Let $f:B_r(0) \to \mathbb C$ holomorphic, $r>0$. Show for $n \in \mathbb N$
$|f^{(n)}(0)| \le \frac {n!}{2r^n} \sup_{|z|=r} |f(z)-f(e^{\pi i /n} z)|$
I know how to derive the usual Cauchy inequality $|f^{(n)}(0)| \le \frac {n!}{r^n} \sup_{|z|=r} |f(z)|$ via a parametrization of the domain but I can't see where $f(e^{\pi i /n} z)$ comes from. I already got the hint to substitute $x \to e^{\pi i /n}x$ but afterwards I couldn't see the next step any better.
Thanks for your help!
This is a weird question, but I think I understand what the hint is saying.
Usually, Cauchy's integral formula is stated as $$ f^{(n)}(0) = \frac{n!}{2\pi i } \oint_{|z| = r} \frac{f(z)}{z^{n+1}} dz.$$
But if we perform the substitution $z_{\rm old} = e^{i \pi /n} z_{\rm new}$, we get another formula for $f^{(n)}(0)$: $$ f^{(n)}(0) = - \frac{n!}{2\pi i } \oint_{|z| = r} \frac{f(e^{i \pi /n}z)}{z^{n+1}} dz.$$ (Yes, there is meant to be a minus sign - please check!)
We can write down yet another formula for $f^{(n)}(0)$ by taking the average of the previous two: $$ f^{(n)}(0) = \frac{n!}{4\pi i } \oint_{|z| = r} \frac{f(z) - f(e^{i \pi / n}z)}{z^{n+1}} dz.$$ Then apply the usual argument...