is a scalar multiple of either $z-a$ or $(z-a)^{-1}$ (because we want an expansion about $a$) with $|w|<1$ (so that $1+w+w^2+\cdots$ converges). In your example, $$ \frac{1}{z-1}=\begin{cases} -\frac{1}{1-w} &\text{on the annulus $0<|z|<1$, where $w=z$,}\\ \frac{w}{1-w} &\text{on the annuli $1<|z|<2$ and $|z|>2$ where $w=\frac{1}{z}$,}\\ \frac{1}{z-1} &\text{on the annuli }0<|z-1|<1 \text{ and } |z-1|>1,\\ \end{cases} $$ $$ \frac{1}{z-2}=\begin{cases} -\frac12\frac{1}{1-w} &\text{on the annuli $0<|z|<1$ and $|z|<2$, where $w=\frac{z}{2}$,}\\ \frac{1}{2}\frac{w}{1-w} &\text{on the annulus $|z|>2$, where $w=\frac{2}{z}$,}\\ -\frac{1}{1-w} &\text{on the annulus $0<|z-1|<1$, where $w=z-1$,}\\ \frac{w}{1-w} &\text{on the annulus $|z-1|>1$, where $w=\frac{1}{z-1}$.}\\ \end{cases} $$