Bound for Analytic Function and Derivatives

Lemma
Let $f$ be a complex function.

Let $z_0$ be a point in $\C$.

Let $r$ be a real number in $\R_{>0}$.

Let $\Gamma$ be a circle in $\C$ with center at $z_0$ and radius $r$.

Let $f$ be analytic on $\Gamma$ and its interior.

Let $t \in \C$ be such that $\cmod {t - z_0} < r$.

Then a real number $M$ exists such that, for every $n \in \N$:
 * $\displaystyle \cmod {f^{\paren n} \paren t} \le \frac {M r \, n!} {\paren {r - \cmod {t - z_0}}^\paren {n + 1}}$

Proof
We have:
 * $f$ is analytic on $\Gamma$ and its interior
 * $t$ is in the interior of $\Gamma$

Therefore:
 * $\displaystyle f^{\paren n} \paren t = \frac {n!} {2 \pi i} \int_\Gamma \frac {f \paren z} {\paren {z - t}^{\paren {n + 1}}} \rd z$ by Cauchy's Integral Formula for Derivatives

where $\Gamma$ is traversed counterclockwise.

We have that $f$ is bounded on $\Gamma$ by the lemma.

Therefore, there is a positive real number $M$ that satisfies:
 * $M \ge \cmod {f \paren z}$ for every $z$ on $\Gamma$

We have $\cmod {t - z_0} < r$.

Therefore:
 * $0 < r - \cmod {t - z_0}$

We observe that $r - \cmod {t - z_0}$ is the minimum distance between $t$ and $\Gamma$.

Therefore:
 * $\paren {r - \cmod {t - z_0}} \le \cmod {z - t}$ for every $z$ on $\Gamma$

We get:

Lemma (Analytic Function Bounded on Circle)
Let $f$ be a complex function.

Let $z_0$ be a point in $\C$.

Let $\Gamma$ be a circle in $\C$ with center at $z_0$ and radius in $\R_{>0}$.

Let $f$ be analytic on $\Gamma$.

Then $f$ is bounded on $\Gamma$.

Proof
Let:
 * $f_{\Re} \paren z = \Re \paren {f \paren z}$
 * $f_{\Im} \paren z = \Im \paren {f \paren z}$

Let $\closedint {a} {b}$, $a < b$, be a real interval.

Let $p$ be a continuous complex-valued function defined such that:
 * $\Gamma = \left\{{p \paren u : u \in \closedint {a} {b}}\right\}$

$f$ is continuous on $\Gamma$ as $f$ is analytic on $\Gamma$ by the definition of analytic.

Also, Real and Imaginary Part Projections are Continuous.

Therefore, $f_{\Re}$ and $f_{\Im}$ are continuous by Composite of Continuous Mappings is Continuous.

Observe that $f_{\Re}$ and $f_{\Im}$ are real-valued functions that are continuous.

Also, $p$ is a continuous function defined on a set of real numbers.

Therefore, $f_{\Re} \paren {p \paren u}$ and $f_{\Im} \paren {p \paren u}$ are continuous real functions by Composite of Continuous Mappings is Continuous.

$f_{\Re} \paren {p \paren u}$ and $f_{\Im} \paren {p \paren u}$ are bounded on $\closedint {a} {b}$ by Continuous Real Function is Bounded.

Therefore, $f \paren {p \paren u}$ is bounded on $\closedint {a} {b}$ as $f \paren {p \paren u} = f_{\Re} \paren {p \paren u} + i f_{\Im} \paren {p \paren u}$ where $i = \sqrt{-1}$.

Accordingly, $f$ is bounded on $\Gamma$ as $\Gamma = \left\{{p \paren u : u \in \closedint {a} {b}}\right\}$.