### Radius of convergence

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. $\blacksquare$

# complex-analysis

# power-series

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