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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 $|z|>0$?

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

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  • $\begingroup$ 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 $|z|>1$? $\endgroup$
    – jimmy
    Commented Apr 16, 2015 at 23:51
  • $\begingroup$ Correct, I had forgotten to read where you wished it to converge. Your solution seems fine. $\endgroup$
    – kmeis
    Commented Apr 16, 2015 at 23:58
  • $\begingroup$ 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? $\endgroup$
    – jimmy
    Commented Apr 17, 2015 at 0:01
  • $\begingroup$ 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. $\endgroup$
    – kmeis
    Commented Apr 17, 2015 at 2:46
  • $\begingroup$ I used partial fractions to get $\frac{1}{i-2}\bigg(\frac{1}{z-i}-\frac{1}{z-2}\bigg)$ therefore $\frac{1}{z-i}=i\bigg(\frac{1}{1-\frac{z}{i}}=i\sum_{n=0}^\infty \bigg(\frac{z}{i}\bigg)^n\bigg)$ $\endgroup$
    – jimmy
    Commented Apr 17, 2015 at 2:53

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