Fundamental Theorem of Algebra via Liouville

Problem

Prove that every non-constant polynomial $p(z) \in \mathbb{C}[z]$ has at least one root in $\mathbb{C}$ .

You may use Liouville's Theorem: every bounded entire function is constant.

Field

Complex Analysis

Why It's Beautiful

The Fundamental Theorem of Algebra is a statement about polynomials — pure algebra. Yet the cleanest proof is analytic, using a theorem about entire functions. The proof is essentially three lines, and the key move is a clever inversion: consider $1/p(z)$ instead of $p(z)$ .

This is a canonical example of complex analysis reaching into algebra in a way that elementary methods cannot match.

Key Idea / Trick

Suppose $p(z)$ has no root. Then $1/p(z)$ is entire. Since $|p(z)| \to \infty$ as $|z| \to \infty$ , the function $1/p(z)$ is bounded. By Liouville, $1/p(z)$ is constant — contradicting that $p$ is non-constant.

Difficulty

2 / 5

Fundamental Theorem of Algebra via Liouville — Answer

Liouville's Theorem (given)

Every bounded entire function $f: \mathbb{C} \to \mathbb{C}$ is constant.

Proof

Let $p(z)$ be a non-constant polynomial. Suppose for contradiction that $p(z) \neq 0$ for all $z \in \mathbb{C}$ .

Step 1. Since $p$ has no roots, the function $f(z) = \frac{1}{p(z)}$ is holomorphic on all of $\mathbb{C}$ (i.e., entire).

Step 2. Since $p$ is a non-constant polynomial, $|p(z)| \to \infty$ as $|z| \to \infty$ .

Therefore $|f(z)| = 1/|p(z)| \to 0$ as $|z| \to \infty$ .

In particular, there exists $R > 0$ such that $|f(z)| \leq 1$ for all $|z| > R$ .

On the compact disk $|z| \leq R$ , a continuous function attains its maximum, so $|f(z)| \leq M$ for some $M < \infty$ .

Now $\mathbb{C}$ is covered by these two regions: $\mathbb{C} = \underbrace{\{|z| \leq R\}}_{\text{bounded by }M} \cup \underbrace{\{|z| > R\}}_{\text{bounded by }1}$

Hence $|f(z)| \leq \max(M, 1)$ for all $z \in \mathbb{C}$ , so $f$ is bounded on all of $\mathbb{C}$ .

Step 3. By Liouville's Theorem, $f$ is constant. But then $p = 1/f$ is also constant — contradicting our assumption that $p$ is non-constant. $\blacksquare$

Why $|p(z)| \to \infty$ ?

For $p(z) = a_n z^n + \cdots + a_0$ with $a_n \neq 0$ : $|p(z)| = |z|^n \left| a_n + \frac{a_{n-1}}{z} + \cdots + \frac{a_0}{z^n} \right| \xrightarrow{|z|\to\infty} \infty$ since the parenthesized expression $\to a_n \neq 0$ .

Why This Proof is Surprising

Elementary proofs of FTA (e.g., topological, via winding numbers) require more machinery or case analysis. This proof reduces everything to one theorem and one observation:

A polynomial grows without bound. Its reciprocal is therefore bounded and entire. Liouville kills it.

The algebra–analysis bridge is the real surprise: a fact about roots of polynomials (algebra) follows from a fact about bounded holomorphic functions (analysis).

Bonus: Full FTA (all $n$ roots)

Liouville gives existence of one root $z_0$ . Then $p(z) = (z - z_0) q(z)$ for some degree- $(n-1)$ polynomial $q$ (polynomial long division). Apply induction: a degree- $n$ polynomial has exactly $n$ roots (counted with multiplicity) in $\mathbb{C}$ .