(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

We have

[tex]f(z) = \frac{1}{z^2 - 2z - 3}[/tex]

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.

2. Relevant equations

[tex]\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}}[/tex]

3. 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 [itex]D_1 = \{z \in C : |z| < 1 \}[/itex], the annulus [itex]D_2 = \{z \in C : 1 < |z| < 3 \}[/itex] and in the area [itex]D_3 = \{z \in C: 3 < |z|\}[/itex].

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

I will try to find the Laurent series, and do so by first rewriting the function.

[tex]f(z) = \frac{1}{4}\bigg(\frac{1}{z-3} - \frac{1}{z+1}\bigg)[/tex]

My first problemis that for [itex]|z| > 1[/itex] I know that

[tex]\frac{1}{1+z} = \sum_{n=0}^{\infty}\frac{(-1)^n}{z^{n+1}}[/tex]

but I don't know why! I can only find information when [itex]|z| < 1[/itex]. Can someone please explain why?

Regardless, I end up with the Laurent series:

[tex]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)[/tex]

The second problemis 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!

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# Homework Help: Laurent Series

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