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I'm struggling on how to compute laurent expansion, for example in the following question:

$$ f(z)= \frac {1}{z(z-1)(z-2)} $$

in the Annulus $(0;0,1)$ i.e. the disc centred at 0 of radius 1 and punctured at 0.

A walkthrough of the solution to this problem would be greatly appreciated so I can apply it in future problems.

Thanks!

juper
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We have here that

$$|z|<1\;,\;\;\left|\frac z2\right|<1\;,\;\;\text{so first partial fractions:}$$

$$\frac1{z(z-1)(z-2)}=\frac1{2z}-\frac1{z-1}+\frac1{2(z-2)}=\frac1{2z}+\frac1{1-z}-\frac1{4\left(1-\frac z2\right)}=$$

$$=\frac1{2z}+\frac12\left(1+z+z^2+\ldots\right)-\frac14\left(1+\frac z2+\frac{z^2}4+\ldots\right)=$$

$$\frac1{2z}-\frac12+\frac38z+\frac7{16}z^2+\ldots$$

Can you come up with the general expression...?

DonAntonio
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  • Should the second expansion have a $\frac {1}{4}$ outside the bracket? So the next step is to try to think of how to express the last line as a series? – juper Jun 02 '17 at 15:46
  • @juper Good catch. Typo corrected, thanks. – DonAntonio Jun 02 '17 at 15:48
  • so it's the expression: $\frac {1}{2z} + \frac {1}{2} \sum_{n=0}^\infty z^n - \frac {1}{4} \sum_{n=0}^\infty (\frac {z}{2})^n$ ? – juper Jun 02 '17 at 17:11
  • @juper Well, yes...obviously. I meant a closed, simpler expression, perhaps even with one single sum... – DonAntonio Jun 02 '17 at 17:58
  • can i just combine the series into one? I'm not really sure – juper Jun 02 '17 at 18:49
  • @juper Yes, you can combine the two series in one, and it isn't that hard. By the way, checking all your questions I see you don't use to upvote and/or accept many of hte answers you get... – DonAntonio Jun 02 '17 at 20:09
  • yes sorry, I usually say thanks and forget to upvote. I assume I can just substract the two sums into one? I have gone to my previous questions and selected the correct answer now. – juper Jun 02 '17 at 20:26
  • @juper Exactly that: observe that there's a very simple formula for the substraction of terms with equal power of $;x;$ ... – DonAntonio Jun 02 '17 at 20:32
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    yes as i thought. Thank you for your time and help on this question. It's been very useful! – juper Jun 02 '17 at 20:35