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I am trying to find two distinct Laurent expansions of $f(z)=\frac{1}{z^2(1-z)},$ in powers of $z$.

My attempt:

I believe I have found one Laurent expansion of the region $0<|z|<1$. \begin{align} f(z)&=\frac{1}{z^2(1-z)} \\ &=\frac{1}{z}+\frac{1}{z^2}+\frac{1}{1-z} \\ &=\sum_{n=-2}^{\infty} z^n \end{align} But I cannot find a second expansion. I know the region for this expansion is $|z|>1$. The answer provided is $\sum_{n=-\infty}^{-3} (-1)^nz^n$, but I don't understand the method of getting to this answer.

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1 Answer 1

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$f(z)=-\frac 1 {z^{3}} \frac 1 {1-\frac 1 z}=-\frac 1 {z^{3}} (1+\frac1 z +\frac 1 {z^{2}}+...)$ for $|z| >1$.

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  • $\begingroup$ For $|z|>1$, do we only consider $\frac{1}{1-z}$? $\endgroup$
    – user557493
    Sep 24, 2018 at 11:51
  • $\begingroup$ for instance, if we have $$f(z)=\frac{3}{(z-2)(z+1)},$$ for the region $|z-1|<1$, we only consider $\frac{1}{z-2}$. Why is that? $\endgroup$
    – user557493
    Sep 24, 2018 at 11:54

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