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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 $P(z) = z^n$.

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.


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up vote 7 down vote accepted

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...

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Short and straight to the point, +1. – 1015 Mar 10 '13 at 21:10

Hint: Schwarz lemma and maximum prinicple

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Schwarz Lemma says $P(z)\le 1$ for all the points $|z|\le 1$, this one says $|z|=1$. – Yimin Mar 10 '13 at 21:02
@yimin for this i need the maximum priniciple :) – Dominic Michaelis Mar 10 '13 at 21:05
I got it, thanks. :D – Yimin Mar 10 '13 at 21:06
Why and how do you use Schwarz lemma? – Davide Giraudo Mar 10 '13 at 21:14
@DavideGiraudo schwarz lemma gives you $|f'(0)|\leq 1 $ and in general $|f^{n}(0)|\leq n! $ – Dominic Michaelis Mar 10 '13 at 21:25

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 $\varepsilon \to 1$.

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