TITLE: How is Cauchy's estimate derived?
QUESTION [20 upvotes]: Cauchy's integral formula says
$$
f^{(n)}(z)=\frac{n!}{2\pi i}\int_C\frac{f(\zeta)d\zeta}{(\zeta-z)^{n+1}}.
$$
If we let $C$ be the circle of radius $r$, such that $|f(\zeta)|\leq M$ on $C$, then taking $z=a$, one obtains Cauchy's estimate that 
$$
|f^{(n)}(a)|\leq Mn!r^{-n}.
$$
How is this derived? I see instead
$$
|f^{(n)}(a)|\leq\frac{n!}{2\pi}\int_C \frac{|f(\zeta)||d\zeta|}{|\zeta-a|^{n+1}}\leq Mn!\int_C\frac{|d\zeta|}{|\zeta-a|^{n+1}}
$$
but I don't see how this eventually gets to Cauchy's estimate.

REPLY [33 votes]: By Cauchy's integral formula you have given, we have 
$$f^{(n)}(a)=\frac{n!}{2\pi i}\int_C\frac{f(\zeta)d\zeta}{(\zeta-a)^{n+1}}$$
where $C$ is a circle of radius $r$ centered at $a$. Therefore, $C$ can be parametrized as $\zeta=a+re^{i\theta}$, $0\leq \theta\leq 2\pi$, which implies 
$$|f^{(n)}(a)|=\left|\frac{n!}{2\pi i}\int_0^{2\pi}\frac{f(a+re^{i\theta})rie^{i\theta}d\theta}{(re^{i\theta})^{n+1}}\right|\leq\frac{n!}{2\pi }\int_0^{2\pi}\left|\frac{f(a+re^{i\theta})rie^{i\theta}}{(re^{i\theta})^{n+1}}\right|d\theta$$
$$=\frac{n!}{2\pi }\int_0^{2\pi}\frac{|f(a+re^{i\theta})|}{r^n}d\theta\leq \frac{n!}{2\pi }\int_0^{2\pi}\frac{M}{r^n}d\theta=\frac{Mn!}{r^n}$$
where the last equality follows from $|e^{i\theta}|=1$ and $|i|=1$, and the last inequality follows from the assumption that $|f|\leq M$ on $C$.