Laurent expansion for a complex function with 3 singularites

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SUMMARY

The discussion focuses on finding the Laurent expansions for the function f(z)=30/((1+z)(z-2)(3+z)) around the origin, valid in three specified regions: Region 1 (-1<|z|<2), Region 2 (-3<|z|<-1), and Region 3 (|z|>2). The user successfully decomposed the function into partial fractions: f(z)=2/(z-2)+3/(3+z)-5/(1+z) and derived the expansion for Region 3 as f(z)_{z>2}=Σ_{n=1}^{∞}(2/z)^{n}. However, they require assistance in expanding for Regions 1 and 2 and understanding the convergence of the Taylor series.

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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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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?
 

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