Laurent series of $f(z)=\frac{1}{z^2-1}$

Question

Find the Laurent series of $f(z)=\frac{1}{z^2-1}$ in the annulus $1<|z-2|<3$

So I use partial fractions and get

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

I asked some question earlier today about these manipulations to geometric series but I still don't get it really. This is what I am doing so far

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

and then I just switch it over to a geometric series, but is this even right? Im so lost when it comes to the radius of convergence and how to manipulate the terms correctly. How should I think?

Answer 1

First we introduce $-2+2$ around $z$: $$-\frac{1}{1-z}=-\frac{1}{1-(z-2+2)}$$

Now lets get the $+2$ out of the brackets: $$=-\frac{1}{-1-(z-2)}=\frac{1}{1+(z-2)}$$

Now we use the geometric series $$=1-(z-2)+(z-2)^2-(z-2)^3\pm \cdots$$

Lets look at the general case. We want the Laurent series in the anulus $b-1<|z-b|<b+2$, where b is positive:

$$\frac{1}{a-z}=\frac{1}{a-(z-b+b)}=\frac{1}{a-b-(z-b)}=\frac{1}{a-b}\frac{1}{1-\frac{z-b}{a-b}}=\frac{1}{a-b}\sum_{n=0}^{\infty}\frac{(z-b)^n}{(a-b)^n}$$

Answer 2

Since $|z-2|\gt1$, we expand $\frac1{z-1}$ as a series in $\frac1{2-z}$.
Since $|z-2|\lt3$, we expand $\frac1{z+1}$ as a series in $\frac{2-z}3$. $$ \begin{align} \frac1{z^2-1} &=\frac12\left(\frac1{z-1}-\frac1{z+1}\right)\\ &=\frac12\left(-\frac1{(2-z)-1}-\frac1{3-(2-z)}\right)\\ &=-\frac12\underbrace{\sum_{k=-\infty}^{-1}(2-z)^k}_{\text{converges: $|2-z|\gt1$}} -\frac16\underbrace{\sum_{k=0}^\infty\left(\frac{2-z}3\right)^k}_{\text{converges: $|2-z|\lt3$}} \end{align} $$