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} + a_{1}z^{ n -1} + ... + a_{n -1}z + a_{n}

where n is an integer > 0, and the a_{i} 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
*z _{k}* such that

*f(z*= 1/

_{k})*k*(

*k*= 1,2...). Since

*f(z)*has a pole at infinity, |

*f(z)*| >1 uniformly outside some circle

*C*. The points z

_{k}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.

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