Am Math Monthly 42(1935), 501-502

A Proof of the Fundamental
Theorem of Algebra:
Standing on the shoulders of giants

R. P. Boas, Jr.

This note gives a proof, believed to be new, of the fundamental theorem of algebra; it is obtained by the use of the classical theorem of Picard: If there are two distinct values which a given entire function never assumes, the function is a constant. The proof is extremely simple and may be of interest as an application of Picard's theorem.

The fundamental theorem of algebra may be formulated as follows: An arbitrary polynomial,

f(z) = z n + a1z n -1 + ... + an -1z + an

where n is an integer > 0, and the ai are constants, has at least one zero (in the complex plane). We shall use in addition to Picard's theorem only the facts that f(z) is an entire function - hence, in particular, continuous - and that f(z) has a pole at infinity.

The proof is indirect. Suppose that f(z) is never zero. I say then that f(z) also fails to take on one of the values 1/k (k = 1,2,...). In fact, suppose that there are points zk such that f(zk ) = 1/k (k = 1,2...). Since f(z) has a pole at infinity, |f(z)| >1 uniformly outside some circle C. The points zk all lie within C, and hence have at least one limit point Z within C. Since f(z) is continuous,

This contradiction allows us to conclude that for some integer k, f(z) fails to take on the value 1/k. By Picard's theorem, f(z), never assuming the distinct values 0 and 1/k, must be constant, contrary to the hypothesis that the degree of f(z) was at least 1. This contradiction shows that f(z) must have at least one zero, and the proof is complete.

|Contact| |Front page| |Contents| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny