Complex Analysis- Finding Laurent Series I'm a bit confused about how to calculate all the laurent series about a given point in the complex plane. 
I have the complex function $$ f(z)=\frac{1}{z^2(z-3)}$$ I need to find all the Laurent series about $z=0$.
I understand $f(z)$ has singularities at $z=3$ and $z=0$, but I do not understand if the singularity at $z=0$ is special because of the $z^2$?
I also don't know which regions I should be looking at to determine the Laurent series, and which regions are Taylor series... and does finding all the Laurent series also mean finding the Taylor series...?
I'm pretty lost overall!
Please help :/
 A: *

*Yes, the singulaty at $z = 0$ is due ot $z^2$

*Try to draw the regions $0 < |z| < 3$ and $|z| > 3$. And also notice that $\frac{3}{|z|} < 1$ so we may use Taylor series around the point zero. 
The series are as follow: 
For $0 < |z| <3 $ we have that 
$$f(z)=\frac{1}{z^2(z-3)} = \frac{1}{z^2}\frac{1}{(z-3)} = -\frac{1}{3z^2}\frac{1}{(1-\frac{z}{3})} = -\frac{1}{3z^2} \sum_{n=0}^{\infty} \frac{z^n}{3^n} =- \sum_{n=0}^{\infty} \frac{z^{n-2}}{3^{n+1}}$$
For $|z| > 3$ we have that 
$$f(z)=\frac{1}{z^2(z-3)} = \frac{1}{z^2}\frac{1}{(z-3)} = \frac{1}{z^3}\frac{1}{(1-\frac{3}{z})} = \frac{1}{z^3} \sum_{n=0}^{\infty} \frac{3^n}{z^n} = \sum_{n=0}^{\infty} \frac{3^n}{z^{n+3}}$$
Any questions leave in the comments. 
A: The $z^2$ simply indicates that the laurent series will have the lowest order term of $z^{-2}$. Note that the Taylor series of a function will be convergent in a disc to the closest singularity, so the function you have does not have a taylor series about $0$. As for finding the laurent series, consider that
$$\frac{1}{z^2(z-3)}=-\frac1{3z^2}-\frac1{9z}+\frac{1}{9(z-3)}$$
Now the third term does have a Taylor series about $z=0$. We can find it by
$$\frac{1}{9(z-3)}=-\frac{1}{27\left(1-\frac z3\right)}=-\frac 1 {27} \sum_{k=0}^\infty \frac{z^n}{3^n}$$
and by the properties of geometric series, the radius of convergence i s $\left|\frac z3\right|<1\implies|z|<3$.
So we have 
$$\frac{1}{z^2(z-3)}=-\frac1{3z^2}-\frac1{9z}-\frac 1 {27} \sum_{k=0}^\infty \frac{z^n}{3^n}=-\frac1{27}\sum_{k=-2}^\infty\frac{z^k}{3^k}=-\sum_{k=0}^\infty\frac{z^{k-2}}{3^{k+1}}$$
for $|z|<3$.
As for the other region it can converge on ($|z|>3)$, consider that 
$$\frac{1}{9(z-3)}=\frac{1}{9z}\cdot\frac1{1-\frac3z}=\frac1{9z}\sum_{k=0}^\infty\frac{3^k}{z^k}=\sum_{k=0}^\infty\frac{3^{k-2}}{z^{k+1}}$$
which has a radius of convergence $\left|\frac3z\right|<1\implies |z|>3$. So we have 
$$\frac{1}{z^2(z-3)}=-\frac1{3z^2}-\frac1{9z}+\sum_{k=0}^\infty\frac{3^{k-2}}{z^{k+1}}=\sum_{k=2}^\infty\frac{3^{k-2}}{z^{k+1}}=\sum_{k=0}^\infty\frac{3^{k}}{z^{k+3}}$$
for $|z|>3$. Since the two annuli above cover the whole complex plane, we have found all the different Laurent series of the function.
