Holomorphic Function is Analytic

Theorem
Let $a \in \C$ be a complex number.

Let $r > 0$ be a real number.

Let $f$ be a function holomorphic on some open ball $D = \map B {a, r}$.

Then $f$ is complex analytic on $D$.

Proof
Let $z \in D$.

Then:

Note that for all $t \in \partial D$, we have:
 * $\cmod {z - a} < \cmod {t - a} = r$.

Therefore:
 * $\cmod {\dfrac {z - a} {t - a} } < 1$

so Sum of Infinite Geometric Progression may be applied.

This gives:

From Continuous Function on Compact Space is Bounded, there exists a real $M \ge 0$, such that:
 * $\forall t \in \partial D: \cmod {\map f t} \le M$

As $\cmod {\dfrac {z - a} {t - a} } < 1$, there exists a real $0 \le N < 1$ such that:
 * $\cmod {\dfrac {z - a} {t - a} } \le N$

Then:


 * $\displaystyle \cmod {\dfrac {\paren {z - a}^n} {\paren {t - a}^{n + 1} } \map f t} < \cmod {\dfrac {\paren {z - a}^n} {\paren {t - a}^n} \map f t} \le M N^n$

As $N < 1$, we have that $\displaystyle \sum_{n \mathop = 0}^\infty M N^n$ converges by Sum of Infinite Geometric Progression.

Therefore by the Weierstrass M-Test, we have that:


 * $\displaystyle \sum_{n \mathop = 0}^\infty \map f t \frac {\paren {z - a}^n} {\paren {t - a}^{n + 1} }$

converges uniformly on $D$.

Therefore:


 * $\displaystyle \frac 1 {2 \pi i} \int_{\partial D} \sum_{n \mathop = 0}^\infty \map f t \frac {\paren {z - a}^n} {\paren {t - a}^{n + 1} } \rd t = \sum_{n \mathop = 0}^\infty \paren {z - a}^n \paren {\frac 1 {2 \pi i} \int_{\partial D} \frac {\map f t} {\paren {t - a}^{n + 1} } \rd t}$

Let:


 * $\displaystyle c_n = \frac 1 {2 \pi i} \int_{\partial D} \frac {\map f t} {\paren {t - a}^{n + 1} } \rd t$

Then:


 * $\displaystyle \map f z = \sum_{n \mathop = 0}^\infty c_n \paren {z - a}^n$

so $f$ is complex analytic on $D$.