Find the Laurent series for $p(4/z)$

Question

Find the Laurent series for $p(4/z)$ given that $p(z)=(z-3)^3$

My attempt: if the Taylor series for $p(z)$ looks like

$$\frac{-27}{0!}+ \frac{27z}{1!}-\frac{18z^2}{2!}+ \frac{6z^3}{3!}+0+0+0...$$

Then the Laurent series is probably the same thing but with a $4$ over the numerator. But then I'm confused about the point we expand about.

Answer 1

Let $p(z)=(z-3)^3$. Then,

$$\begin{align} p(4/z)&=\frac{(4-3z)^3}{z^3}\\\\ &=\frac{-27z^3+108z^2-144z+64}{z^3}\\\\ &=-27+108\frac1z-144\frac1{z^2}+64\frac1{z^3} \end{align}$$

Answer 2

You just need to find Taylor series, in this case the series is Laourent also because the series includes terms of negative degree.

$$p(z)=-27+27z-9z^2+z^3=\sum_{n=0}^{3}z^n(-3)^{3-n}\binom {3}{n}\\ p(4/z)=\frac{64}{z^3}-\frac{144}{z^2}+\frac{108}{z}-27$$