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Homework Help: Algebra and Complex Numbers This one is tough

  1. Jul 14, 2013 #1
    Prove that all polynomials with real coefficients, having complex roots can occur in complex conjugates only.

    It's easy to prove in a quadratic equation....
    ## ax^{2} + bx + c = 0 ##

    ## \displaystyle x = \frac{-b \pm \sqrt(b^2 - 4ac)}{2a} ##

    But how to prove the same in general?
    Please help, wasted a lot of time thinking about it!
    Last edited: Jul 14, 2013
  2. jcsd
  3. Jul 14, 2013 #2
    You want to show that if [itex] P [/itex] is a real polynomial, then [itex] P(z)=0 [/itex] implies [itex] P(\bar{z})=0 [/itex]. You will find this is not too hard to do if you use the fact that [tex] (\bar{z})^n=\bar{(z^n)} [/tex]

    Edit: Please reply to your other thread.
  4. Jul 14, 2013 #3


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    Every polynomial with real coefficients can be factored, in terms of real numbers to linear or quadratic factors.
  5. Jul 14, 2013 #4
    How do you know that without assuming both the fundamental theorem of algebra (which is a far more sophisticated result than the statement the OP wants to prove) and the statement that the OP is trying to prove. It seems to me (and I could be wrong) that the only way to conclude that you could factor like this is to assume in the first place that for each linear (complex) factor in the factorization, the linear factor corresponding to the conjugate is also in the factorization, which is what allows you to create real, quadratic factors.
    Last edited: Jul 14, 2013
  6. Jul 14, 2013 #5
    You might not know it, but this exercise (or a generalization of it) is the very reason that Galois theory works.

    OK, never mind me, just wanted to say why this exercise is important :biggrin:
  7. Jul 15, 2013 #6
    Mentor.... i solved many quadratic, cubic and quartic equations and whenever i got a complex solution, one of the other solutions was always a complex conjugate... This made me think that the result was true for all polynomial and hence the question... Thanks to all of you, i have finally been able to produce a valid proof.... Actually, this question is about the complex conjugate root theorem.
    Last edited: Jul 15, 2013
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