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

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 $q(z)=z^{n}p(1/z)$.

• i think the degree has to be bigger than 1? – jorst Mar 2 '14 at 18:25
• Or (to clarify) it doesn't has to, but then: if $n=0$ nothing is to do, if $n=1$ this is just translating the unit circle, so the circle remains itself only if the translation constant is $0$ – jorst Mar 2 '14 at 18:47
• I like your last idea. You have $q(0)=1$, so there has to be a point $z'$ on the unit circle for which $|q(z')|\ge1$. What are the cases of equality in the maximum modulus principle? – Greg Martin Mar 2 '14 at 22:08

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, $$\int_0^{2\pi} f(\mathrm{e}^{it})\,\mathrm{e}^{-int}\,dt\in\mathbb R\,$$ and $$1=\frac{1}{2\pi }\int_0^{2\pi} f(\mathrm{e}^{it})\,\mathrm{e}^{-int}\,dt\le \frac{1}{2\pi }\int_0^{2\pi} |f(\mathrm{e}^{it})|\,dt\le 1, \tag{1}$$ since $$\lvert\, f(z)\rvert\le 1$$, for all $$\lvert z\rvert=1$$. Therefore, $$\mathrm{Re}\,\big(f(\mathrm{e}^{it})\, \mathrm{e}^{-int}\big)=|f(\mathrm{e}^{it})|=1,$$ for all $$\,t\in[0,2\pi]$$, and thus $$\,\mathrm{Im}\,\big(f(\mathrm{e}^{it})\,\mathrm{e}^{-int}\big)=0,$$ for all $$\,t\in[0,2\pi]$$. Consequently, $$f(\mathrm{e}^{it})\,\mathrm{e}^{-int}=1,\quad\text{for all t\in[0,2\pi],}$$ or equivalently $$f(z)=z^n,$$ for $$\lvert z\rvert=1$$, which implies that $$f(z)=z^n$$.
• Thanks for the great answer. Why $|f(z)|\leq 1$ for all $|z|=1$ implies that $|f(e^{it})e^{-int}| =1$? – Spock Dec 7 '14 at 9:36