$|p(z)| \leq M$ for $|z| \leq 1$ Show that $|p(z)| \leq M|z|^n$ for $|z| \geq 1$
Solution 1:
Consider the function $f(z)=\sum_{k=0}^n a_k z^{n-k}=z^n p(1/z)$. We have for $|z|=1$,$|f(z)| = |p(1/z)| \leq M$.
By maximum modulus principle, we have $|f(z)| \leq M, \forall |z|\leq 1$.
Thus $|z^np(1/z)| \leq M, \forall z| \leq 1$, i.e. $$|p(1/z)|\leq M|\frac{1}{z}|^n, \forall 0< \left|z\right| \leq 1$$
So $|p(z)| \leq Mz^n, \forall |z|\ge 1$