$p(z) = 2z^n + a_{n-1}z^{n-1}+...+a_0$ is a polynomial, need to show $|z|\le1$ and $|p(z)|>1$

Question

Let $p(z) = 2z^n + a_{n-1}z^{n-1}+...+a_0$ be a polynomial. Show that there is a point $z$ such that $|z| \le 1$ and $|p(z)| > 1$.

I am new in complex-analysis and honestly I do not know where to start, I would really appreciate any hint or help.

Answer 1

By contradiction, suppose $|p(z)|\leq 1$ for $|z|\leq1$. Then by Cauchy $$2n!=|f^{(n)}(0)|=\left|\frac{n!}{2\pi i}\int_{|z|=1}\frac{f(\zeta)d\zeta}{\zeta^{n+1}}\right|\leq n!,$$ contradiction.