Cauchy's differentiation formula consequences

Question

I have just started complex analysis. I understand Cauchy's integral formula and its differentiation formula, but I do not understand this consequence my professor listed: "The Cauchy differentiation formula $f^{n}(z)=\frac{n!}{2\pi i} \oint \frac{f(\zeta)}{(\zeta -z)^{n+1}}d\zeta$ implies that $|f^{(n)}(z)| \leq n! \frac{M(R)}{R^n}$, where R is any number such that f is analytic anywhere in the disk $| \zeta -z| \leq R$ and M is the maximal value of $|f|$ on the boundary of this disk".

I understand where the $n!$ comes from. I know that $| \zeta -z| \leq R$ so I would guess that once the integral is done we would have a denominator $(\zeta-z)^n$ which is $\leq R^n$ by what we have said above. For M(R) I am not sure. Once we solve the integral wouldn't we have $f(\zeta)$ elevated to some number? I am a bit confused on this last bit.

Answer 1

Essentially, put the absolute value inside the integral, then in the denominator, $|\zeta-z|=R$, and in the numerator we can bound it by $M_R$, and lastly, integrating around a circle of radius $R$ gives $2\pi R$, hence we have the bound of $\left|\frac{n!}{2\pi i}\right|\cdot \frac{M_R}{R^{n+1}}\cdot 2\pi R=\frac{n!M_R}{R^n}$.

With more details, here it is: you would be integrating along the circle of radius $R$ centered at $z$, so \begin{align} |f^{(n)}(z)|&=\left|\frac{n!}{2\pi i}\int_{|\zeta-z|=R}\frac{f(\zeta)}{(\zeta-z)^{n+1}}\,d\zeta\right|\\ &=\left|\frac{n!}{2\pi i}\int_0^{2\pi}\frac{f(z+Re^{it})}{(Re^{it})^{n+1}}Rie^{it}\,dt\right|\\ &\leq \frac{n!}{2\pi}\int_0^{2\pi}\left|\frac{f(z+Re^{it})}{(Re^{it})^{n+1}}Rie^{it}\right|\,dt\\ &\leq\frac{n!}{2\pi}\int_0^{2\pi}\frac{M_R}{R^n}\,dt\\ &=\frac{n!}{2\pi}\cdot 2\pi \frac{M_R}{R^n}\\ &=\frac{n!M_R}{R^n}. \end{align} The crux of the matter is putting the absolute value inside the integral and using that $M_R$ is an upper bound of $|f|$ on the boundary circle.

Answer 2

With $\gamma(t) = z + Re^{it}$ we have $$f^{(n)}(z)=\frac{n!}{2\pi i} \oint \frac{f(\zeta)}{(\zeta -z)^{n+1}}d\zeta = \frac{n!}{2\pi i}\int_0^{2\pi} \frac{f(\gamma(t))\gamma'(t)}{(\gamma(t) -z)^{n+1}}dt = \frac{n!}{2\pi i}\int_0^{2\pi R^n} \frac{f(\gamma(t))iRe^{it}}{(Re^{it})^{n+1}}dt \\= \frac{n!}{2\pi R^n }\int_0^{2\pi} \frac{f(\gamma(t))}{e^{int}}dt$$ and therefore $$\lvert f^{(n)}(z) \rvert \le \frac{n!}{2\pi R^n}\int_0^{2\pi} \left\lvert\frac{f(\gamma(t))}{e^{int}}\right\rvert dt = \frac{n!}{2\pi R^n}\int_0^{2\pi} \lvert f(\gamma(t)) \rvert dt \\ \le \frac{n!}{2\pi R^n}\int_0^{2\pi} M(R) dt = n! \frac{M(R)}{R^{n}} .$$