Laurent series for 1/(z²-2z-3) in annulus 1<|z|<3

[MrGandalf]

Homework Statement

We have
$$f(z) = \frac{1}{z^2 - 2z - 3}$$
For this function, we want to find the Laurent Series around z=0, that converges when z=2 and we want to find the area of convergence.

Homework Equations

$$\frac{1}{z-3} = -\frac{1}{3}\bigg( \frac{1}{1 - \frac{z}{3}}\bigg) = -\frac{1}{3}\sum_{n=0}^{\infty}\frac{z^n}{3^n} = -\sum_{n=0}^{\infty}\frac{z^n}{3^{n+1}}$$

The Attempt at a Solution

There are two things I'm having trouble with.
I know that the function is analytic (holomorph) at the disk $D_1 = \{z \in C : |z| < 1 \}$, the annulus $D_2 = \{z \in C : 1 < |z| < 3 \}$ and in the area $D_3 = \{z \in C: 3 < |z|\}$.

Am I correct when I focus my attention to the area $D_2$? Since we are interested in the area where f is convergent for $z=2$?

I will try to find the Laurent series, and do so by first rewriting the function.
$$f(z) = \frac{1}{4}\bigg(\frac{1}{z-3} - \frac{1}{z+1}\bigg)$$

My first problem is that for $|z| > 1$ I know that
$$\frac{1}{1+z} = \sum_{n=0}^{\infty}\frac{(-1)^n}{z^{n+1}}$$
but I don't know why! I can only find information when $|z| < 1$. Can someone please explain why?

Regardless, I end up with the Laurent series:
$$f(z) = \frac{1}{4}\bigg( -\sum_{n=0}^{\infty}\frac{z^n}{3^{n+1}} -\sum_{n=0}^{\infty}\frac{(-1)^n}{z^n+1}\bigg)$$

The second problem is to find the area of convergence. I usually do the ratio test, but I don't really know where to start with this one.

Any help will be very appreciated!

 

[Dick]

The series in your first problem comes from writing 1/(1+z) as (1/z)*(1/(1+1/z)). Now do a geometric series expansion on the second factor. As for determining the region of convergence, a Laurent series generally converges on some sort of an annulus, r<|x|<R. You get the inner radius from testing the negative powers in your expansion and the outer radius from testing the positive powers.