The Fundamental Theorem of Algebra Revisited

Back to Contents

According to the fundamental theorem of algebra, if we have a non-constant polynomial function of one variable and if that function has complex coefficients, then there exists at least one complex root for that function, a non-zero number that zeroes out the function. In order for that proposition to stand true to mathematics the infinite set of the complex numbers must contain all of the numbers necessary to carry out any algebraic calculation. Now we want to prove and verify that twin proposition.

With any complex numbers ak and z we generate the general polynomial,

(Eq’n 1)

through the arithmetic processes of exponentiation, multiplication, and addition. We thus expect to use some combination of subtraction, division, and root-extraction to disassemble the polynomial in finding a solution. That expectation leads us to believe that we will use negative numbers, fractions, and imaginary numbers, in addition to the positive integers, in dissolving the equation P(z)=0. We thus have the entire infinite set of the complex numbers as necessary to the solution of all polynomial equations in one variable. Further, the three analytic processes named above necessitate the creation of only three kinds of new numbers – negative, fractional, and imaginary – in addition to the natural numbers. There exist no other kinds of numbers, so the infinite set of the complex numbers is sufficient for the solution of any polynomial equation.

We want to prove and verify the proposition that any polynomial P(z) with arbitrary complex coefficients ak has a solution in the set of the complex numbers; that is, that there exists some z0 such that P(z0)=0. To that end we use the polar representation

(Eq’ns 2)

The k-th term of the polynomial thus has the form

(Eq’n 3)

in which the subscript k is an index and the superscript k is an exponent. If we take that term by itself, we see that as θ goes from 0 to 2π the point described by Pk traces out a circle of radius rkrk k times.

The straight line extending from the center of a circle, as described above, to the point on the circle specified by θ corresponds to a vector in the complex plane. Thus we can represent our polynomial as a vector sum, with the vectors corresponding to the various terms resting against each other head to tail, with the tail of the first vector resting on the point a0 and with the head of the last vector resting on the point P(z). As θ goes from 0 to 2π the vectors turn counterclockwise and the head of the last vector whirls and gyrates around the complex plane and then comes back to its beginning location. Can we guarantee that there exists at least one value of z that will make the head of the last vector touch the origin of the complex plane?

We now have our arbitrary polynomial as an algebraic function of two independent variables, which each have an infinite domain (to say that θ has an infinite domain means that is goes "around the clock" more than once, as many times as we desire). We thus expect the polynomial to have infinite range, covering each and every point on the complex plane, including, relevantly, the origin. If that proposition did not stand true to mathematics, then there would exist at least one forbidden point that the polynomial could not cover.

Assume that such a forbidden point exists (if we have a forbidden region on the complex plane, we want to look at a point on the boundary between that region and the allowed area of the complex plane). There exists an infinitesimally small distance between that forbidden point and the nearest allowed point. That statement implies the non-existence of any infinitesimal change in the values of r and θ that will shift the polynomial onto the forbidden point and thus necessitates the existence of a discontinuity in the polynomial. But an algebraic function, a finite combination of the three fundamental processes of arithmetic, that has the values of real numbers (with their infinitely extended fractional parts) as its domain can have no discontinuities. Therefore, our assumption of the existence of forbidden points must stand false to mathematics and we must assert that our arbitrary polynomial can reach and cover each and every point on the complex plane. Thus, there must exist at least one pair of values (r,θ) for which P(z)=0.

QED

habg

Back to Contents