## Franklin's Notes

Given a complex power series which converges in some neighborhood of the origin, we define the radius of convergence of the power series to be the supremum of all $r>0$ such that the power series converges in the open ball $B_r(0)$. A function defined by such a power series is called analytic .

Given a power series like this, it is natural to ask about the region consisting of all values of $z$ for which the series converges. As it happens, this region is always a circle. To see why this is true, suppose that the series converges for some value $z=\zeta$, with radius $r=|\zeta|$ from the origin. For any $z$ with $|z|<|\zeta|$, we have that the terms of the series satisfy In order for the sum to converge when evaluated at $\zeta$, it is necessary that $|a_n\zeta^n|\to 0$ as $n\to\infty$. However, this means that the magnitude of $|a_nz^n|$ is dominated by $|z/\zeta|^n$, and the sums of $(z/\zeta)^n$ converge geometrically and therefore converge absolutely. Hence, for any $z$ with magnitude $|z|<|\zeta|$, the series converges absolutely, implying that if the series converges at $\zeta$, it converges everywhere in the open ball of radius $|\zeta|$. This means that the maximal (open) region consisting of points of convergence is an open ball centered at the origin, whose radius is equal to the radius of convergence!

Proposition 1. If $f$ is meromorphic and defined at $z_0$, then the radius of convergence of its Taylor series about $z=z_0$ is equal to the distance from $z_0$ to the nearest pole.

Proof. Using the Cauchy integral formula , we can show that if $f$ is holomorphic in the ball $B_r(z_0)$ of radius $r>0$ about $z_0$, then $|f^{(n)}(z_0)| < n!M/r^{n+1}$ where $M$ is the maximum magnitude of $f$ on the circle $|z-z_0| = r$. This means that the magnitude of the terms of its Taylor series are bounded as follows: and therefore the Taylor series converges absolutely for $|z-z_0|<r$. This implies that the radius of convergence $R$ is at least $r$ for each $r$ such that $f$ is holomorphic in $B_r(z_0)$, hence $R$ is greater than or equal to the distance from $z_0$ to the nearest pole. But clearly $R$ cannot be greater than this distance, for this pole would then have to lie inside of the circle of convergence, whereas it is impossible for the series to converge to a value at that point. Hence, $R$ is precisely the distance from $z_0$ to the nearest pole. $\blacksquare$