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.

Dear professor Bogomolny:

I have visited your web site on mathematics. I found it very interesting. I am enclosing a latex file of my paper "On the Fundamental Theorem of Algebra" (Colloquium Mathematicum, Vol. 73, No. 2 (1997), 193-194) in which I show that the clasical proof of the Fundamental Theorem of Algebra via Liouville can be substantialy simplified. Perhaps you can include this proof at your web site.

Best regards,

Diego Vaggione


VOL. 731997NO. 2




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 {zC: |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 limz→a f(z)/(z - a)k exists, it is 0.

Proof. Suppose limz → 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 cC.) Let {zn: n≥ 1} be a sequence such that limn→∞zn = a and b(zn - a)k is a negative real number, for every n ≥ 1. Thus we have

1= limn→ε f(z)/b(zn - a)k
= Re limn→ε f(z)/b(zn - a)k
= limn→ε Re f(z)/b(zn - a)k ≤ 0,

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

limz→a (R(z) - R(a))/(z - a)k0.

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 zC. Since limn → ∞ |p(z)| = ∞, there exists aC such that |p(a)||p(z)| for all zC. 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

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