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!
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*}