The Fundamental Theorem of
Algebra

We claim that with the conception of the complex numbers the set of all numbers achieves closure with respect to arithmetic; that is, we claim that the set of all numbers contains enough elements to satisfy any calculation we might perform with the operations of arithmetic. The set of all numbers, as we have seen, includes negative numbers, so that we can complete all possible subtractions; fractional numbers, so that we can complete all possible divisions; and imaginary numbers, so that we can complete all possible extractions of roots. Now we want to prove and verify the proposition that the set of complex numbers (that is, the set of all numbers) gives us enough numbers that we can carry to completion any calculation comprising the operations of arithmetic put together in any number and in any sequence. We call that proposition the Fundamental Theorem of Algebra.

For the sake of the proof we use arithmetic extended and abstracted into algebra, hence the name of the theorem. To simplify further, I will note that Taylor's theorem tells us that we can represent any smooth, multiply differentiable function of a complex number Z with a polynomial in Z. We now claim that every polynomial equation having complex coefficients and degree greater than or equal to one has at least one complex root. That statement gives us the essence of the Fundamental Theorem of Algebra. The proof of this theorem closes the set of numbers: we now have all the numbers that we could ever need and no more numbers exist for us to discover in the realm of Mathematics.

We have the n-th order
polynomial

(Eq'n 1)

in which the coefficients can represent
complex numbers. We now claim for our proof that for that equation there exists
at least one number Z such that

(Eq'n 2)

I will present an
elegant proof that was devised by Diego Vaggione of the Faculty of Mathematics,
Astronomy and Physics at the University of Cordoba in Argentina. Professor
Vaggione presented his proof in the paper "On the Fundamental Theorem of
Algebra" in Colloquium Mathematicum, Vol. 73, No. 2 (1997) pp 193-194 and it has
been reprinted on the website at URL
http://www.cut-the-knot.org/fta/Vaggione.shtml. The proof proceeds in three
steps:

Step 1

We establish the truth
to mathematics of the following Lemma: Given a domain D=D(A,
ρ) in
the complex plane containing the points p_{D} within a radius
ρ of a point A, let f represent a
function on the complex plane such that f(p_{D}) is contained in the
half plane whose defining straight line contains the point Z=0. Let k≥1. Then if
the limit exists, we have as true to mathematics

(Eq'n 3)

To prove and verify
that statement we suppose that its opposite comes true to mathematics; that is,
we postulate

(Eq'n 4)

We lose no generality in our result if we
assume into our premises the statement that we have contained f(p_{D})
in the half of the complex plane in which the real parts of all the points in
the half plane are positive numbers. Define a sequence Z_{n} for n≥1
such that

(Eq'n 5)

and that for every n≥1 the function b(Z_{n}-A)^{k}
produces a negative real number. If we divide Equation 4 by b, we get

(Eq'n 6)

which is absurd.

Note that in Equation
6 in the first line I replaced the limit as Z goes to A in Equation 4 with the
limit as n goes to infinity. I was able to make that replacement by way of the
sequence defined, in part, by Equation 5. In the second line I brought in the
Real Number function to acknowledge that, because the number one on the left
side of the equation is a real number, whatever we have on the right side must
also be a real number. In the third line I exploited the fact that the Real
Number operator and the Limit operator commute with each other. That third line
then yields a positive real number (Ref(Z) by assumption) divided by a negative
real number (by definition of the sequence Z_{n}), so the equation tells
us that a positive real number equals a negative real number or zero, which is
absurd.

Thus, we must dismiss
the premise behind Equation 4 and declare the Lemma proven and verified as true
to mathematics. Q.E.D.

Step 2

We establish the truth to mathematics of the Maximum Modulus Theorem for Rational Functions: Define a rational function of a complex number Z as R(Z)=p(Z)/q(Z), in which p(Z) and q(Z) represent complex polynomials that have no common factors. If there exists a complex number A such that q(A)≠0 and |R(Z)|≤|R(Z)| for every Z in the domain D(A, ρ), with ρ>0, then R is a constant function.

We begin our proof by
supposing that R is not a constant function on the given domain. We thus have a
non-zero function

(Eq'n 7)

that goes to zero at Z=A. We can see that
f(Z) is the difference between the products of pairs of complex polynomials, so
we know that there exists some integer k≥1 and a polynomial c(Z) such that c(A)≠0
and

(Eq'n 8)

So now we write

(Eq'n 9)

And we take the limit of that equation as
Z goes to A and get

(Eq'n 10)

But |R(Z)|≤|R(A)| for every Z in the domain D(A, ρ), so R(Z)-R(A) satisfies the hypothesis of the above Lemma. On that basis we expect the limit to go to zero and thus we have a contradiction.

