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 $n \in \N$:
 * $\displaystyle \cmod {\map {f^{\paren n} } t} \le \frac {M r \, n!} {\paren {r - \cmod {t - z_0} }^\paren {n + 1} }$

Lemma (Analytic Function Bounded on Circle)
We have:
 * $f$ is analytic on $\Gamma$ and its interior
 * $t$ is in the interior of $\Gamma$

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

where $\Gamma$ is traversed counterclockwise.

We have that $f$ is bounded on $\Gamma$ by Lemma (Analytic Function Bounded on Circle).

Therefore, there is a positive real number $M$ that satisfies:
 * $M \ge \cmod {\map f z}$ 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: