Fundamental Theorem of Algebra

From Liouville’s Theorem to Polynomial Roots

A nonconstant polynomial can grow without bound, so Liouville’s Theorem does not say that the polynomial itself is constant. But if a polynomial had no root, its reciprocal would be entire. Polynomial growth would then make that reciprocal bounded outside a sufficiently large disk; continuity would bound it on the disk as well. Liouville’s Theorem would force the reciprocal to be constant, which is impossible for a nonconstant polynomial.

This argument proves the Fundamental Theorem of Algebra: every nonconstant polynomial with complex coefficients has a complex root. We will also see how this existence result gives a complete factorization into linear factors. The proof relies on Liouville’s Theorem from the previous tutorial, as well as the Extreme Value Theorem: a continuous real-valued function on a compact set attains its maximum and minimum.

A nonzero constant polynomial has no roots, so the theorem concerns polynomials of degree at least one. Its conclusion is about complex roots: a polynomial with real coefficients need not have any real roots. For example, \(x^2+1\) has no real roots, but it does have complex roots.

Polynomial Growth at Infinity

The leading term of a polynomial controls its size when \(|z|\) is large. The reverse triangle inequality lets us make that observation precise, even when the other terms point in directions that might partially cancel the leading term.

Proof. Since \(a_n\neq0\), the reverse triangle inequality gives

For \(r=|z|>0\), divide the sum of lower-order terms by \(r^n\):

For each \(k<n\), the exponent \(k-n\) is negative, so \(r^{k-n}\to0\) as \(r\to\infty\). There are only finitely many terms in the sum. Therefore the sum tends to zero, and we can choose \(R>0\) such that it is at most \(|a_n|/2\) for every \(r\geq R\). Multiplying that inequality by \(r^n\) and using the displayed lower bound gives

The right-hand side tends to infinity with \(|z|\), which proves the lemma. \(\square\)

This estimate gives more than unboundedness: it supplies a lower bound that can be inverted. The next example makes that bound explicit for a particular polynomial.

The Fundamental Theorem of Algebra

Proof. Let \(p\) be a polynomial of degree \(n\geq1\). Suppose, for a contradiction, that \(p(z)\neq0\) for every \(z\in\mathbb{C}\). Define

The polynomial \(p\) is holomorphic on \(\mathbb{C}\), and by the supposition it never vanishes. The Reciprocal of a Holomorphic Function theorem therefore shows that \(g\) is holomorphic on \(\mathbb{C}\), so \(g\) is entire.

By the Growth of a Polynomial Lemma, there is an \(R>0\) such that \(|p(z)|\geq (|a_n|/2)|z|^n\) whenever \(|z|\geq R\). Consequently,

On the closed disk \(\overline{D(0,R)}\), \(g\) is continuous, because it is holomorphic there. The disk is compact, so the Extreme Value Theorem implies that \(|g|\) has a finite maximum on it. Combining this bound with the bound outside the disk shows that \(g\) is bounded on all of \(\mathbb{C}\).

Liouville’s Theorem now says that \(g\) is constant. But \(g(z)=1/p(z)\) is never zero, so its constant value must be nonzero. It follows that \(p(z)=1/g(z)\) is constant, contradicting the assumption that \(p\) has degree \(n\geq1\). The supposition that \(p\) has no root is false; hence \(p\) has a complex root. \(\square\)

The proof uses both parts of the domain argument. Polynomial growth bounds the reciprocal outside a disk, while compactness and continuity bound it inside. A bound only on the exterior would not, by itself, establish that the reciprocal is bounded on the whole plane.

From One Root to a Complete Factorization

The Fundamental Theorem of Algebra guarantees one root. The algebraic factor theorem turns that root into a linear factor, and the same existence result can then be applied to the remaining polynomial. Repeating this process produces a factor for every degree.

Proof. We use induction on the degree \(n\). For \(n=1\), write \(p(z)=a_1z+a_0\), where \(a_1\neq0\). Then

which is the required factorization. Now suppose \(n\geq2\) and the result holds for polynomials of degree \(n-1\). By the Fundamental Theorem of Algebra, \(p\) has a root \(r_1\). Polynomial division by \(z-r_1\) gives a quotient \(q\) and a constant remainder. Substituting \(z=r_1\) shows that the remainder is \(p(r_1)=0\). Therefore \(p(z)=(z-r_1)q(z)\), where \(q\) has degree \(n-1\) and leading coefficient \(a_n\). By the induction hypothesis, \(q(z)=a_n\prod_{j=2}^{n}(z-r_j)\) for some complex numbers \(r_2,\ldots,r_n\). Multiplication by \(z-r_1\) gives the stated factorization for \(p\). This completes the induction. \(\square\)

The list of roots in this theorem includes repetitions when a root occurs more than once as a factor. A degree-\(n\) polynomial therefore has exactly \(n\) roots counted with multiplicity, even if it has fewer than \(n\) distinct roots.

What the Theorem Does—and Does Not—Say

The Fundamental Theorem of Algebra guarantees roots in \(\mathbb{C}\), not necessarily in \(\mathbb{R}\). For instance, \(z^2+1\) has roots \(i\) and \(-i\), since \(i^2+1=(-1)+1=0\) and \((-i)^2+1=(-1)+1=0\). Thus a real polynomial can have no real root while still satisfying the theorem.

The theorem also does not guarantee that all roots are distinct. For example, \(z^2-2z+1=(z-1)^2\) has only one distinct root, \(1\), but has two roots when counted with multiplicity. The linear factorization theorem makes this counting convention precise. Together, the two results say that over the complex numbers, a polynomial’s degree accounts for all its linear factors and roots.

The proof of root existence is a useful example of a global conclusion obtained from Liouville’s Theorem. A hypothetical absence of roots would produce an entire reciprocal; growth at infinity and compactness would make that reciprocal bounded; and Liouville’s Theorem would then contradict the polynomial’s nonconstant degree.