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

Σ_j=0ⁿc_jz^j

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 c_j be denoted as c_j*. Let

p(z) = Σc_jz^j
and
q(z) = Σc_k*z^k

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

Then

w(z) = p(z)q(z) = ΣΣc_jc_k*z^j+k

But

w(z) = Σ_m=0²ⁿd_mz^m
where
d_m = Σ_h=0^mc_hc_m-h*

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

d_m* = Σ_h=0^mc_h*c_m-h

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

d_m* = Σ_g=0^mc_m-gc_g*

which is identical except for the labeling of the summation index to the expression for d_m. Thus d_m=d_m* 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 Σ_j=0ⁿr_jz^j. 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 z^j 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.)