Does there exist $\epsilon>0$ such that if a polynomial $p$ satisfies $|p(z)|<\epsilon$ for $|z| \le 1$, then $p(a) < 1$?

Question

This is an old qual question that I'm stuck on.

For any complex number $a$ not in the unit disk, does there exist $\epsilon>0$ such that if a polynomial $p$ satisfies $|p(z)|<\epsilon$ for $|z| \le 1$, then $|p(a)| < 1$?

Does anyone have a hint? Thanks in advance!

Answer 1

Suppose that the statement is true. And let $\epsilon>0$ such that the statement holds.

Then choose $M$ such that $\frac{1}{M}<\epsilon$, and $n$ such that $|a|^{n}>M$.

Then for $p(z) = \dfrac{z^{n}}{M}$ we have $|p(z)|< \epsilon$ on the unit disk.

However $$|p(a)|= \frac{|a|^{n}}{M}>1,$$ a contradiction.

Answer 2

I observe that $p_n(z) = \left(\frac{2z}{|a|-1} \right)^n$, $n = 1, 2, \dots$ is a sequence of polynomials getting closer and closer to $0$ inside the unit circle, having magnitude $1$ on the circle of radius half-way between $1$ and $|a|$, and is getting larger and larger at $a$.

This is two ideas: the "advanced calculus" idea the $x^n$ goes to zero on $[0,1)$, is $1$ on $\{1\}$, and blows up to the right of $1$ together with the observation that we can put the "1" between the unit circle and $a$. (I picked half-way between in the above, but anywhere between would do.)