existence of power series

In this entry we shall demonstrate the logical equivalence of the holomorphicand analytic concepts. As is the case with so many basic results incomplex analysis, the proof of these facts hinges on the Cauchyintegral theorem, and the Cauchy integral formula.

Holomorphic implies analytic.

Theorem 1

Let $U\mathrm{\subset }\mathrm{C}$ be an open domain that contains the origin, andlet $f\mathrm{:}U\mathrm{\to }\mathrm{C}\mathrm{,}$be a function such thatthe complex derivative

$$f^{\prime }(z)=\underset{\zeta \to 0}{lim}\frac{f(z+\zeta )-f(z)}{\zeta }$$

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

for asufficiently small radius of convergence $R\mathrm{>}\mathrm{0}$.

Note: it is just aseasy to show the existence of a power series representation aroundevery basepoint in $z_0\in U$; one need only consider the holomorphicfunction $f(z-z_0)$.

Proof. Choose an $R>0$ sufficiently small so that thedisk $\parallel z\parallel \le R$ is contained in $U$.By the Cauchy integral formula we havethat

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

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

where

$$a_k=\frac{1}{2\pi i}{\oint }_{\parallel \zeta \parallel =R}\frac{f(\zeta )}{{\zeta }^{k+1}},$$

as desired. QED

Analytic implies holomorphic.

Theorem 2

Let

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

$$f^{\prime }(z)=\underset{\zeta \to 0}{lim}\frac{f(z+\zeta )-f(z)}{\zeta }$$

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

Note: this theorem generalizes immediately to shifted power series in$z-z_0,z_0\in ℂ$.

Proof. For every $z_0\in D$, the function $f(z)$ can be recastas a power series centered at $z_0$. Hence, without loss ofgenerality it suffices to prove the theorem for $z=0$. The powerseries

$$\sum _{n=0}^{\mathrm{\infty }}{a}_{n+1}{\zeta }^n,\zeta \in D$$

converges, and equals$(f(\zeta )-f(0))/\zeta$ for $\zeta \ne 0$. Consequently, the complexderivative $f^{\prime }(0)$ exists; indeed it is equal to $a_1$. QED