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Complex Numbers ~ Factors

  1. Mar 18, 2009 #1

    Mentallic

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    1. The problem statement, all variables and given/known data
    Resolve [itex]z^5-1[/itex] into real linear and quadratic factors.

    Hence prove that [tex]cos\frac{2\pi}{5}+cos\frac{4\pi}{5}=-\frac{1}{2}[/tex]


    2. Relevant equations
    [tex]z=cis\theta[/tex]

    [tex]z\bar{z}=cis\theta.cis(-\theta)=cos^2\theta+sin^2\theta=1[/tex]

    [tex]z+\bar{z}=cis\theta+cis(-\theta)=2cos\theta[/tex]


    3. The attempt at a solution
    I was able to show that the the roots of [itex]z^5-1=0[/itex] are:

    [tex]z=1,cis\frac{2\pi}{5},cis\frac{-2\pi}{5},cis\frac{4\pi}{5},cis\frac{-4\pi}{5}[/tex]

    And hence, the real factors are:

    [tex](z-1)(z^2-2z.cos\frac{2\pi}{5}+1)(z^2-2z.cos\frac{4\pi}{5}+1)=0[/tex]

    But now I'm stuck and not sure how to start proving that last equation.
     
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  3. Mar 18, 2009 #2

    Dick

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    If you multiply your factored form back out again, then the identity you are trying to prove is the coefficient of z^4. It's also basically the sum of all of the five roots.
     
    Last edited: Mar 18, 2009
  4. Mar 18, 2009 #3

    Mentallic

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    Aha and the coefficient of z4 is 0, so:

    [tex]1+cis\frac{2\pi}{5}+cis\frac{-2\pi}{5}+cis\frac{4\pi}{5}+cis\frac{-4\pi}{5}=0[/tex]

    Therefore, [tex]1+2cos\frac{2\pi}{5}+2cos\frac{4\pi}{5}=0[/tex]

    [tex]cos\frac{2\pi}{5}+cos\frac{4\pi}{5}=\frac{-1}{2}[/tex]

    Thanks :smile:
     
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