RE: Why do complex zeros come in complex conjugate pairs? /

I was responding to @voice-of-apollo in the question

https://steemit.com/math/@voice-of-apollo/if-a-polynomial-has-real-coefficients-then-why-do-its-zeros-come-in-complex-conjugate-pairs

and when my response reached 100 lines, I decided to make it a post.

He asks for an insight of why a complex polynomial always has solutions that are merely mirrored over the real number line.

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The complex numbers are by far not the only mathematical structure with such a symmetry.
A deeper foundation for a "why" can be found in Galois theory.

Here we find that at no point in the algebraic theory of complex number does it matter whether we work with a+bi, or a-bi, as long as we are consistent about it.
It would be fun to talk about the mathematical revolutionary Évariste Galois, who in 1830 de facto founded abstract algebra, but then got shot over a girl before reaching the age of 22...

My favorite technical history book on the subject is Modern Algebra and the Rise of Mathematical Structures by the historian Leo Corry, which spans 200 years from Galois to category theory in the 1970's.

But I digress...

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We could say that as multiplication by a complex numbers of absolute value |a+bi|=1 amounts to a rotation in the complex plane, the symmetry comes down to the fact that the world would be effective the same if you changed left and right along some axis. But that's a bit loaded.
But before we'd go symmetries in general, I'll be more hands on here and take a particular equation apart.

Consider the equation

(- 52) + 3 x - x^3 = 0

This is a polynomial of third order, so by the fundamental theorem of algebra it has three solutions. Those three solutions happen to be
x = a + b i = -4
x = a + b i = +2 + 3 i
x = a + b i = +2 - 3 i

Let's go back a step and assume we don't know the solutions yet.
Any complex number is of the form a+b i, so we know that any solution looks like

x = a+b i

for some real numbers a and some b.
Let's plug in this generic expression into the equation

• 52 + 3 x - x^3 = 0
  ==>
• 52 + 3 (a+b i) - (a+b i)^3 = 0

Not let's expand the third term, keeping in mind that i^2 reduces to -1.

(-52 + a (3 + 3 b^2 - a^2)) + b (3 - 3 a^2 + b^2) i = 0

This is still the same equation as above, except that we expressed it in terms of two real variables a and b, instead of one complex x.
Put yet differently, note that this is again of the form

A + B i = 0

where
A = -52 + a (3 + 3 b^2 - a^2)
and
B = b (3 - 3 a^2 + b^2)

For the equation A + B i = 0 to be solved, since complex and real part don't mix, we need both A=0 and B=0.
In other words, the expression

B = b (3 - 3 a^2 + b^2)

must be set to zero and solved for a and b, simultaneously with A=0.
And because of the square here at the end of the bracket, b^2, whatever a solution b is, so that B=0, the negative of b will also be zero. So for example, at the beginning I told you
x = a + b i = +2 + 3 i
would be a solution.

Indeed
3 - 3 (2)^2 + (-3)^2 = 3 - 12 + 9 = 0
But also
3 - 3 (2)^2 + (+3)^2 = 3 - 12 + 9 = 0

The point is that the condition that a complex polynomial be zero always comes down to a condition of two real polynomials be zero, and those are always such that the b-component has powers to that the sign doesn't matter.

To see this in for the general case, we must take a generic polynomial, the sum over coefficient a_n and monomials x^n
a_0 + a_1 · x + a_2 · x^2 + ...,
then plug in x=a+bi and systematically apply the binomial formula, taking into account that i^2 always reduces to -1. We'll be left with even powers in b.

Hope that helps
@qed