# What is meant by “constant” in Liouville's Theorem?

Liouville's Theorem states that:

Every holomorphic* function for which there exists a positive number M such that $|f(z)| \le M$ for all $z \in \mathbb C$ is constant.

I'm using this to prove the Fundamental Theorem of Algebra and I understand that for the proof to be complete:

It is sufficient to show any $p(z)$ has one root, for by division we can then write $p(z)=(z-z_0)g(z)$, with $g$ of lower degree. Note that if $$p(z)=a_n z^n + a_{n-1} z^{n-1} + ... + a_0,$$ then as $|z| \to \infty$, $|p(z)| \to \infty$. \cite{FTANY} This follows as: $$p(z)= z^n \left| a_n + \frac{a_{n-1}}{z} + ... + \frac{a_0}{z^n} \right|.$$ Assume $p(z) \ne 0,$ so $1/p(z)$ is bounded for $|z| \le R$ by continuity. Thus, $1/p(z)$ is bounded, entire function, which must be constant. Thus, $p(z)$ is constant, a contradiction which implies $p(z)$ must have a zero (our assumption.)

What I want to understand is why showing that $f$ is constant shows that $f$ has a root?

• It shows that if $f$ has no roots, then it is constant. This is eqivalent to stating that if $f$ is not constant, then it has at least one root. – user228113 Apr 29 '15 at 17:49

1) Let $p(z)$ be a nonconstant polynomial.
2) Assume $p(z)$ has no root.
3) Conclude $p$ is constant.