Laurent series expansion of function $f(z) =\frac{z^2+1}{z(z-3)}$

Question

I am given the problem to calculate the Laurent series expansion fo $f(z) = \frac{z^2+1}{z(z-3)}$ in the regions $0<|z|<3$ and $3<|z|< \infty$. My question is around what center should I do the Laurent expansion?

Answer 1

For $0<|z|<3$, we find the expansion around $z=0$ as

$$\begin{align} \frac{z^2+1}{z(z-3)}&=-\frac13(z+z^{-1})\left(\frac{1}{1-z/3}\right)\\\\ &=-\frac13(z+z^{-1})\sum_{n=0}^\infty\frac{z^{n}}{3^{n}}\\\\ &=-\frac13\sum_{n=0}^\infty\frac{z^{n+1}}{3^{n}}-\frac13\sum_{n=0}^\infty\frac{z^{n-1}}{3^{n}}\\\\ &=-\frac{1}{3z}-\frac19-\frac{10}{9}\sum_{n=0}^\infty\left(\frac{z}{3}\right)^{n+1} \end{align}$$

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For $|z|>3$, we can facilitate our finding the expansion around $z=3$ by introducing a new variable $w=z-3$ and expanding around $w=0$. Proceeding, we can write

$$\begin{align} \frac{z^2+1}{z(z-3)}&=((w+3)^2+1)\left(\frac{1}{w(w+3)}\right)\\\\ &=\frac13(w+6+10w^{-1})\left(\frac{1}{1+w/3}\right)\\\\ &=\frac13(w+6+10w^{-1})\sum_{n=0}^\infty\frac{(-w)^{n}}{3^{n}}\\\\ &=-\sum_{n=0}^\infty \left(\frac{-w}{3}\right)^{n+1}+2\sum_{n=0}^\infty \left(\frac{-w}{3}\right)^{n}-\frac{10}{9}\sum_{n=0}^\infty \left(\frac{-w}{3}\right)^{n-1}\\\\ &=\frac{10}{3(z-3)}+\frac89-\frac19\sum_{n=0}^\infty\left(\frac{-(z-3)}{3}\right)^{n+1} \end{align}$$