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Math Help - Complex nunber question

  1. #1
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    Complex nunber question

    \alpha=cos\frac{2\pi}{5}+isin\frac{2\pi}{5}.
    Find the quadratic equation with roots \alpha+\alpha^4 and \alpha^2+\alpha^3.
    Hence show that:
    cos\frac{2\pi}{5}=\frac{\sqrt5-1}{4}.

    So far all I can come up with for the quadratic is:
    (x-\alpha-\alpha^4)(x-\alpha^2-\alpha^3), but I cannot think of how that will help with the second part. Help would be much appreciated!
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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  3. #3
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    Yes I know de Moivre's formula its what all of this sheet has been about. I just cannot see how this helps to find cos\frac{2\pi}{5}
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  4. #4
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    I will use a for $alpha$ and c for $cos(2*pi/5)$

    If you use DeMoivre's and some trig, then you can show that
    root1 = a + a^4 = 2cos(2*pi/5) = 2c
    and root2 = a^2 + a^3 = 2cos(4*pi/5) = 2(2cos^2(2*pi/5) - 1) = 2(2c^2 - 1)
    All the imaginary parts disappear.

    So your quadratic is (x - root1)(x - root2) = (x - a - a^4)(x - a^2 - a^3)
    Now, the constant parts of the quadratic eq must be equal, so
    root1 * root2 = a^2 + a^4 + a^6 + a^7

    We can show that a^2 + a^4 + a^6 + a^7 = -1

    This means that
    root1 * root2 = 2c * 2(2c^2-1) = -1

    Now, solve for c. This will give you 3 solutions, but only one should make sense for cos(2*pi/5), and the problem already tells you the answer
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  5. #5
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    Awesome, thank you so much!
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