Find Laurent Series & Area of Convergence for f(z)

In summary, the conversation is about finding the Laurent series for the function f(z) = \frac{1}{z^2 - 2z - 3} around z=0, which converges when z=2 and determining the area of convergence. The conversation also includes the attempt at finding the Laurent series and the problem of finding the region of convergence.
  • #1
MrGandalf
30
0

Homework Statement


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.

Homework 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]

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 problem is 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 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!
 
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  • #2
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.
 

1. What is a Laurent series and how is it different from a Taylor series?

A Laurent series is a representation of a complex function as an infinite sum of powers of z, both positive and negative. It is similar to a Taylor series, but it also includes negative powers of z, giving it a wider range of convergence.

2. How do you find the Laurent series for a given function?

To find the Laurent series for a function, you can use the Cauchy integral formula or the residue theorem. These methods involve finding the coefficients of the series by evaluating integrals or residues at points within the domain of convergence.

3. What is the area of convergence for a Laurent series?

The area of convergence for a Laurent series is the region in the complex plane where the series converges. This can be a annulus or a disk, depending on the function and the location of its singularities.

4. How do you determine the area of convergence for a given function?

The area of convergence can be determined by analyzing the singularities of the function within the complex plane. If the function has isolated singularities, the area of convergence will be an annulus. If the function has a branch point or essential singularity, the area of convergence will be a disk.

5. Can a function have multiple Laurent series and areas of convergence?

Yes, a function can have multiple Laurent series and areas of convergence. This can occur when a function has multiple singularities within its domain, or when the function has a branch point. In these cases, each singularity will have its own corresponding Laurent series and area of convergence.

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