Determine complex polynomial

Question

Problem

Let $P(z) = z^n + a_{n−1}z^{n−1} + \cdots + a_1z + a_0$ be a polynomial of degree $n > 0$. Show that if $\lvert P(z) \lvert \le 1$ whenever $\lvert z \rvert = 1$ then $P(z) = z^n$.

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I have tried to see $\dfrac{P(z)}{z^n}$, but nothing happens. I wonder which theorems should I use to solve this.

I think hints are enough.

Thanks.

Answer 1

Hint: let $Q(z):=z^nP(z^{-1})$, which is a polynomial. Then $|Q(z)|\leqslant 1$ when $|z|=1$ and $Q(0)=1$ so by maximum modulus principle...

Answer 2

Hint: Schwarz lemma and maximum prinicple

Answer 3

Another approach. By Rouché's theorem $z^n - \varepsilon\, P(z)$ has $n$ roots in the unit disc for all $\varepsilon \in (0, 1)$. This implies that

$$0 \leq \frac{\varepsilon \,|a_k|}{1-\varepsilon} \leq {n \choose k}$$

for $k \in \{0, \dotsc, n-1\}$ or in other words

$$0 \leq |a_k| \leq \frac{1-\varepsilon}{\varepsilon}{n \choose k}.$$

Now let $\varepsilon \to 1$.