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Preparations for Proving
The Fundamental Theorem of Algebra

The Fundamental Theorem of Algebra says that an n-th degree polynomial having complex coefficients has, counting multiplicities, n complex roots. It was first successfully proven by the great German mathematician Karl Friederich Gauss.

The task of proving this theorem can be greatly simplified. If it can be proven that such a polynomial p(z) has at least one complex root, say r, then the polynomial can be divided by (z−r), where r is the root, to obtain a polynomial of degree one less. Repeated operations of this sort results in n roots counting multiplicities.

Reduction to Polynomials with Real Valued Coefficients

Let p(z) be polynomial in z of degree n

Σj=0ncjzj

Let q(z) be the polynomial created by replacing all of the coefficients of p(z) with their complex conjugate.

Then w(z)=p(z)q(z) is a polynomial of degree 2n with all real coefficients.

Proof:

Let the conjugate of cj be denoted as cj*. Let

p(z) = Σcjzj
and
q(z) = Σck*zk

where the indices j and k run from 0 to n.

Then

w(z) = p(z)q(z) = ΣΣcjck*zj+k

But

w(z) = Σm=02ndmzm
where
dm = Σh=0mchcm-h*

Now consider the conjugate of a coefficient dm of w(z),

dm* = Σh=0mch*cm-h

But changing the summation index from h to g=m-h gives

dm* = Σg=0mcm-gcg*

which is identical except for the labeling of the summation index to the expression for dm. Thus dm=dm* and hence the coefficients of w(z) are real.

If w(z0)=0 then either p(z0)=0 or q(z0)=0.

Polynomials of Degree n with Real Coefficients

Let p(z) be Σj=0nrjzj. Let z=x+iy where i is the square root of negative one and both x and y are real. Suppose all of the powers of zj are expanded into terms of powers of x and iy. All of the terms involving even powers of iy will be real and all terms involving odd powers of iy will be imaginary. Thus

p(x+iy) = R(x,y) + iI(x,y)

All terms in R(x,y) will involve even powers of y, including possibly a constant which involves y0. All terms in I(x,y) involve y to an odd power, and thus y may be factored out leaving only terms with an even power of y. Therefore

p(x+iy) = F(x, y2) + iyG(x, y2)

Roots

A root of p(z) is an x+iy such that

F(x, y2) = 0
and
yG(x, y2) = 0

The second condition requires either that y=0 and hence the root is a real number, or that G(x, y2) = 0.

If y is not zero and some value x0 and y02 such that F=0 and G=0 then x0+iy0 is a root and also x0−iy0.

A Degree 2 Polynomial, a Quadratic

Let p(z)=z²+bz+c. Then

p(x+iy) = (x+iy)² + b(x+iy) + c
= x² + 2ixy −y² + bx + iby + c
= x² + bx + c −y² + i(2xy + by)
= x² + bx + c −y² + iy(2x + b)

A root would involve

either y=0 or 2x+b=0
and
x² + bx + c −y² = 0

If y=0 then x²+bx+c=0. This latter equation can be expressed as

(x+b/2)²+(c−(b/2)²)=0
and thus
(x+b/2)=±((b/2)²−c)½.

This is the case of two real roots.

If 2x+b=0 then x=−b/2 and y²=x²+bx+c. The solution is straightforward but it is insightful to view the solution as the intersection of the two functions, x=−b/2 and y²=x²+bx+c, as shown below for b=1 and c=1.

Taking the square root of y² gives a display in terms of x and y.

A Degree 3 Polynomial, a Cubic

Let p(z)=z³+bz²+cz+d. Then

p(x+iy) = (x+iy)³ + b(x+iy)² + c(x+iy) + d

which reduces to
p(x+iy) = x³ −3xy² + bx² −by² + cx +d
+iy(3x² + 2bxy + c −y²)

Thus

either y=0 or y²=3x²+2bx+c
and
(3x+b)y²=x³+bx²+cx+d.

Ignoring the case of y=0, the two functions for y² for the case of b=1, c=1 and d=1 are plotted below.

There is a singularity at x=−b/3 and the graph line at the singularity is an artifact of the graphing program and not truly a part of the function.

A Degree 4 Polynomial, A Quartic

Let p(z)=z4+bz³+cz²+dz+e. Then

p(x+iy) = (x+iy)4 + b(x+iy)³ + c(x+iy)² + dz + e

which reduces to
p(x+iy) = x4 + bx³ − (6y²−c)x² + dx −by² + e + y4
+iy(4x³ + 3bx² − 2cy² + d)

Then

either y=0 or y² = (4x³ + 3bx² + 2cx + d)/(4x+1)

and there is a singularity at x=−1/4. The other condition for a root is

y4 + (6x²+3bx+c)y² = x4 + bx³ + cx² + dx + e

This is a quadratic equation in y² and is easily solved. The results are:

(To be continued.)

Real Roots for Real Coefficient
Polynomials of an Odd Degree

Suppose p(z) is polynomial of degree n, where n is an odd number, with real coefficients and the coefficient of zn is unity. Then p(z) can be represented as b(1/z)zn where b(1/z) for z real has a limit of 1 as real z increases without bound in either the positive direction or the negative direction.

For some sufficiently large value of real z,say z+, p(z) is positive. For some sufficiently large negative real value of z, say z-, p(z) is negative. Therefore, by the Intermediate Value Theorem, there is some value of z between z- and z+ there is a value of z, say z0, such that p(z0)=0.

(To be continued.)


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