This is a question from a previous qualifying exam that I am working through to study for my own upcoming qual. I'm pretty sure that this one is straightforward and that I did it correctly, but can someone please confirm whether or not it is correct and complete?
Problem:
Consider the function $f(z) = \frac{z+6}{z^2-2z-3} $.
(a) Expand $f(z)$ in a Taylor Series about $z=0$ and find its radius of convergence.
(b) Expand $f(z)$ in a Laurent Series in the annulus $1<|z|<3$.
My Solution:
We first write $f(z)$ as a sum of partial fractions, such that $$ f(z) = \frac{z+6}{z^2-2z-3} = \frac{z+6}{(z-3)(z+1)} = \frac{\frac{9}{4}}{z-3} + \frac{-\frac{5}{4}}{z+1}. $$
Next, we rewrite each term such that it can take advantage of the known geometric series $\frac{1}{1-r} = \sum_{k=0}^\infty r^k$ for $|r|<1$ as follows.
$$ f(z) = \frac{\frac{9}{4}}{z-3} + \frac{-\frac{5}{4}}{z+1} = \frac{-\frac{3}{4}}{1-\frac{z}{3}} + \frac{-\frac{5}{4}}{1-(-z)}. $$
(a) We can now write $$ \frac{-\frac{3}{4}}{1-\frac{z}{3}} = -\frac{3}{4}\sum_{k=0}^\infty \left(\frac{z}{3}\right)^k = \sum_{k=0}^\infty \left(-\frac{3^{1-k}}{4} \right) z^k $$
for $\left\vert\frac{z}{3}\right\vert < 1 \implies |z|<3$ and
$$ \frac{-\frac{5}{4}}{1-(-z)} = -\frac{5}{4} \sum_{k=0}^\infty (-z)^k = \sum_{k=0}^\infty \left(\frac{5}{4}\right)(-1)(-1)^k z^k = \sum_{k=0}^\infty \left(\frac{5}{4}\right)(-1)^{k+1} z^k $$
for $|z|<1$. Thus, by combining these two terms, the function $f(z)$ can be written as the Taylor Series
$$ f(z) = \sum_{k=0}^\infty \left(-\frac{3^{1-k}}{4} \right) z^k + \sum_{k=0}^\infty \left(\frac{5}{4}\right)(-1)^{k+1}z^k = \sum_{k=0}^\infty \left(\frac{-3^{1-k}+5(-1)^{k+1}}{4} \right)z^k $$
with a radius of convergence of $|z|<1$.
(b) For the Laurent series, we want a series that will be valid on the annulus $1<|z|<3$. From above, we already have the analytic part of this series, such that
$$ \frac{-\frac{3}{4}}{1-\frac{z}{3}} = \sum_{k=0}^\infty \left(-\frac{3^{1-k}}{4} \right) z^k $$
for $\left\vert\frac{z}{3}\right\vert < 1 \implies |z|<3$.
We now need to consider $|z|>1 \implies \left\vert \frac{1}{z}\right\vert < 1$. We can obtain the principal part of the series by rewriting
$$ \begin{align*} \frac{-\frac{5}{4}}{z+1} &= \left(-\frac{5}{4}\right) \left(\frac{1}{z}\right)\frac{1}{1-\left(-\frac{1}{z}\right)} \\ &= \left(-\frac{5}{4}\right) \left(\frac{1}{z}\right) \sum_{k=0}^\infty \left(-\frac{1}{z}\right)^k \\ &= \sum_{k=0}^\infty \left(\frac{5}{4}\right)(-1)^{k+1} \frac{1}{z^{k+1}} \\ &= \sum_{k=1}^\infty \left(\frac{5}{4}\right)(-1)^{k} z^{-k} \end{align*} $$
for $|z|>1$. We combine the two parts to obtain the desired Laurent series $$ f(z) = \sum_{k=0}^\infty \left(-\frac{3^{1-k}}{4} \right) z^k + \sum_{k=1}^\infty \left(\frac{5}{4}\right)(-1)^{k} \frac{1}{z^{k}} $$
which is valid on the annulus $1<|z|<3$, as desired.