TITLE: Complex polynomial and the unit circle
QUESTION [10 upvotes]: Given a polynomial $ P(z) = z^n + a_{n-1}z^{n-1} + \cdots + a_0 $, such that
  $\max_{|z|=1} |P(z)| = 1 $
Prove: $ P(z) = z^n $
Hint: Use cauchy derivative estimation 
  $$ |f^{(n)} (z_0)| \leq \frac{n!}{r^n} \max_{|z-z_0|\leq r} |f(z)| $$ and look at the function $ \frac{P(z)}{z^n} $  

It seems to be maximum principle related, but I can't see how to use it and I can't understand how to use the hint.

REPLY [4 votes]: Another way to do it is to use the orthogonality of the exponentials to see
$$\frac{1}{2\pi}\int_0^{2\pi}|P(e^{it})|^2\,dt = 1^2 +|a_{n-1}|^2 + \cdots + |a_0|^2.$$
Since $|P|\le 1$ on the circle, the expression on the left is $\le 1.$ It follows that $a_k=0$ for $k=0,\dots ,n-1.$