Radius of Convergence of Laurent Series

Question

How does one prove that if a function as $n$ poles, and domain the complex plane except the poles, then the radius of convergence of the Laurent series at each pole is the distance to the nearest pole.

Answer 1

Let $f$ be your function. If the poles are $z_0,z_1,\ldots,z_n$, consider the Laurent series$$\sum_{n=-N}^\infty a_n(z-z_0)^n\tag1$$of $f$ about $z_0$, where $N$ is the order of the pole. Assume the no pole of $f$ is closest to $z_0$ than $z_1$. If $(1)$ converges in a region $D(z_0,R)\setminus\{z_0\}$ with $R>\lvert z_1-z_0\rvert$, then $\lim_{z\to z_1}\sum_{n=-N}^\infty a_n(z-z_0)^n$ exists, which is impossible, since $z_1$ is a pole of $f$.

On the other hande, consider $F(z)=(z-z_0)^Nf(z)$. There is a standard Complex Analysis theorem that says that the radius of convergence of theTaylor series of $F$ about $z_0$ is at least the distance from $z_0$ to the closest point outside the domain of $f$, that is, $\lvert z_0-z_1\rvert$. Since $f(z)=\frac{F(z)}{(z-z_0)^N}$…