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### Complex Conjugate Roots of Real Polynomials

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Date: 01/11/2001 at 14:07:07
From: Matt Swank
Subject: Complex conjugate roots of real polynomials

I'm trying to prove that if a polynomial p(x) with real coefficients
has a complex number as a root, then its complex conjugate must also
be a root.

This is easy enough to prove for second degree polynomials, and I can
prove it for any polynomial that has a real polynomial of lesser
degree as a factor. However, I can't seem to convince myself that
given q(x) such that q(x) has complex coefficients, and
p(x) = (x-c)q(x), that either p(x) must have complex coefficients, or
the complex conjugate of c must be a root of q(x).
```

```
Date: 01/11/2001 at 15:07:49
From: Doctor Schwa
Subject: Re: Complex conjugate roots of real polynomials

Hi Matt,

Here's an argument, not quite a proof, but it sure convinces me: i is
defined as the sqrt(-1). That is, the definition of i is "that thing
so that i^2 = -1." But of course, (-i)^2 = -1 also.

So, unless you already have i written into your equation somewhere,
how can the equation (containing only real numbers) possibly tell the
difference between i and -i?  If you switched i with -i everywhere, it
couldn't possibly change the value of the polynomial.

And that leads to the proof:

If z is a complex number, and z bar its conjugate, first you need to
prove that conjugation distributes over multiplication and addition:

(z1 * z2) bar = (z1 bar) * (z2 bar)
(z1 + z2) bar = (z1 bar) + (z2 bar)

That leads to a proof that p(z bar) = (p(z)) bar

Then, finally, if p(a + bi) = 0, then p(a - bi) = the conjugate of 0,
which is still 0.

- Doctor Schwa, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Imaginary/Complex Numbers
High School Polynomials

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