# 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:

 $\ds \map f z$ $=$ $\ds \frac 1 {2 \pi i} \int_{\partial D} \frac {\map f t} {t - z} \rd t$ Cauchy's Integral Formula $\ds$ $=$ $\ds \frac 1 {2 \pi i} \int_{\partial D} \frac {\map f t} {\paren {t - a} \paren {1 - \frac {z - a} {t - a} } } \rd t$

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 Sequence may be applied.

This gives:

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

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:

$\ds \cmod {\dfrac {\paren {z - a}^n} {\paren {t - a}^{n + 1} } \map f t} = \cmod {\frac 1 {t - a} } \cmod {\dfrac {\paren {z - a}^n} {\paren {t - a}^n} } \cmod {\map f t} \le \frac 1 r M N^n$

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

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

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

converges uniformly on $D$.

Therefore:

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

Let:

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

Then:

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

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

$\blacksquare$