I know that
$$ f(z)=(z-3)\sum^{\infty}_{n=0}\frac{(-1)^n}{(2n+1)!(z+2)^{2n+1}} $$
But I am not sure how to find the region of convergence $f(z)$. Is there a way to find the region of convergence for $f(z)$?
2026-04-02 00:17:58.1775089078
How to find the region of convergence of the Laurent series of $f(z)=(z-3)\sin\frac{1}{z+2}$ at $z=-2$?
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