# existence of power series

## Holomorphic implies analytic.

###### Theorem 1

Let $U\subset\mathbb{C}$ be an open domain that contains the origin, and let $f:U\rightarrow\mathbb{C},$ be a function such that the complex derivative  $f^{\prime}(z)=\lim_{\zeta\rightarrow 0}\frac{f(z+\zeta)-f(z)}{\zeta}$

exists for all $z\in U$. Then, there exists a power series  representation

 $f(z)=\sum_{k=0}^{\infty}a_{k}z^{k},\quad\|z\|

for a sufficiently small radius of convergence  $R>0$.

Note: it is just as easy to show the existence of a power series representation around every basepoint in $z_{0}\in U$; one need only consider the holomorphic function $f(z-z_{0})$.

Proof. Choose an $R>0$ sufficiently small so that the disk $\|z\|\leq R$ is contained in $U$. By the Cauchy integral formula we have that

 $f(z)=\frac{1}{2\pi i}\oint_{\|\zeta\|=R}\frac{f(\zeta)}{\zeta-z}\,d\zeta,\quad% \|z\|

where, as usual, the integration contour is oriented counterclockwise. For every $\zeta$ of modulus $R$, we can expand the integrand as a geometric power series in $z$, namely

 $\frac{f(\zeta)}{\zeta-z}=\frac{f(\zeta)/\zeta}{1-z/\zeta}=\sum_{k=0}^{\infty}% \,\frac{f(\zeta)}{\zeta^{k+1}}\,z^{k},\quad\|z\|

The circle of radius $R$ is a compact set; hence $f(\zeta)$ is bounded   on it; and hence, the power series above converges uniformly with respect to $\zeta$. Consequently, the order of the infinite  summation and the integration operations  can be interchanged. Hence,

 $f(z)=\sum_{k=0}^{\infty}a_{k}z^{k},\quad\|z\|

where

 $a_{k}=\frac{1}{2\pi i}\oint_{\|\zeta\|=R}\frac{f(\zeta)}{\zeta^{k+1}},$

as desired. QED

## Analytic implies holomorphic.

###### Theorem 2

Let

 $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n},\quad a_{n}\in\mathbb{C},\quad\|z\|<\epsilon$

be a power series, converging in $D=D_{\epsilon}(0)$, the open disk of radius $\epsilon>0$ about the origin. Then the complex derivative

 $f^{\prime}(z)=\lim_{\zeta\rightarrow 0}\frac{f(z+\zeta)-f(z)}{\zeta}$

exists for all $z\in D$, i.e. the function $f:D\rightarrow\mathbb{C}$ is holomorphic.

Note: this theorem generalizes immediately to shifted power series in $z-z_{0},\;z_{0}\in\mathbb{C}$.

Proof. For every $z_{0}\in D$, the function $f(z)$ can be recast as a power series centered at $z_{0}$. Hence, without loss of generality it suffices to prove the theorem for $z=0$. The power series

 $\sum_{n=0}^{\infty}a_{n+1}\zeta^{n},\quad\zeta\in D$

converges  , and equals $(f(\zeta)-f(0))/\zeta$ for $\zeta\neq 0$. Consequently, the complex derivative $f^{\prime}(0)$ exists; indeed it is equal to $a_{1}$. QED

Title existence of power series ExistenceOfPowerSeries 2013-03-22 12:56:27 2013-03-22 12:56:27 rmilson (146) rmilson (146) 5 rmilson (146) Result msc 30B10