Radius of a convergence of power series of holomorphic function about a point

Question

I observed that the radius of convergence of power series in the complex plane is basically the distance to nearest singularity of the function.I want to know whether my observation is correct or not.I have tried for a proof but didn't succeed.Any help will be appreciable.(If it is correct then either provide the proof or a reference from book) Thanks in advance!

Answer 1

Let $f:U \to \mathbb C$ holomorphic with $U$ maximal domain for $f$ and $w \in U$ with $r_w=d(w, \partial U)$ or equivalently $r_w=\sup_r D(w,r) \subset U$; then using the Cauchy formula on the closed disc $D(w, r-\epsilon)$ and the usual power series expansion trick of $1/(w-\zeta), |\zeta-w|=r-\epsilon$ shows immediately that the power series of $f$ at $w$ has radius at least $r-\epsilon$; this immediately implies that the power series has radius at least $r_w$ and the maximality of $U$ as domain of holomorphicity for $f$ implies the reverse inequality.

(edit later as requested a little more detail)

If $\overline {D(w,\rho)} \subset U$ we can apply Cauchy and get $$f(z)=\frac{1}{2\pi i}\int_{|w-\zeta|=\rho}\frac{f(\zeta)\,\mathrm d\zeta}{\zeta-z},\quad |z-w|<\rho$$ but now $$\frac{1}{\zeta-z}=\frac{1}{\zeta-w}\frac{1}{1-(z-w)/(\zeta-w)}=\sum_{n \ge 0}\frac{(z-w)^n}{(\zeta-w)^{n+1}},$$ hence by absolute convergence we can switch sumand integral and we get $$ f(z)=\sum_{n \ge 0}\left(\frac{1}{2\pi i}\int_{|w-\zeta|=\rho} \frac{f(\zeta)\,\mathrm d\zeta}{(\zeta-w)^{n+1}} \right) (z-w)^n $$ for all $z$ in the open disc around $w$ of radius $\rho < r_w$; hence the result holds for all $z$ in the open disc of radius $r_w$ so we get one implication ($f$ has a power series around $w$ of radius at least $r_w=r_{w,U}$; the other implication (the radius is exactly $r_{w,U}$) depends on some form of maximality of $U$ and is tied in with the notion of analytic continuation, so it is trickier to express explicitly but rigurously, but informally it can be expressed as the OP required as the "distance from $w$ to the nearest singularity".

Answer 2

More precisely, the radius of convergence of the series about $z=a$ is the largest $r$ such that there is an analytic function in the open disk of radius $r$ about $a$ which has the given power series as its Taylor series around $z=a$. In particular, removable singularities don't count, and branch cuts don't count if they can be moved out of the way.

EDIT: To prove this, you can proceed as follows:

1. If $f$ is analytic in the open disc of radius $r$ about $a$, use Cauchy's estimates to show that for any $r_1 < r$ the radius of convergence of the Taylor series of $f$ around $z=a$ is at least $r_1$. Thus the radius of convergence is at least $r$.
2. Conversely, if the series $\sum_{k} c_k (z-a)^k$ has radius of convergence $\ge r$, then this series converges uniformly on compact subsets of the disk of radius $r$ about $a$, and the sum of the series is therefore an analytic function on that disk; moreover (using Cauchy's formula) the Taylor series of this function around $z=a$ is the given series.