Proof of the Fundamental Theorem of Algebra

We prove that every non-constant polynomial with complex coefficients has at least one complex root. This proof uses only real analysis and basic properties of continuous functions.

Proof Using Minimum Modulus

Proof

Let $p(z) = a_n z^n + a_{n-1} z^{n-1} + \cdots + a_1 z + a_0$ with $a_n \neq 0$ and $n \geq 1$ . We prove $p$ has a root by showing that $|p(z)|$ attains its minimum at some $z_0 \in \mathbb{C}$ , and this minimum must be zero.

Step 1: $|p(z)|$ attains a minimum.

For large $|z|$ , we have

$|p(z)| = |z|^n \left|a_n + \frac{a_{n-1}}{z} + \cdots + \frac{a_0}{z^n}\right| \geq |z|^n \left(|a_n| - \frac{|a_{n-1}|}{|z|} - \cdots - \frac{|a_0|}{|z|^n}\right).$

For $|z|$ sufficiently large (say $|z| \geq R$ for some $R > 0$ ), the term in parentheses is at least $|a_n|/2$ , so

$|p(z)| \geq \frac{|a_n|}{2} |z|^n \to \infty \quad \text{as } |z| \to \infty.$

Thus there exists $R > 0$ such that $|p(z)| > |p(0)|$ for all $|z| \geq R$ . Therefore, the minimum of $|p(z)|$ over all $\mathbb{C}$ is attained in the compact set $\{|z| \leq R\}$ . By compactness, $|p(z)|$ attains its minimum at some $z_0 \in \mathbb{C}$ .

Step 2: This minimum is zero.

Suppose for contradiction that $p(z_0) \neq 0$ . Translate coordinates so that $z_0 = 0$ (replace $p(z)$ by $p(z + z_0)$ ). Write

$p(z) = c_0 + c_k z^k + c_{k+1} z^{k+1} + \cdots + c_n z^n$

where $c_0 = p(0) \neq 0$ and $k \geq 1$ is the smallest index with $c_k \neq 0$ (if no such $k$ exists, $p$ is constant, contradiction).

Write $c_k = r e^{i\alpha}$ for some $r > 0$ and $\alpha \in \mathbb{R}$ . Choose $\theta \in \mathbb{R}$ such that $e^{ik\theta} = -e^{i\alpha}$ , i.e., $k\theta = \alpha + \pi$ (mod $2\pi$ ). For small $t > 0$ , let $z = t e^{i\theta}$ . Then

$p(z) = c_0 + c_k t^k e^{ik\theta} + O(t^{k+1}) = c_0 + r t^k e^{i\alpha} e^{ik\theta} + O(t^{k+1}) = c_0 - r t^k + O(t^{k+1}).$

For sufficiently small $t > 0$ , the $O(t^{k+1})$ term is negligible compared to $r t^k$ , so

$|p(z)| \approx |c_0 - r t^k| < |c_0| = |p(0)|$

for small enough $t > 0$ . This contradicts the assumption that $|p(0)|$ is the minimum of $|p(z)|$ .

Therefore, $p(z_0) = 0$ .

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Key Ideas

RemarkWhy this proof works

Compactness: The function $|p(z)|$ attains its minimum on $\mathbb{C}$ because it tends to infinity at infinity.
Local improvement: If $p(z_0) \neq 0$ , we can find a direction in which $|p(z)|$ decreases, contradicting minimality.
Polynomial behavior: The key is that $p(z) = c_0 + c_k z^k + \text{higher order terms}$ , and the $z^k$ term dominates for small $|z|$ .

This proof is entirely elementary, using only continuity, compactness, and polynomial algebra.

Alternative Proofs

RemarkOther approaches

Liouville's theorem: If $p$ has no zeros, then $1/p$ is entire and bounded, hence constant by Liouville's theorem. This implies $p$ is constant, a contradiction. This is the most elegant proof but requires complex analysis machinery.
Topological proof: Use the winding number of $p$ on a large circle. The degree of $p$ as a map $\mathbb{C} \to \mathbb{C}$ is $n \geq 1$ , so the image of a large circle winds $n$ times around the origin. By continuity, $p$ must pass through zero.
Rouché's theorem: Write $p(z) = z^n (a_n + a_{n-1}/z + \cdots + a_0/z^n)$ . On a large circle, $|z^n a_n| > |a_{n-1} z^{n-1} + \cdots + a_0|$ , so by Rouché's theorem, $p$ has $n$ zeros (counted with multiplicity).

Historical Note

RemarkGauss's contribution

Gauss gave four proofs of the fundamental theorem:

(1799) A topological argument using the winding number (not fully rigorous by modern standards).
(1816) An improved version of the first proof.
(1816) An algebraic proof reducing to the case of real polynomials.
(1849) A proof using the location of roots and Descartes' rule of signs.

The proof given here (using the minimum modulus) is closest to the third proof and is the most accessible.

Corollary: Complete Factorization

RemarkFactorization into linear factors

By induction on the degree, every polynomial $p(z)$ of degree $n \geq 1$ factors as

$p(z) = a_n (z - z_1)(z - z_2) \cdots (z - z_n)$

where $z_1, \ldots, z_n \in \mathbb{C}$ are the roots (counted with multiplicity). This is immediate from the fundamental theorem: if $p(z_1) = 0$ , then $p(z) = (z - z_1) q(z)$ for some polynomial $q$ of degree $n-1$ . Repeat until $q$ has degree $0$ .

Summary

RemarkSignificance

The fundamental theorem of algebra is called "fundamental" because:

It shows $\mathbb{C}$ is algebraically closed (every polynomial equation has a solution).
It implies every polynomial factors completely into linear factors.
It is the foundation for the theory of algebraic equations and many results in complex analysis.
The proof here is elementary but the result is deep: it requires analysis (compactness, continuity) to prove an algebraic statement.

This interplay between algebra and analysis is a hallmark of complex analysis.