Show that complex conjugate is also a root of polynomial with real coefficients

  1. 1. The problem statement, all variables and given/known data
    Suppose that f(x) is a polynomial of degree n with real coefficients; that is,

    f(x)=a_n x^n+ a_(n-1) x^(n-1)+ …+a_1 x+ a_0, a_n,… ,a_0∈ R(real)

    Suppose that c ∈ C(complex) is a root of f(x). Prove that c conjugate is also a root of f(x)

    2. Relevant equations

    (a+bi)*(a-bi) = a^2 + b^2 where a, and b are always reals?
    Not really sure if this helps or not.

    3. The attempt at a solution

    I'm really clueless on how to start approaching this. I was thinking perhaps the fundamental theorem of algebra might be of some use, or perhaps the fact that a number of complex form multiplied by it's conjugate is a real number, but I'm really not sure.

    Could anyone give me a nudge in the right direction?
    Any help would be greatly appreciated!! Thanks!
     
  2. jcsd
  3. Dick

    Dick 25,810
    Science Advisor
    Homework Helper

    The root is where f(c)=0. Take the complex conjugate of that equation.
     
  4. Never mind! I found the actual theorem on the web, and I think this is pretty much what I was looking for anyway

    Consider the polynomial
    f(x)=a_n x^n+ a_(n-1) x^(n-1)+ …+a_1 x+ a_0, a_n,… ,a_0∈ R(real)
    where all a*x are real. The equation f(x) = 0 is thus
    a_n x^n+ a_(n-1) x^(n-1)+ …+a_1 x+ a_0, a_n = 0
    Given that all of the coefficients are real, we have
    a_n x^n(conjugate) = a_n x^n(x is conjugate)
    Thus it follows that
    a_n x^n(conj)+ a_(n-1) x^(n-1)(conj)+ …+a_1 x(conj)+ a_0, a_n = 0(conj) = 0
    and thus that for any root ζ its complex conjugate is also a root.
     
  5. Hahah, thanks Dick! I caught on a little late, but thanks a bunch for your reply!
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?