For \(z_0 \in D_R(z_0) \subset \Omega\), setting \(M \coloneqq\sup_{z\in \gamma}{\left\lvert {f(z)} \right\rvert}\) so \({\left\lvert {f(z)} \right\rvert}\leq M\) on \(\gamma\) \begin{align*} {\left\lvert { f^{(n)} (z_0) } \right\rvert} \leq \frac{n !}{2 \pi} \int_{0}^{2 \pi} \frac{ M } {R^{n+1}} R \,d\theta = \frac{M n ! }{R^n} .\end{align*}
The \(n\)th Taylor coefficient of an analytic function is at most \(\sup_{{\left\lvert {z} \right\rvert} = R} {\left\lvert {f} \right\rvert}/R^n\). \begin{align*} {\left\lvert {c_k} \right\rvert} \ll {{\left\lVert {f} \right\rVert}_\infty \over R^k} .\end{align*}
- Given \(z_0\in \Omega\), pick the largest disc \(D_R(z_0) \subset \Omega\) and let \(C = {{\partial}}D_R\).
- Then apply the integral formula.
\begin{align*} \left|f^{(n)}(z_0)\right| &= {\left\lvert { \frac{n !}{2 \pi i} \int_{C} \frac{f(\zeta) }{(\zeta-z_0)^{n+1}} \,d\zeta} \right\rvert} \\ &=\left|\frac{n !}{2 \pi i} \int_{0}^{2 \pi} \frac{f\left(z_0 + r e^{i \theta}\right) r i e^{i \theta} }{\left(r e^{i \theta}\right)^{n+1}} \,d\theta\right| \\ &\leq \frac{n !}{2 \pi} \int_{0}^{2 \pi}\left|\frac{f\left( z_0 +r e^{i \theta}\right) r i e^{i \theta}}{\left(r e^{i \theta}\right)^{n+1}}\right| \,d\theta\\ &=\frac{n !}{2 \pi} \int_{0}^{2 \pi} \frac{\left|f\left(z_0 +r e^{i \theta}\right)\right|}{r^{n}} \,d\theta\\ &\leq \frac{n !}{2 \pi} \int_{0}^{2 \pi} \frac{M}{r^{n}} \,d\theta\\ &=\frac{M n !}{r^{n}} .\end{align*}
Exercises
Show that if \(f\) is entire and \({\left\lvert {f(z)} \right\rvert} \in { \mathsf{O}}({\left\lvert {z} \right\rvert}^p)\) for \({\left\lvert {z} \right\rvert}\) sufficiently large, then \(f\) is a polynomial of degree at most \({\left\lfloor p \right\rfloor}\).
The basic idea: \begin{align*} {\left\lvert {c_k} \right\rvert} &\leq {k!\over 2\pi}\int_{{\left\lvert {z} \right\rvert} = R} {\left\lvert {f(\xi) \over (\xi - 0)^{k+1}} \right\rvert}\,d\xi\\ &\leq {k! \over 2\pi}\int_{{\left\lvert {z} \right\rvert} = R}{ {\left\lvert {\xi} \right\rvert}^{p} \over {\left\lvert {\xi} \right\rvert}^{k+1} }\,d\xi\\ &= {k! \over 2\pi}\int_{{\left\lvert {z} \right\rvert} = R} {1\over {\left\lvert {R} \right\rvert}^{k+1-p}} \,d\xi\\ &= {k! \over 2\pi} {1\over {\left\lvert {R} \right\rvert}^{k+1-p}} \cdot 2\pi R \\ &= { \mathsf{O}}(1/R^{k-p}) ,\end{align*} which converges to \(0\) as \(R\to \infty\) provided that \(k-p>0\), so \(k>p\). So any coefficient \(c_k\) for \(k\geq {\left\lfloor p \right\rfloor}\) vanishes.