The Cauchy Estimate The Cauchy Estimate is:
If a function $f:D\to\mathbb{C}$ is differentiable for $|z-z_0|<R$, $0<r<R$ and $|f(z)|\leqslant M$ for all $|z-z_0|=r$, then
$$
|f^{(n)}(z_0)|\leqslant\frac{Mn!}{r^n}.
$$
Now here is a proposition:
The Cauchy Estimate is an equality if and only if $f(z)=Kz^n$ for some $K\in\mathbb{C}$, $n=1,2,\dots$.
I want to prove this but I do not quite understand how this works so I could not find a clue. Could someone help me with this please? 
 A: It helps to check the steps in the proof of Cauchy's inequality. Usually it is done using Cauchy's integral formula:
$$
\begin{align*}
|f^{n}(z_0)| 
  & = \left| \frac{n!}{2\pi i} \int_{|z-z_0| = r} \frac{f(z)}{(z-z_0)^{n+1}} \, dz \right| \\
  & \le \frac{n!}{2\pi r^{n+1}} \int_{|z-z_0|=r} |f(z)| \, dz \\
  & \le \frac{n!}{2\pi r^{n+1}}2\pi r M = \frac{M n!}{r^n}
\end{align*}
$$
In order for equality to hold, both inequalities have to be equalities. The first one, the triangle inequality for integrals, is an equality iff the integrand has constant argument, where the argument of $dz$ is the argument of the tangent vector, i.e., in this case $\arg dz = \pi/2+ \arg (z-z_0)$. This is equivalent $\arg \frac{f(z)}{(z-z_0)^{n}} \equiv C$ for some constant $C$ on $|z-z_0|=r$. The second inequality is an equality iff $|f(z)| \equiv M$ for $|z-z_0|=r$. Combine these two and get that $\frac{f(z)}{(z-z_0)^{n}} \equiv K$ for $|z-z_0|=r$, where $K$ is some complex constant. Then $f(z) = K(z-z_0)^n$ for $|z-z_0|=r$, and by the uniqueness principle this is also true for $|z-z_0| < r$.
