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?

  • $\begingroup$ 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. $\endgroup$ – user228113 Apr 29 '15 at 17:49

It is a proof by contradiction.

1) Let $p(z)$ be a nonconstant polynomial.

2) Assume $p(z)$ has no root.

3) Conclude $p$ is constant.

Point 3) contradicts point 1). The conclusion isn't that 3) holds, it is that 2) fails.

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  • $\begingroup$ I don't understand what is meant by constant. Is it constant in the traditional sense? i.e. f(z) = 9 $\endgroup$ – Siyanda Apr 29 '15 at 18:05
  • $\begingroup$ Yes. It is exactly that. $\endgroup$ – Krishan Bhalla Apr 29 '15 at 18:15

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