If $p$ is a polynomial, then either $\,p(z)=z^n\,$ or $\,\max_{|z|=1}|p(z)|>1$.

Question

Let $p(z)=z^{n}+a_{n-1}z^{n-1}+...+a_{0}$ be a polynomial of degree $n\geq1$.

How can you prove that either $p(z)=z^{n}$ or there exists $z'$ with $|z'|=1$ such that $|p(z')|>1$.

Maybe we can use the maximum modulus principle and consider $q(z)=z^{n}p(1/z)$.

Answer 1

Let us assume that $\lvert f(z)\rvert\le 1$, for all $\lvert z\rvert=1$, and show that in such case $f$ has to be equal to $z^n$.

Cauchy Integral formula implies that $$ n!=f^{(n)}(0)=\frac{n!}{2\pi i}\int_{|z|=1}\frac{f(z)\,dz}{z^{n+1}}=\frac{n!}{2\pi }\int_0^{2\pi} f(\mathrm{e}^{it})\,\mathrm{e}^{-int}\,dt, $$ and thus, $$ 1=\frac{1}{2\pi }\int_0^{2\pi} f(\mathrm{e}^{it})\,\mathrm{e}^{-int}\,dt. \tag{1} $$ But we have assumed that $\lvert\, f(z)\rvert\le 1$, for all $\lvert z\rvert=1$, and hence $$ \lvert\, f(\mathrm{e}^{it})\,\mathrm{e}^{-int}\rvert \le 1. \tag{2} $$ The only way for $(1)$ and $(2)$ to be satisfied simultaneously is if $$ f(\mathrm{e}^{it})\,\mathrm{e}^{-int}=1,\quad\text{for all $t\in[0,2\pi]$,} $$ or equivalently $$ f(z)=z^n, $$ for $\lvert z\rvert=1$, which implies that $f(z)=z^n$.