Does this qualify as a Laurent series?

Question

Evaluate the Laurent series around the singularity at $ z_0 = 3$.

$$ \frac{1}{z^2(z-3)} $$

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I can apply the geometric series as follows:

$$\frac{1}{z^2}\cdot\frac{1}{z-3}=-\frac{1}{3z^2}\cdot\frac{1}{1+\frac{z}{3}}=\frac{1}{3z^2}\sum_{n=0}^{\infty}(-1)^n\left(\frac{z}{3}\right)^n$$

though I'm not sure that if the final answer, since the Laurent series involves negative powers of $ n $.

What do you think?

Answer 1

Hint

It doesn't work.

$$\frac{1}{z^2(z-3)}=\frac{1}{z-3}\cdot \frac{1}{z^2}$$

You just have to develop $\frac{1}{z^2}$ around $z=3$ to get the result.

Answer 2

As Mariano commented you have a Laurent series around $z=0$. A Laurent series around $z=3$ would have powers of $z-3$.

To represent $\frac{1}{z^2}$ around $z=3$ consider $$ z=3\Bigl(1+\frac{1}{3}(z-3)\Bigr) $$ and use this to rewrite $\frac{1}{z^2}$ and then use $\frac{1}{(1-w)^2}=1+2w+3w^2+\dotsb$

Answer 3

For a Laurent series about the point $z_0=3$, you want the series to be in powers of $z-3$. $$ \begin{align} \frac1{z^2(z-3)} &=\frac1{((z-3)+3)^2(z-3)}\\ &=\frac1{9\left(1+\frac{z-3}3\right)^2}\cdot\frac1{z-3}\\ &=\frac19\frac1{z-3}\left(1-2\frac{z-3}3+3\frac{(z-3)^2}9-4\frac{(z-3)^3}{27}+\dots\right)\\ &=\frac1{9(z-3)}-\frac2{27}+\frac{3(z-3)}{81}-\frac{4(z-3)^2}{243}+\dots\\ &=\sum_{k=0}^\infty\frac{k+1}{(-3)^{k+2}}(z-3)^{k-1} \end{align} $$