$P(z)$ be a monic polynomial with complex co-efficients of degree $n$ such that $|P(z)| \le 1, \forall |z| \le 1$ ; then is $P(z)=z^n$?

Question

Let $P(z)$ be a monic polynomial with complex co-efficients of degree $n$. Since by Cauchy integral formula bound, $n!=P^{(n)} (0) \le n! Sup_{|z|=1} |P(z)|$ , so $Sup_{|z|=1} |P(z)| \ge 1$ .

So by Maximal modulus principle, $Sup _{|z| \le 1} |P(z)|=Sup_{|z|=1} |P(z)| \ge 1$.

Now if $|P(z)| \le 1, \forall |z| \le 1$, then is it true that $P(z)=z^n$ ?

Answer 1

Yes. If $f(z) = \sum_{k=0}^n a_k z^k$, we have $$ 1 \ge \frac{1}{2\pi} \int_{0}^{2\pi} |f(e^{i\theta})|^2\; d\theta = \sum_{k=0}^n |a_k|^2$$ But $a_n = 1$, so all other $a_k$ must be $0$.