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Complex Conjugate Roots of Real Polynomials


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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