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C O L L O Q U I U M M A T H E M A T I C U M

VOL. 73 1997 NO. 2

ON THE FUNDAMENTAL THEOREM OF ALGEBRA

BY

DIEGO V A G G I O N E (C ´ ORDOBA)

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.

P r o o f. Suppose lim z→a f (z)/(z − a) k = b 6= 0. Without loss of generality we can suppose that b = 1 and that f (D(a, ε)) is contained in the half plane {z : Re(z) ≥ 0} (take e f = cf for a suitable c ∈ C). Let {z n : n ≥ 1} be a sequence such that lim n→∞ z n = a and (z n − a) k is a negative real number, for every n ≥ 1. Thus we have

1 = lim

n→∞ f (z n )/(z n − a) k = Re lim

n→∞ f (z n )/(z n − a) k

= lim

n→∞ Re f (z n )/(z n − a) k ≤ 0, which is absurd.

Maximum Modulus Theorem for Rational Functions. Let R(z)

= p(z)/q(z), with p, q complex polynomials without common factors. Suppose

1991 Mathematics Subject Classification: Primary 12D10.

[193]

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194

D. V A G G I O N E

there exists a ∈ C such that q(a) 6= 0 and |R(z)| ≤ |R(a)| for every z ∈ D(a, ε), with ε > 0. Then R is a constant function.

P r o o f. Suppose that R is not constant. Since p 1 (z) = q(a)p(z)−p(a)q(z) has a zero at z = a, there exist an integer k ≥ 1 and a polynomial c(z) such that p 1 (z) = (z − a) k c(z) and c(a) 6= 0. Thus

R(z) − R(a)

(z − a) k = p 1 (z)

q(a)q(z)(z − a) k = c(z) q(a)q(z) and therefore

z→a lim

R(z) − R(a) (z − a) k 6= 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.

Fundamental Theorem of Algebra. A polynomial with no zeros is constant.

P r o o f. Suppose that p(z) is not constant and p(z) 6= 0 for every z ∈ C.

Since lim z→∞ |p(z)| = ∞, there exists a ∈ C such that |p(a)| ≤ |p(z)| for every z ∈ C. Thus applying the Maximum Modulus Theorem to the rational function 1/p(z), we arrive at a contradiction.

Acknowledgements. I would like to thank Mar´ıa Elba Fasah for her assistance with the linguistic aspects of this note.

Facultad de Matem´ atica, Astronom´ıa y F´ısica (FAMAF) Universidad Nacional de C´ ordoba

Ciudad Universitaria C´ ordoba 5000, Argentina

E-mail: vaggione@mate.uncor.edu

Received 8 August 1996

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