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Given $f(z)=\frac{1}{z^2(1-z)}$ I am to find two Laurent series expansions. There are two singularities, $z=0$ and $z=1$. So for the first expansion, I used the region $0<|z|<1$ and I got $\sum_{n=0}^\infty z^n+\frac{1}{z}+\frac{1}{z^2}$. The second expansion is for the region $1<|z|<\infty$. I don't know how to approach this, the explanation in my book is confusing. Any help?

Thanks

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The problem is that for $|z|>1$ the absolute value of $z$ is greater than $1$ (ok that sound ridicolus but it is really the problem) as the geometric sequence wont work in the given form.

We just change it a bit. $$\frac{1}{z^2(1-z)} =\frac{1}{z^2} \cdot \frac{1}{1-z} = \frac{1}{z^2} \cdot \frac{1}{\frac{z}{z}(1-z)}=\frac{1}{z^2} \cdot \frac{1}{z (\frac{1}{z}-1)}$$

This is equivalent to $$\frac{1}{z^2(1-z)} =-\frac{1}{z^3} \cdot \frac{1}{1-\frac{1}{z}}$$ note that $\left|\frac{1}{z}\right|<1$ hence $$\frac{1}{z^2(1-z)}=-\frac{1}{z^3} \cdot \sum_{k=0}^\infty \left(\frac{1}{z}\right)^k $$ This is the same as $$-\sum_{k=0}^\infty \left(\frac{1}{z}\right)^{k+3}=-\sum_{n=3}^\infty \left(\frac{1}{z}\right)^n =-\sum_{n=3}^\infty z^{-n}=-\sum_{m=-\infty}^{-3} z^m$$

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