Complex polynomial of degree $n$

Question

I've got following problem. Let $z$ be a complex number. Let $P(z)$ be a polynomial of degree $n$. Suppose that $|P(z)| \le M$ for $|z| \le 1$. Prove that for $|z| \ge 1$ we have. $|P(z)| \le M \cdot |z|^n$. Can somebody help?

Answer 1

This is a special case of the Bernstein-Walsh inequality, but there's a simple proof here:

write $P(z) = z^nQ(1/z)$ where $Q$ is a polynomial. Restricting to $|z| = 1$ we see that $\|P\| = \|Q\|$, where $\| f \|$ is the max of $|f|$ on the unit circle $|z| = 1$. Now just note that if $|z| > 1$, $|1/z| < 1$, so $|Q(1/z)| \leq \|Q\|$ by the maximum principle.