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I am having difficulties finding Laurent series of the above function, around these two domains

a) $1<|z|$

b) $1<|z-1|$

For a) I do $$ \sum_{n=0}^\infty \frac{1}{z^{n+1}} - \frac{1}{z} $$

Is this correct? And how can I solve b)?

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1 Answer 1

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Since$$f(z)=\frac1z\cdot\frac1{1-z}$$and since, when $|z|>1$, you have$$\frac1{1-z}=-\sum_{n=-\infty}^{-1}z^n,$$you have, in same same region\begin{align}f(z)&=\frac1z\left(-\sum_{n=-\infty}^{-1}z^n\right)\\&=-\sum_{n=-\infty}^{-1}z^{n-1}\\&=-\sum_{n=-\infty}^{-2}z^n.\end{align}In the other region, use the fact that$$f(z)=\frac1{z-1}\cdot\frac1{1+(z-1)}$$

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  • $\begingroup$ $\begin{align}f(z)&=\frac1{z-1}\left(-\sum_{n=-\infty}^{-1}(z+1)^n\right)\\end{align}$ for the oher region?Thank you very much. $\endgroup$ Jun 5, 2020 at 10:53
  • $\begingroup$ Yes, that is correct. $\endgroup$ Jun 5, 2020 at 10:56

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