Holomorphic function on open disk admits globally convergent power series

Statement

Suppose $U\subseteq \mathbb {C}$ is an open disk with center $z_{0}$ and radius $R>0$. Suppose $f:U\to \mathbb {C}$ is a holomorphic function. Then, there exists a power series:

$\sum _{n=0}^{\infty }a_{n}(z-z_{0})^{n}$

with radius of convergence at least $R$, and such that for all $z\in U$:

$f(z)=\sum _{n=0}^{\infty }a_{n}(z-z_{0})^{n}$.

Related facts

• Cauchy integral formula
• Cauchy integral formula for derivatives: The formula obtained for the coefficients here is the same as the formula suggested by the Cauchy integral formula for derivatives, although that formula does not directly show that we actually get a holomorphic function.

Applications

• Holomorphic implies complex-analytic

Facts used

1. Cauchy integral formula
2. A corollary of dominated convergence theorem, allowing for the exchange of an integral and an infinite summation.

Proof

Given: $U\subseteq \mathbb {C}$ is an open disk with center $z_{0}$ and radius $R>0$. Suppose $f:U\to \mathbb {C}$ is a holomorphic function.

To prove: There exists a power series:

$\sum _{n=0}^{\infty }a_{n}(z-z_{0})^{n}$

such radius of convergence at least $R$, and such that for all $z\in U$:

$f(z)=\sum _{n=0}^{\infty }a_{n}(z-z_{0})^{n}$.

Proof: We first show that for any $r<R$, there exists a power series about $z_{0}$ convergent at all points within distance $r$ from $z_{0}$. Then, we argue that this implies the existence of a single power series.

By applying the Cauchy integral formula to the disk $D$ of radius $r$, centered at $z_{0}$, we see that for any point $z$ with $|z-z_{0}|<r$, we have:

$f(z)=\oint _{\partial D}{\frac {f(\xi )}{\xi -z}}\,d\xi$.

To examine the dependence of this integral on $z$, we use the fact that:

$(\xi -z)^{-1}=\xi ^{-1}(1-{\frac {z}{\xi }})^{-1}={\frac {1}{\xi -z_{0}}}(1+{\frac {z-z_{0}}{\xi -z_{0}}}+{\frac {(z-z_{0})^{2}}{(\xi -z_{0})^{2}}}+\dots )$

Plugging this in, we see that $f(z)$ can be expressed as the integral of a power series in $z$, with coefficients depending on $\xi$:

$f(z)=\oint _{\partial D}\left(\sum _{n=0}^{\infty }{\frac {(z-z_{0})^{n}f(\xi )}{(\xi -z_{0})^{n+1}}}\right)\,d\xi$

We now use fact (2) to exchange the summation and the integral, thus obtaining:

$f(z)=\sum _{n=0}^{\infty }(z-z_{0})^{n}\oint _{\partial D}{\frac {f(\xi )}{(\xi -z_{0})^{n+1}}}\,d\xi$.

Thus, we get the required power series for $f$, with coefficients:

$a_{n}={\frac {f(\xi )}{(\xi -z_{0})^{n+1}}}$