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Determine the Laurent expansion about $z_0=0$ for $$g(z)=\frac1{(z-1)(z-3)} \text{ on } \left\{z\in\mathbb C:1<\lvert z\rvert<3\right\}$$

I'm currently trying to solve this question, I have work out the solution to be $$-\frac{1}{2} \bigg(\sum_{n=0}^{\infty} \frac{z^n}{3^{n+1}} + \sum_{n=1}^{\infty} \frac{1}{z^{n}} \bigg) $$ could someone confirm if this is correct.

Extension!

Find the Laurent expansion about $0$ of $$f(z)=\frac1{(z-i)(z-2)}$$ on the following annlui: $0<|z|<1$

I'm not too sure how to solve this, I've found that for $|z|<1$ we have $$i\sum_{n=0}^\infty \bigg(\frac{z}{i}\bigg)^n$$ however, how would I find the summation for $|z|>0$?

Lorenzo B.
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jimmy
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1 Answers1

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I apologize for my mistake, I had misread or forgotten to read the entire question.

kmeis
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  • on the second part where you wrote $\frac{1}{2} \sum_{n=0}^{\infty}z^n$, does this not consider $|z|<1$ whereas we're looking for $|z|>1$? – jimmy Apr 16 '15 at 23:51
  • Correct, I had forgotten to read where you wished it to converge. Your solution seems fine. – kmeis Apr 16 '15 at 23:58
  • awesome! thank you :D, could I add a separate question as an extension to this as I don't fully understand how to find the convergence? – jimmy Apr 17 '15 at 00:01
  • Its been a while since I've done this, but a few questions to clarify the extension. For finding the series that converges on $|z|<1$ for $f(z)$ will include zero. I'm not certain as to why we wish to find it on the punctured disk $0<|z|<1$ since it is analytic at 0. I am not certain if that what was meant or not.

    Also, I have arrived at a different series for $f(z)$ on $|z|<1$. I just wish to clarify the first point before reviewing this part.

     – kmeis Apr 17 '15 at 02:46
  • I used partial fractions to get $\frac{1}{i-2}\bigg(\frac{1}{z-i}-\frac{1}{z-2}\bigg)$ therefore $\frac{1}{z-i}=i\bigg(\frac{1}{1-\frac{z}{i}}=i\sum_{n=0}^\infty \bigg(\frac{z}{i}\bigg)^n\bigg)$ – jimmy Apr 17 '15 at 02:53
  • I agree with the work here, but we still need to find the series for $-\frac{1}{z-2}$ for $|z|<1$ then add it to what we already know. I believe that should do it, aside from the fact that it converges at $z=0$ as well. – kmeis Apr 17 '15 at 03:01
  • of courseeee!!! i was just being silly, thank you! – jimmy Apr 17 '15 at 03:07