Laurent expansion for a complex function with 3 singularites

In summary, the conversation is about finding three Laurent expansions for the function f(z)=\frac{30}{(1+z)(z-2)(3+z)} in different regions. The regions specified are -1<|z|<2, -3<|z|<-1, and |z|>2. The first step taken is to split f(z) into partial fractions and expand for the region |z|>2, resulting in the series f(z)_{z>2}=\Sigma_{n=1}^{\infty}(\frac{2}{z})^{n}. However, the person is unsure if this is correct and needs help expanding for the other regions. They also mention trying to calculate
  • #1
Dixanadu
254
2

Homework Statement


Hey guys,

So I need a bit of help with this question:

Find three Laurent expansions around the origin, valid in three regions you should specify, for the function
[itex]f(z)=\frac{30}{(1+z)(z-2)(3+z)}[/itex]


Homework Equations


None that I know of...just binomial expansion


The Attempt at a Solution



Okay so what I did was first specify the regions. Not sure if they are right though, although I think they are:

Region 1: [itex]-1<|z|<2[/itex]
Region 2: [itex]-3<|z|<-1[/itex]
Region 3: [itex]|z|>2[/itex]

Then I split f(z) into partial fractions:
[itex]f(z)=\frac{2}{z-2}+\frac{3}{3+z}-\frac{5}{1+z}[/itex]

Then I expanded for the region |z|>2, using the first term of the partial fractions (ignoring the other ones...right?) and I got

[itex]f(z)_{z>2}=\Sigma_{n=1}^{\infty}(\frac{2}{z})^{n}[/itex]

So now the problem is... first of all I don't know if that's right. Even if it is, I have no idea how to expand for the other regions...for example, say I wanted to do region 2...I don't even know where to start, do I first expand for -3<|z|, then |z|<-1 and add them...or what?

Really need some help here guys! the fate of the universe hinges on this unfortunate question sheet!
 
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  • #2
Let's start with the most obvious one: the regular Taylor series. Can you calculate it, and for what values of z does it converge?
 

FAQ: Laurent expansion for a complex function with 3 singularites

1. What is a Laurent expansion for a complex function?

A Laurent expansion is a way of representing a complex function as a series of terms involving powers of the variable, both positive and negative. It is similar to a Taylor expansion, but allows for singularities or poles in the function.

2. How many singularities can a complex function have in order for a Laurent expansion to be possible?

A complex function can have up to three singularities in order for a Laurent expansion to be possible. This is because the expansion involves dividing the function into three regions around each singularity.

3. Can a complex function have more than three singularities and still have a Laurent expansion?

No, a complex function can only have up to three singularities in order for a Laurent expansion to be possible. If there are more than three singularities, the function cannot be expanded using the Laurent series.

4. How can I determine the coefficients in a Laurent expansion?

The coefficients in a Laurent expansion can be determined by using the Cauchy integral formula, which involves integrating the function along a contour surrounding the singularity of interest. The coefficients can also be found by using the residues of the function at each singularity.

5. What is the significance of the principal part in a Laurent expansion?

The principal part in a Laurent expansion refers to the terms in the series that have negative powers of the variable. These terms represent the contribution of the singularities to the overall behavior of the function. By examining the principal part, we can gain insight into the behavior of the function near its singularities.

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