Confirming my solution to the expansion of $z/(z-1)(z+3)$ I would just like a confirmation of my work for the following problem.

Represent the function $$ \frac{z}{(z-1)(z+3)} $$ as a Laurent Series around $z=0$ when $1<|z|<3$. 

Using partial fraction decomposition to simplify the problem a bit, 
$$ \frac{1}{(z-1)(z+3)} = \frac{1}{4(z-1)} - \frac{1}{4(z+3)} \\ \Rightarrow  \frac{z}{(z-1)(z+3)} = \frac{z}{4(z-1)} - \frac{z}{4(z+3)}$$
Tackling each fraction seperatly, 
$$ \frac{1}{z-1} = \frac{1}{z}\cdot \frac{1}{1-z^{-1}} = \frac{1}{z} \sum_{n=0}^\infty z^{-n} = \sum_{n=0}^\infty z^{-n-1}= \sum_{n=1}^\infty z^{-n} $$
$$ \Rightarrow \frac{z}{4} \cdot \frac{1}{z-1} = \frac{z}{4} \sum_{n=1}^\infty z^{-n} 
 = \sum_{n=1}^\infty \frac{z^{-(n-1)}}{4} = \sum_{n=0}^\infty \frac{z^{-n}}{4} \\ $$
$$ \frac{1}{3+z} = \frac{1}{3z} \cdot \frac{1}{1-(-3z)^{-1}} = \frac{1}{3z} \sum_{n=0}^\infty \frac{(-1)^n z^{-n}}{3} = \sum_{n=0}^\infty  \frac{(-1)^n z^{-n-1}}{9} = \sum_{n=1}^\infty \frac{(-1)^{n-1} z^{-n}}{9} \\ \Rightarrow \frac{z}{4}\cdot \frac{1}{3+z} = \frac{z}{4} \sum_{n=1}^\infty \frac{(-1)^{n-1} z^{-n}}{9} = \sum_{n=1}^\infty \frac{(-1)^{n-1} z^{-(n-1)}}{36} = \sum_{n=0}^\infty \frac{(-1)^{n} z^{-n}}{36} \\ $$
Now adding them back together,
$$ \frac{z}{(z-1)(z+3)} = \frac{z}{4(z-1)} - \frac{z}{4(z+3)} = \sum_{n=0}^\infty \frac{z^{-n}}{4} - \sum_{n=0}^\infty \frac{(-1)^{n} z^{-n}}{36} = \sum_{n=0}^\infty \frac{(9-(-1)^n)}{36} z^{-n} $$
Therefore, 
$$ \frac{z}{(z-1)(z+3)} = \sum_{n=0}^\infty \frac{(9-(-1)^n)}{36} z^{-n} \ \
 \ (1<|z|<3) $$
Does everything look correct?
Thank you!

The function
\begin{align*}
 f(z)&=\frac{z}{(z-1)(z+3)}\\
 &=\frac{1}{4(z-1)}+\frac{3}{4(z+3)}\\
\end{align*}
 has two simple poles at $z=1$ and $z=-3$.

Since we want to find a Laurent expansion with center $0$, we look at the poles $1$ and $-3$ and see they determine three regions.
\begin{align*}
 |z|<1,\qquad\quad
1<|z|<3,\qquad\quad
 3<|z|
 \end{align*} 
  
  
*
  
*The first region $ |z|<1$ is a disc with center $0$, radius $1$ and the pole $1$ at the boundary of the disc. In the interior of this disc all two fractions with poles $1$ and $-3$  admit a representation as power series at $z=0$.
  
*The second region $1<|z|<3$ is the annulus with center $0$, inner radius $1$ and outer radius $3$. Here we have a representation of the fraction with the pole at $1$ as principal part of a Laurent series at $z=0$, while the fraction with pole at $-3$ admits a representation as power series.
  
*The third region $|z|>3$ containing all points outside the disc with center $0$ and radius $3$ admits for all fractions a representation as principal part of a Laurent series at $z=0$.

A power series expansion of $\frac{1}{z+a}$ at $z=0$ is
\begin{align*}
\frac{1}{z+a}&=\frac{1}{a}\cdot\frac{1}{1+\frac{z}{a}}\\
&=\sum_{n=0}^{\infty}\frac{1}{a^{n+1}}(-z)^n
\end{align*}
The principal part of $\frac{1}{z+a}$ at $z=0$ is
\begin{align*}
\frac{1}{z+a}&=\frac{1}{z}\cdot\frac{1}{1+\frac{a}{z}}=\frac{1}{z}\sum_{n=0}^{\infty}\frac{a^n}{(-z)^n}
=-\sum_{n=0}^{\infty}\frac{a^n}{(-z)^{n+1}}\\
&=-\sum_{n=1}^{\infty}\frac{a^{n-1}}{(-z)^n}
\end{align*}

We can now obtain the Laurent expansion of $f(z)$ at $z=0$ for all three regions. Here we have to consider 
  
  
*
  
*Region 2: $1<|z|<3$
  
  
  \begin{align*}
f(z)&=\frac{1}{4(z-1)}+\frac{3}{4(z+3)}\\
&=-\frac{1}{4}\sum_{n=1}^\infty\frac{(-1)^{n-1}}{(-z)^n}+\frac{3}{4}\sum_{n=0}^\infty \frac{1}{3^{n+1}}(-z)^n\\
&=\frac{1}{4}\sum_{n=1}^\infty\frac{1}{z^n}+\frac{1}{4}\sum_{n=0}^\infty\left(- \frac{1}{3}\right)^nz^n\\
\end{align*}
The Laurent expansion for the other regions can be calculated similarly.

