I'm stuck with the next demo:
Let f an analytic function in $\mathbb{C}$ and $|f(z)|\leq M$ for z in $|z|=R$, where R is fixed. Show for $k\in\mathbb{N}, 0\leq r<R$:
$$|f^{(k)}(re^{i\theta})|\leq \frac{k!M}{(R-r)^{k}}$$
My attempt:
By Cauchy's Integral Formula for Derivatives we have:
$$f^{k}(a)=\frac{k!}{2\pi i}\int_{\gamma}\frac{f(z)}{(z-a)^{k+1}}dz$$
Where a is a point where the f is analytic, so we can try to take the absolute value in both sides and $a=re^{i\theta}$, so:
$$|f^{k}(re^{i\theta})|=\left|\frac{k!}{2\pi i}\int_{\gamma}\frac{f(z)}{(z-re^{i\theta})^{k+1}}dz\right|$$ Developing RHS:
$$\dots=\frac{k!}{2\pi }\left|\int_{\gamma}\frac{f(z)}{(z-re^{i\theta})^{k+1}}dz\right|\leq \frac{k!}{2\pi }\int_{\gamma}\left|\frac{f(z)}{(z-re^{i\theta})^{k+1}}dz\right| \leq\frac{k!M}{2\pi }\int_{\gamma}\frac{|dz|}{|z-re^{i\theta}|^{k+1}}$$
So, here I've tried by the following parametrization:
Let $z=Re^{i\theta}$
if $z=Re^{i\theta}\rightarrow dz=iRe^{i\theta}d\theta$, where, $0\leq\theta\leq2\pi$ thus:
$$\dots=\frac{k!M}{2\pi }\int_{0}^{2\pi}\frac{|iRe^{i\theta}|d\theta}{|Re^{i\theta}-re^{i\theta}|^{k+1}}=\frac{k!M}{2\pi }\int_{0}^{2\pi}\frac{Rd\theta}{(R-r)^{k+1}}=\frac{k!M2\pi R}{2\pi(R-r)^{k+1}}=\frac{k!M R}{(R-r)^{k+1}}\leq\frac{k!M R}{(R-r)^{k}}$$ the last one inequality I think is correct beacuase we are working on positive numbers by hypotesis, so if the denominator is a small number then the total fraction will be a large number. The final step is eliminate the R on the top, but idk how to do it and I am not sure if the procedure is correct. Thanks for ur time!