Find the Laurent series for the function $g(z)=\frac{1}{3-z}$ for the annulus $0 < |z|<3$ and $|z|>3$. I understand for the first case, the Laurent series would be $\sum_{n \geq0} \frac{z^n}{3^{n+1}}$, $0 < |z|<3$. Why for the second case, the Laurent series have to be different, i.e. $-\sum_{n \geq0} 3^n z^n$, $|z|>3$. Is there someone who could explain to me?
2026-03-31 23:31:03.1774999863
$g(z)=\frac{1}{3-z}$ - Laurent series for two differents annulus
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