I want to find the Laurent Series for $$\frac{1}{(z-2)(z-4)}$$

valid for $|z-2|>2$.

I'm confused as to the radius of convergence I don't know how to expand the series valid for $\frac{2}{|z-2|}<1$. I used partial fraction decomposition and obtained $$\frac{-1/2}{z-2}+\frac{1/2}{z-4}$$. My next step that I want to do is to try and rewrite these fractions as a form of $ \large \large \large \frac{1}{1-\frac{2}{z-2}}$ so that I can use the definition of $\frac{1}{1-z}=\sum_{k=0}^{\infty} z^n$. Here's where I'm stuck. Can I get some help here? Thank you.

  • $\begingroup$ This is a question on a previous exit exam that I'm preparing. I just want to make sure that the radius of convergence works out for $|z-2|>2$ because if it was $|z-2|>1$ then I could easily do this. $\endgroup$ – User69127 Apr 28 '14 at 20:18
  • $\begingroup$ About which z do you expand? Your function has poles in z=2 and z=4. But your condition keeps z away from them. Setting ζ=z-3 you have analyticity for |ζ|<1. $\endgroup$ – Urgje Apr 28 '14 at 20:44

I'll assume your partial fraction is correct, the n we advance as

$$ \frac{-1/2}{z-2}+\frac{1/2}{z-4} = \frac{-1/2}{z-2}+\frac{1/2}{(z-2)-2}=-\frac{1}{2}\frac{1}{z-2}+\frac{1}{2(z-1)}\frac{1}{1-\frac{2}{z-2}} $$

$$ = -\frac{1}{2}\frac{1}{z-2}+\frac{1}{2(z-2)}\sum_{k=0}^{\infty} \frac{2^k}{(z-2)^k},\quad |z-2|>2 .$$

Now, just rearrange things to get a better form for the solution.

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  • $\begingroup$ How is this portion true? $\frac{1/2}{(z-2)-2}=\frac{1}{2(z-1)}\frac{1}{1-\frac{2}{z-2}} $? $\endgroup$ – User69127 Apr 30 '14 at 9:04

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