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, $D = B \left({a, r}\right)$.

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

Proof
Let $z \in D$, then:

Note that for all $t \in \partial D$, we have $\left\vert{z - a}\right\vert < \left\vert{t - a}\right\vert = r$.

Therefore $\left\vert{\dfrac {z - a} {t - a} }\right\vert < 1$, so Sum of Infinite Geometric Progression may be applied, giving:

From Continuous Function on Compact Space is Bounded, there exists a real $M \ge 0$, such that $\left\vert{f\left({t}\right)}\right\vert \le M$ for all $t \in \partial D$.

As $\left\vert{\dfrac {z - a} {t - a} }\right\vert < 1$, there exists a real $0 \le N < 1$ such that $\left\vert{\dfrac {z - a} {t - a} }\right\vert \le N$.

Then:


 * $\displaystyle \left\vert{\dfrac {\left({z - a}\right)^n} {\left({t - a}\right)^{n + 1} } f \left({t}\right)}\right\vert < \left\vert{\dfrac {\left({z - a}\right)^n} {\left({t - a}\right)^n} f\left({t}\right)}\right\vert \le M N^n$

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:


 * $\displaystyle \sum_{n \mathop = 0}^\infty f \left({t}\right) \frac {\left({z - a}\right)^n} {\left({t - a}\right)^{n + 1} }$

converges uniformly on $D$.

Therefore:


 * $\displaystyle \frac 1 {2 \pi i} \int_{\partial D} \sum_{n \mathop = 0}^\infty f \left({t}\right) \frac {\left({z - a}\right)^n} {\left({t - a}\right)^{n + 1} } \mathrm dt = \sum_{n \mathop = 0}^\infty \left({z - a}\right)^n \left({\frac 1 {2 \pi i} \int_{\partial D} \frac {f \left({t}\right)} {\left({t - a}\right)^{n + 1} } \mathrm dt}\right)$

Let:


 * $\displaystyle c_n = \frac 1 {2 \pi i} \int_{\partial D} \frac {f \left({t}\right)} {\left({t - a}\right)^{n + 1} } \mathrm dt$

Then:


 * $\displaystyle f \left({z}\right) = \sum_{n \mathop = 0}^\infty c_n \left({z - a}\right)^n$

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