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Find the Laurent series of $$f(z)=\frac{3}{(1-z)(z+2)}$$ centered at $0$ with the three domains $|z|<1$, $1<|z|<2$, $2<|z|$. I know you use partial fractions and I got $\frac{1}{1-z} + \frac{1}{z+2}+\dots$. If I could get help with one of them on how to find one of them that would be awesome.

Mhenni Benghorbal
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bob
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1 Answers1

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For example:

$$1<|z|<2\implies \frac{1}{1-z}+\frac{1}{2+z}=-\frac{1}{z}\frac{1}{1-\frac{1}{z}}+\frac{1}{2}\frac{1}{1+\frac{z}{2}}=$$

$$=-\frac{1}{z}\left(1+\frac{1}{z}+\frac{1}{z^2}+\ldots\right)+\frac{1}{2}\left(1-\frac{z}{2}+\frac{z^2}{2^4}-\frac{z^3}{2^3}+\ldots\right)$$

Complete the above, explain the steps and then do something similar for the other domains.

DonAntonio
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  • can you explain why you chose to split it up how you did? – bob Apr 19 '13 at 03:35
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    Because $$|z|>1\iff \frac{1}{|z|}<1\implies \frac{1}{1\pm\frac{1}{z}};;\text{has a convergent series}$$ By the same token, $$|z|<2\iff \frac{|z|}{2}<1\implies\frac{1}{1+\frac{z}{2}};;\text{has a convergent series}\ldots$$ – DonAntonio Apr 19 '13 at 12:26
  • that makes sense. thank you for the help! – bob Apr 21 '13 at 23:42