We must now discard
our supposition that R is not a constant function and declare the theorem proven
and verified as true to mathematics. Q.E.D.

Step 3

We establish the truth to mathematics of the Fundamental Theorem of Algebra: A polynomial with no zeros is constant.

To begin the proof we
assume into our premises the statement that the polynomial P_{n}(Z) is
not a constant function and that P_{n}(Z)≠0 for all of the complex
numbers Z. Next we rewrite Equation 1 to calculate |P_{n}(Z)| and obtain

(Eq'n 11)

in which b_{i}=a_{i}/a_{n}.
That equation tells us that

(Eq'n 12)

From that fact we infer the statement
that if we extend a line across the complex plane and calculate the value of |P_{n}(Z)|
at every point on that line, the value that we calculate at one point will be
less than any of all the other values that we calculate; that is, it will be a
minimum. If we then scan that line across the entire complex plane, we will find
that there exists a complex number A such that |P_{n}(A)|≤|P_{n}(Z)|
for all possible values of Z.

Define the rational
function R(Z)=1/P_{n}(Z). Then we have |R(Z)|≤|R(A)| and P_{n}(A)≠0
(by hypothesis), so R(Z) must conform to the Maximum Modulus Theorem; that is,
R(Z) is a constant function, one that maps all values of Z onto a single number.
But if R(Z) is a constant function, then P_{n}(Z) must also be a
constant function.

That inference
contradicts our assumption, so we must dismiss that assumption and assert its
opposite as true to mathematics. We have two possibilities. We can say that P_{n}(Z)
is a constant function and that P_{n}(Z)≠0 for all Z (which gives us our
Q.E.D.) or we can say that P_{n}(Z) is a variable function and that P_{n}(Z)=0
for at least one value of Z.

That latter statement tells us that the set of the complex numbers, represented by the complex plane, guarantees us a solution of any complex polynomial. And that statement, in its turn, gives us what we set to prove and to verify. So now we know that we have enough numbers available to us that we can carry out any algebraic calculation. This shows us that the set of complex numbers gives us all of the numbers that we will ever need for our descriptions of Mathematical Reality and, by extension, of Physical Reality.

Now if we let Z_{1}
represent a number that, when substituted for Z, makes the polynomial in
Equation 1 equal zero, then we know that Z-Z_{1} is a factor of P_{n}(Z);
that is, there exists a polynomial P_{n-1}(Z) of order n-1 for which

(Eq'n 13)

(I include a sketch of the proof of that
proposition below as an appendix). Now P_{n-1}(Z) represents a complex
polynomial that has a solution Z_{2}, so we can write

(Eq'n 14)

We can continue that process right down
to the polynomial of first order. So now we know that we can represent our n-th
order complex polynomial as a product,

(Eq'n 15)

Multiply that product
out, gather together the terms in the same powers of Z, and compare the result,
term by term, with the polynomial expression in Equation 1. Equating the
coefficients of identical powers of Z then leads us to Vieta's root theorem,
originally devised by François Viéte (Franciscus Vieta, 1540-1603): for an n-th
order complex polynomial we have as true to mathematics

(Eq'ns 16)

Those equations can often offer clues in
solving the polynomial equation.

Appendix

We want to show how to prove and verify the theorem:

If U represents a number that makes a polynomial P(Z) equal zero when we substitute it for Z, then P(Z) has Z-U as a factor; that is, there exists a polynomial Q(Z) of lower degree than P(Z) such that P(Z)=(Z-U)Q(Z).

We know that we can
divide any polynomial P(Z) by Z-U for any number U. We get a quotient function
Q(Z) one degree lower than P(Z) and a remainder R one degree lower than Z-U.
That remainder must be a constant. Thus we can represent the original polynomial
as

(Eq'n 17)

But if substituting U for Z makes P(Z)=0,
then that equation becomes

(Eq'n 18)

which necessarily entails R=0. We must
therefore rewrite Equation 17 as

(Eq'n 19)

Q.E.D.

Consider a trivial
example. Let's divide Z-U into Z^{2}+2Z+1. We have

We have the remainder R=U^{2}+2U+1,
which equals zero if U=-1. That means that the divisor and the quotient above
both come to Z+1, which are the factors of Z^{2}+2Z+1.

efefefaaabbbefefef