Let $p(z)=z^n+a_{n-1}z^{n-1}+a_{n-2}z^{n-2}+...+a_0$ be a complex polynomial satisfying $|p(z)| \leq 1$ for $|z|\leq 1$, then prove that $p(z)=z^n$

Question

Let $p(z)=z^n+a_{n-1}z^{n-1}+a_{n-2}z^{n-2}+...+a_0$ be a complex polynomial satisfying $|p(z)| \leq 1$ for $|z|\leq 1$, then prove that $p(z)=z^n$

I tried with Cauchy integral formula for $n$th derivative to get $$n!=\frac{n!}{2\pi}\int_0^{2\pi}p(e^{i\theta})e^{-in\theta}d\theta$$ implies $$1\leq \frac{1}{2\pi}\int_0^{2\pi}|p(e^{i\theta})e^{-in\theta}|d\theta$$ and thereby $|p(e^{i\theta})e^{-in\theta}|=1$. (since $|p(e^{i\theta})e^{-in\theta}|$ is a non negative real continuous function $\leq 1$)

ie., $|p(z)|=|z^n|$ for $z=e^{i\theta}$, i stucked here and how can we procced?

Answer 1

Let $q(z)=z^{n}p(\frac 1 z)$. Then $q$ is a polynomial and $|q(z)| =|z|^{n}|p(\frac 1 z)| \leq 1$ for $|z|=1$. By MMP we get $|q(z)| \leq 1$ for $|z| \leq 1$. But $q(0)=1$ so $|q|$ attains its maximum inside the unit disk. This implies that $q$ is a constant, say $c$ and this gives $p(\frac 1 z)=\frac c {z^{n}}$ or $p(z)=cz^{n}$. Clearly, $c$ must be $1$.