Holomorphic functions are analytic

Statement

If $f$ is holomorphic on an open set $U\subseteq\mathbb C$ , then at every $a\in U$ there is $r>0$ such that

f(z)=\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(z-a)^n \qquad (|z-a|<r).

One may take any $r$ for which the closed disc centered at $a$ lies in $U$ . Thus one-variable complex differentiability implies complex analyticity.

Proof mechanism

Apply the Cauchy integral formula on a circle around $a$ , expand

\frac1{\zeta-z} =\frac1{\zeta-a}\sum_{n\ge0} \left(\frac{z-a}{\zeta-a}\right)^n,

and integrate term by term. The coefficient formula is therefore forced by the boundary values of $f$ .

Contrast with real analysis

A real $C^\infty$ function need not equal its Taylor series. The equivalence of holomorphic and analytic behavior is a rigidity specific to complex analysis and explains why the analytic clause in holomorphic map is a theorem, not an independent definition.

References

Lars V. Ahlfors, Complex Analysis, 3rd ed., McGraw–Hill, 1979. Relevant: Chapter 4, §2.