Finding all Laurent series of a function I have to find all of the Laurent series for the function
$$ f(z)=\frac{z}{(z+1)(z-2)} $$ about $z=0$
I'm a little confused about the regions that I'm dealing with. I started with the partial fraction decomposition: $$ f(z)=\frac{1}{3(z+1)}+\frac{2}{3(z-2)} $$
Coming out from $z=0$ there's a Taylor series up until $z=-1$ that I guess works for both singularities? i.e. for $|z|<1$ the Laurent series is: $$\frac{1}{3}\sum_{n=0}^{\infty}((-1)^n-(\frac{1}{2})^n)z^n$$
Is the above right? If not how do you go about finding the rest of the Laurent series for $f(z)$? What I really need to understand is how you need to look at the regions because I'm finding it really confusing for this problem.
Thank you!

As @DanielFischer told you, the regions to consider are $(1)$ $\lvert z\rvert<1$, $(2)$ $1<\lvert z\rvert < 2$, and $(3)$ $\lvert z\rvert >2$. For $(1)$, your solution is correct.
For $(2)$, we want Geometric series with $\lvert r\rvert <1$ so $1<\lvert z\rvert\Rightarrow \frac{1}{\lvert z\rvert} < 1$ and $\frac{\lvert z\rvert}{2}<1$. Then
\begin{align}
\frac{1/3}{z+1}+\frac{2/3}{z-2} &=\frac{1}{z}\frac{1/3}{1+\frac{1}{z}}-\frac{1/3}{1-\frac{z}{2}}\\
&=\frac{1}{3z}\sum_{n=0}^{\infty}(-1)^n\Bigl(\frac{1}{z}\Bigr)^n-\frac{1}{3}\sum_{n=0}^{\infty}\Bigl(\frac{z}{2}\Bigr)^n\\
&=\frac{1}{3}\sum_{n=0}^{\infty}\biggl[(-1)^n\Bigl(\frac{1}{z}\Bigr)^{n+1}-\Bigl(\frac{z}{2}\Bigr)^n\biggr]
\end{align}
For $(3)$, we have $\lvert z\rvert >2$ so $\frac{1}{\lvert z\rvert} < 1$ which we found in $(2)$ and $\frac{2}{\lvert z\rvert}<1$. We only need to determine the Laurent series for $\frac{2}{\lvert z\rvert}<1$.
\begin{align}
\frac{1}{3z}\sum_{n=0}^{\infty}(-1)^n\Bigl(\frac{1}{z}\Bigr)^n+\frac{2/3}{z-2}&=\frac{1}{3z}\sum_{n=0}^{\infty}(-1)^n\Bigl(\frac{1}{z}\Bigr)^n+\frac{1}{z}\frac{2/3}{1-\frac{2}{z}}\\
&=\frac{1}{3z}\sum_{n=0}^{\infty}(-1)^n\Bigl(\frac{1}{z}\Bigr)^n+\frac{2}{3z}\sum_{n=0}^{\infty}(-1)^n\Bigl(\frac{2}{z}\Bigr)^n\\
&=\frac{1}{3}\sum_{n=0}^{\infty}(-1)^n\biggl[\Bigl(\frac{1}{z}\Bigr)^{n+1}+\Bigl(\frac{2}{z}\Bigr)^{n+1}\biggr]
\end{align}
