Laurent Series given some annuli

Question

The question is to represent $g(z) := \frac{1}{(z-3)(z+1)}$ by a Laurent series, with regards to the annuli $A_1:= \{z \in \mathbb{C} : 1 < |z| < 3\}$ and $A_2:= \{z \in \mathbb{C} : 3 < |z|\}$

I'm really struggling with understanding how to turn stuff into Laurent series, especially with regards to Annuli. I have worked out that $g(z)$ can be converted into partial fractions, resulting in $g(z) = \frac{\frac{1}{4}}{z-3}-\frac{\frac{1}{4}}{z+1}$. It's at this point that I'm stuck.

Answer 1

Hint

This one is about as straightforward as they come.

Look at $\dfrac 1{z-3}$ two ways:

1. $\dfrac 1{z-3}=-\frac 13\dfrac 1{1-\frac z3}=-\frac 13\sum(\dfrac z3)^n$ on $\lvert z\rvert \lt3$
2. $\dfrac 1{z-3}=\frac 1z\dfrac 1{1-\frac 3z}=\frac 1z\sum( \dfrac 3z)^n$ for $\lvert z\rvert \gt3.$

And of course $\dfrac 1{z+1}=\frac 1z\dfrac 1{1-(-\frac 1z)}=\frac 1z\sum(-\dfrac 1z)^n$ for $\lvert z\rvert \gt1.$

Now put them together. In other words, on $1\lt\lvert z\rvert \lt3$, get $$\frac 14(\dfrac 1{z-3}-\dfrac 1{z+1})=\frac 14(\sum(-\frac 13)(\dfrac z3)^n-(\frac1z)(-\dfrac 1z)^n).$$

And on $\lvert z\rvert \gt3,$ we have $$\dfrac 1{4z}\sum((\frac 3z)^n-(-\frac 1z)^n).$$

Answer 2

For $A_1$ $$\begin{align*} g(z)&=\frac{1}{4}\cdot\frac{1}{z-3}-\frac{1}{4}\cdot\frac{1}{z+1} \\ &=-\frac{1}{12}\cdot\frac{1}{1-z/3}-\frac{1}{4z}\cdot\frac{1}{1+1/z} \\ &=-\frac{1}{12}\cdot\left[1+\left(\frac{z}{3}\right)+\left(\frac{z}{3}\right)^2+\mathcal{O}\left(z^3\right)\right]-\frac{1}{4z}\cdot\left[1-\frac{1}{z}+\frac{1}{z^2}+\mathcal{O}\left(\frac{1}{z^3}\right)\right] \\ &=-\frac{1}{12}\cdot\sum_{n=0}^\infty \left(\frac{z}{3}\right)^n-\frac{1}{4}\sum_{n=0}^\infty\frac{(-1)^n}{z^{n+1}} \end{align*}$$ The idea here is that because $A_1$ is defined as $1<|z|<3$, we need the geometric series to converge in that region. Recall that the geometric series, $$\frac{1}{1-z}=\sum_{n=0}^\infty z^n$$ converges when $|z|<1$. So for example the geometric series, $\sum_{n=0}^\infty \left(\frac{z}{3}\right)^n$ is convergent in the region $|z|<3$ and the other series $\sum_{n=0}^\infty\left(-\frac{1}{z}\right)^n$ converges for $|z|>1$. Combining this we get a laurent expansion valid in $A_1$.

For $A_2$ $$\begin{align*} g(z)&=\frac{1}{4}\cdot\frac{1}{z-3}-\frac{1}{4}\cdot\frac{1}{z+1} \\ &=\frac{1}{4z}\cdot\frac{1}{1-3/z}-\frac{1}{4z}\cdot\frac{1}{1+1/z} \\ &=\frac{1}{4z}\cdot\left[1+\left(\frac{3}{z}\right)+\left(\frac{3}{z}\right)^2+\mathcal{O}\left(\frac{1}{z^3}\right)\right]-\frac{1}{4z}\cdot\left[1-\frac{1}{z}+\frac{1}{z^2}+\mathcal{O}\left(\frac{1}{z^3}\right)\right] \\ &=\frac{1}{4}\sum_{n=0}^\infty\frac{3^n}{z^{n+1}}-\frac{1}{4}\sum_{n=0}^\infty\frac{(-1)^n}{z^{n+1}} \\ &=\frac{1}{4}\sum_{n=0}^\infty\frac{3^n-(-1)^n}{z^{n+1}} \end{align*}$$ Notice that we used the form $\sum_{n=0}^\infty\left(\frac{3}{z}\right)^n$ which converges for $|z|>3$, and the other series converges for $|z|>1$. So we know this is the laurent expansion with regards to $A_2$.