Complex polynomial and the unit circle

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.

Solution 1:

Let $g(z):=z^nP\left(\frac 1z\right)$. It's a polynomial whose leading term is $a_0$ and constant coefficient is $1$. We have that $g(0)=1$ and $\max_{|z|=1}|g(z)|=1$, hence by maximum modulus principle, $g$ is constant equal to $1$. This gives the wanted result.

Solution 2:

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 $k=0,\dots ,n-1.$