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There is a thing that always confuse me about laurent series expansions. For example, when expanding $$g(z) = \sum_{k=0}^{\infty} (k+1)z^k$$ for $z$ in the unit disc, we have for $|z|>1$ that $$g(z) = \frac{1}{z^2} \frac{1}{(1-1/z)^2} = - \sum_{m=-\infty}^{-2}(m+1)z^m.$$

Or for $\frac{1}{z^2-1}$ in the region $0 < |z| < 1$ we have $$\frac{1}{z^2-z} = -\frac{1}{z} \frac{1}{1-z} = \dots = - \sum_{n=-1}^{\infty} z^n$$ while for $|z| > 1$ we have $$\frac{1}{z^2-z}=\frac{1}{z^2} \frac{1}{1-1/z} = \dots = \sum_{n=-\infty}^{-2}z^n.$$

I never feal sure of how to know how the different regions affect how I should expand the series. Why it is one way for $|z| > 1$ and another for $0 < |z| < 1$, for example. It has something to do with convergence, but what exactly?

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It is exactly about convergence. In the example you gave, the series $-\sum_{n=-1}^\infty z^n$ converges to an analytic function on the disk $|z| < 1$, while the series $\sum_{n = -\infty}^{-2} z^n$ converges to an analytic function in the region $|z| > 1$. This can be seen by the convergence of geometric series: $\sum_{n=-1}^\infty z^n$ converges absolutely if and only if $|z| < 1$, while $\sum_{n= -\infty}^{-2} z^n = \sum_{k=2}^\infty z^{-k}$ converges if and only if $|z| >1$.

This behavior arises because the function $z \mapsto 1/(z^2 -z)$ has two poles, at $0$ and $1$. If there was only one pole, say with $1/(z^2 - 1)$, then the Laurent series would converge everywhere. The expansion for $|z| < 1$ arises from the Laurent expansion on an annulus with radii $0 < r_1 < r_2 < 1$, while the expansion for $|z| > 1$ arises from the Laurent expansion on an annulus with radii $1 < r_1 < r_2$. Since the pole at $z = 1$ is in a different position with regard to these regions, the expansions are different.