Laurent series in a certain region

Question

The problem: Find the laurent series expansion in powers of $z+1$ of the function:

$g(z)= \frac{1}{(z+2)(z-2)}$

For the region $1<|z-1|<3$

My attempt: I suspect I need to do a change of variables. If I try to solve this problem the "normal" way (working with partial fractions and so on) I get stuck. So far I've done the partial fractioning of $g(z)$ and got:

$ g(z) = \frac{-1/4}{z+2} + \frac{1/4}{z-2} $

What I'd usually do next is work with the denominator to try and use the geometric series identity (either dividing the denominator by z or by a set number, depends on the convergence I'd need). My main problem is the "region" I'm asked to work on. I have trouble figuring out how to make something in the likes of $a_n(z+1)^n$ converge for $1<|z-1|<3$.

Any help will be appreciated .

Side note: Please tell me if this problem makes sense. I could have made a mistake while copying.

Answer 1

The partial fraction decomposition looks correct.

$g(z) = \frac{-\frac {1}{4}}{z+2} + \frac{\frac {1}{4}}{z-2}$

You have a contridiction. Do we want to center around $z = 1$ or $z = -1$

The annulus is centered at $z = 1$ but that would suggest the Laurent series should be in powers of $(z-1)$

$g(z) = -\frac 14 \frac{1}{(z-1)+3} + \frac {1}{4} \frac{1}{(z-1)-1}$

Now we need to use that if $|z| < 1:$

$\frac{1}{1-z} = \sum_\limits{n=0}^{\infty} z^n$

And do what algebra is necessary to get there.

$\frac {1}{4} \frac{1}{(z-1)-1} = -\frac {1}{4} \frac{1}{1-(z-1)}$ but we have that $|z-1| > 1$

$\frac {1}{4(z-1)} \frac{1}{1-\frac {1}{(z-1)}} = \frac 14 \sum_\limits{n=1}^{\infty} (z-1)^{-n}$

$-\frac 14 \frac{1}{(z-1)+3} = -\frac {1}{12} \frac {1}{1+\frac{z-1}{3}} = -\frac {1}{12} \sum_\limits{n=0}^{\infty} (-1)^n\left(\frac {z-1}{3}\right)^n$

$ \frac 14 \sum_\limits{n=-\infty}^{1} (z-1)^{n} - \frac 1{12} \sum_\limits{n=0}^{\infty} (z-1)^{n}$

Answer 2

Note that, $$g(z)=\frac{1}{4}\left(\dfrac{1}{z-2}-\dfrac{1}{z+2}\right)$$

Now we find a laurent expansion in powers of $z+1$, so

$$\dfrac{1}{z-2}=\dfrac{1}{(z+1)-3}=-\frac{1}{3}\left(\dfrac{1}{1-\frac{z+1}{3}}\right)=-\frac{1}{3}\sum_{n=0}^{\infty}{\left(\frac{z+1}{3}\right)^n}$$

And this series is for each $z\in\mathbb{C}$ such that $|z+1|<3$. Furthermore,

$$\dfrac{1}{z+2}=\dfrac{1}{1+(z+1)}=\sum_{n=0}^{\infty}{(-1)^n(z+1)^n}$$. Therefore,

$$g(z)=\frac{1}{(z+2)(z-2)}=\sum_{n=0}^{\infty}{\left[(-1)^n-\frac{1}{3^{n+1}}\right]\dfrac{(z+1)^n}{4}}$$

Answer 3

As was noted, we can do it if, say, the annulus is $1\lt\mid z+1\mid\lt3$.

First $\frac1{(z+2)(z-2)}=\frac14(\frac1{z-2}-\frac1{z+2})$.

Now, $\frac1{z+2}=\frac1{1+(z+1)}=\frac1{z+1}\frac1{1-(-\frac1{z+1})}=\sum_{n=0}^\infty (-1)^n(z+1)^{-n-1}$, for $\mid z+1\mid\gt1$.

And, $\frac1{z-2}=\frac1{-3+(z+1)}=-\frac13\cdot\frac1{1-(\frac{z+1}3)}=-\frac13\sum_{n=0}^\infty (\frac{z+1}3)^n$, for $\mid\frac{z+1}3\mid\lt1$.

So, we get $\frac14\sum_{n=-\infty}^0(-1)^n(z+1)^n-\frac1{12}\sum_{n=0}^\infty(\frac{z+1}3)^n$.