Laurent Series of f On Given Annulus I am attempting to solve the following question and I am encountering some difficulties:
Expand the function:
$$
f(z) = \frac{z}{z^2 + 2z -3}
$$
in powers of z to find a series that is valid for an annulus containing z=2. For what values of z does this series converge?
My attempt at a solution was to write out the function using partial fraction decomposition as:
$$
f(z) = \frac{3/4}{z+3} - \frac{1/4}{1-z}
$$
I then tried using the formula for a power series to write the two terms as:
$$
f(z) = \frac{3}{4}\frac{1}{1-(-\frac{z}{3})} - \frac{1}{4}\frac{1}{1-z} = \frac{3}{4}\sum_{k=0}^{\infty}(-1)^k \cdot (z/3)^k + \frac{1}{4}\sum_{k=0}^{\infty}(z)^k
$$
Though, it is clear that the second series is only valid for $|z| < 1$, so I am stuck. How should I move forward with this? Thank you for your help!
 A: We want to expand a series valid for an annulus containing $z=2$. When looking at
\begin{align*}
\color{blue}{f(z)=\frac{3}{4}\left(\frac{1}{z+3}\right)-\frac{1}{4}\left(\frac{1}{1-z}\right)}\tag{1}
\end{align*}

*

*the left summand of (1) can be expanded as usual. We derive using the geometrics series expansion
\begin{align*}
\color{blue}{\frac{1}{z+3}}=\frac{1}{3}\left(\frac{1}{1+\frac{z}{3}}\right)\color{blue}{=\frac{1}{3}\sum_{k=0}^{\infty}(-1)^k\left(\frac{z}{3}\right)^k\qquad\qquad |z|<3}\tag{2}
\end{align*}
We derive the region of convergence from $\left|\frac{z}{3}\right|<1$ which is a disc with center $0$ and radius $3$.


*Doing the same with the right summand of (1) is not helpful, since
\begin{align*}
\frac{1}{1-z}=\sum_{k=0}^{\infty}z^k\qquad\qquad |z|<1
\end{align*}
gives as region of convergence the unit disc with center $0$ which is not helpful since it doesn't contain $z=2$. But, if we expand $\frac{1}{1-z}$ in terms of $\frac{1}{z}$ instead of $z$ we get the complement of the unit disc as region of convergence (besides the boundary of the unit disc). We obtain
\begin{align*}
\color{blue}{\frac{1}{1-z}}&=\left(-\frac{1}{z}\right)\frac{1}{1-\frac{1}{z}}=\left(-\frac{1}{z}\right)\sum_{k=0}^{\infty}\left(\frac{1}{z}\right)^k\\
&=-\sum_{k=0}^{\infty}\frac{1}{z^{k+1}}\color{blue}{=-\sum_{k=1}^{\infty}\frac{1}{z^k}\qquad\qquad\qquad|z|>1}\tag{3}
\end{align*}
We derive the region of convergence from $\left|\frac{1}{z}\right|<1$ which is the complement of the unit disc with center $0$.


*The region of convergence of $f(z)$ is consequently the annulus
\begin{align*}
 \color{blue}{1<|z|<3}
 \end{align*}
and the series expansion is combining (1) to (3)
\begin{align*}
 \color{blue}{f(z)=\frac{1}{4}\sum_{k=0}^{\infty}(-1)^k\left(\frac{z}{3}\right)^k+\frac{1}{4}\sum_{k=1}^{\infty}\frac{1}{z^k}}
 \end{align*}
