In a recent email Professor Diego Vaggione of the National University of Cordoba, Argentina kindly drew my attention to a note of his that appeared in Colloquium Mathematicum not long ago. The note that presents a short proof of the Fundamental Theorem of Algebra follows (in an HTML rendition) the message from Professor Vaggione.

COLLOQUIUM MATHEMATICUM

In most traditional textbooks on complex variables, the Fundamental Theorem of Algebra is obtained as a corollary of Liouville's theorem using elementary topological arguments.

The difficulty presented by such a scheme is that the proofs of Liouville's theorem involve complex integration which makes the reader believe that a proof of the Fundamental Theorem of Algebra is too involved. even when topological arguments are used.

In this note we show that such a difficulty can be avoided by giving a simple proof of the Maximum Modulus Theorem for rational functions and then obtaining the Fundamental Theorem of Algebra as a corollary. The proof obtained in this way is intuitive and mnemotechnic in contrast to the usual elementary proofs of the Fundamental Theorem of Algebra.

As usual we use C to denote the set of complex numbers. By D(a, ε) we denote the set {z ∈ C: |z - a| < ε}.

LEMMA. Let f be a function such that f(D(a,ε)) is contained in a half plane whose defining straight line contains 0. Let k ≥ 1. Then if the limit lim_z→a f(z)/(z - a)^k exists, it is 0.

Proof. Suppose lim_{z → f(z)/(z - a)^k = b≠}0. Without loss of generality we can suppose that f(D(a, ε)) is contained in the half plane {z: Re(z) ≥ 0} (replace f with cf for a suitable c ∈ C.) Let {z_n: n≥ 1} be a sequence such that lim_n→∞z_n = a and b(z_n - a)^k is a negative real number, for every n ≥ 1. Thus we have

which is absurd. Q.E.D.

MAXIMUM MODULUS THEOREM FOR RATIONAL FUNCTIONS. Let R(z) = p(z)/q(z), with p, q complex polynomials without common factors. Suppose there exists a ∈ C such that q(a) ≠ 0 and |R(z)| ≤ |R(a)|, for every z ∈ D(a,ε), with ε > 0. Then R is a constant function.

Proof. Suppose that R is not constant. Since p₁(z) = q(a)p(z) - p(a)q(z) has a zero at z = a, there exists an integer k ≥ 1 and a polynomial c(z) such that p₁(z) = (z - a)^kc(z) and c(a) ≠ 0. Thus

(R(z) - R(a))/(z - a)^k = p₁(z)/[q(a)q(z)(z - a)^k] = c(z)/[q(a)q(z)]

and therefore

lim_z→a (R(z) - R(a))/(z - a)^k ≠ 0.

Since |R(z)|≤|R(a)| for every z ∈ D(a, ε), f(z) = R(z) - R(a) satisfies the hypothesis of the above lemma (make a picture). Thus we arrive at a contradiction. Q.E.D.

FUNDAMENTAL THEOREM OF ALGEBRA. A polynomial with no zeros is constant.

Proof. Suppose that p(z) is not constant and p(z) ≠ 0 for every z ∈ C. Since lim_{n → ∞} |p(z)| = ∞, there exists a ∈ C such that |p(a)| ≤ |p(z)| for all z ∈ C. Thus applying the Maximum Modulus Theorem to the rational function 1/p(z), we arrive at a contradiction. Q.E.D.

Acknowledgement. I would like to thank María Elba Fasah for her assitance with the linguistic aspects of this note.

Facultad de Mathemática, Astronomía y Física (FAMAF)
Universidad Nacional de Córdoba
Ciudad Universitaria
Córdoba 5000, Argentina

Received 8 August 1996

• Perfect numbers are complex, complex numbers might be perfect
• Fundamental Theorem of Algebra: Statement and Significance
• What's in a proof?
• More about proofs
• Axiomatics
• Intuition and Rigor
• How to Prove Bolzano's Theorem
• Early attempts
• Proofs of the Fundamental Theorem of Algebra
  • Remarks on Proving The Fundamental Theorem of Algebra
    • Sketch of a Proof by Birkhoff and MacLane
    • Sketch of Second Proof (after Cauchy)
    • Details of the Cauchy's Proof
    • Note on the Extreme Value Theorem
    • Sketch of Proof by the methods of the theory of Complex Variables (after Liouville)
    • Maximum Modulus Theorem
  • A Proof of the Fundamental Theorem of Algebra: Standing on the shoulders of giants
  • Yet Another Proof of the Fundamental Theorem of Algebra
  • Fundamental Theorem of Algebra - Yet Another Proof
  • A topological proof, going in circles and counting
  • A Simple Complex Analysis Proof
  • An Advanced Calculus Proof

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