if $f(x)$ is a polynomial with complex coefficients, prove that if $|f(x)|\leq 1$ for all $x\in \mathbb{C}$, then $f$ is a constant polynomial.

Question

Want to prove that if $f(x)$ is a polynomial with complex coefficients, prove that if $|f(x)|\leq 1$ for all $x\in \mathbb{C}$, then $f$ is a constant polynomial.

So far I'm using the fundamental theorem of algebra to say that f(x) has no roots since it is of degree zero, which means that it will hence be a constant polynomial.

However, I am missing the link between how the question implies f(x) is of zero degree, can someone help me fill in this link and figure out why f(x) is of zero dengree?

Answer 1

Suppose it is not constant, then $\;f(z)=a_nz^n+\ldots+a_1x+a_0\;$ , with $\;n\ge 1,\,\,a_n\neq0\;$ , but then

$$|f(z)|=|z|^n\left|\frac{a_n}{z^n}+\ldots+\frac{a_0}{z^n}\right|\;\ldots$$

Can you now show that your polynomial isn't in fact bounded and thus get a straightforward contradiction?

Of course, with Liouville's Theorem the result follows at once.