Complex Conjugate Roots of Real PolynomialsDate: 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/ |
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