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Need help with Proof

  1. Nov 27, 2006 #1
    Need Assistance w/ existence of conjugate roots in polynomial

    given that f(x) = Anx^n + An-1X^n-1+ ....A1X + A0 , and An does not = 0
    => if f(x) has a root of the form (A+Bi), then it must have a root of the form (A-iB). (complex roots)

    So far I've come up with the conjugate roots theorem to help me where it states that if a function f(x) is given with real coefficients then the conjugate must exist also. My professor suggested I try the Remainder Theorem??
    Last edited: Nov 27, 2006
  2. jcsd
  3. Nov 27, 2006 #2
    What exactly are you stuck on? By inspection you know that if (A+Bi) divides your polynomial then (A-Bi) must also divide your polynomial because your polynomial does not have any complex coefficients. So, I'd recommend doing either a direct or indirect proof - though it seems as if a direct proof will be an easier way to go. Assume that (A+Bi) divides your polynomial. It follows then that:

    [tex]A_{n}x^n + A_{n-1}x^{n-1} + ... + A_{1}x + A_{0} = 0 (mod (A+Bi))[/tex]

    The rest should fall out from there...
    Last edited: Nov 27, 2006
  4. Nov 27, 2006 #3
    Appreciate the fast reply, Well first I'm stuck on why my professor told me to use the Remainder Theorem where if a polynomial f(x) is divided by x-c then f(x) is the remainder. Second, I'm not quite sure I can prove that the conjugate exists with only the given statement of the polynomial function not having any complex coefficients. Third, I doubt that our professor would assign such a easy problem for homework. (He only assigned one problem)

    What does (mod(A+Bi)) mean?

    I am assuming that the roots are (x-(A+Bi))= 0 and (x+(A-Bi))= 0? Right?

    Tell me what you think. Regards, Chris
  5. Nov 27, 2006 #4
    Mod is just a fancy way of saying remainder really. Example 4 mod 2 = 0.
  6. Nov 27, 2006 #5
    Let's look at a simple polynomial as an example.

    Say 0 = x^2 + 2x -3
    Your solution would be (x+3)(x-1) = 0.
    So your roots are x= -3, and x = 1.

    That means something * (x+3) gives you x^2 + 2x -3.
    And something * (x-1) gives you x^2 + 2x -3.

    In this case we know what the something is, as we put in all the numbers.
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