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Math Help - Complex Numbers.

  1. #1
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    Complex Numbers.

    \alpha is the complex root (with the smallest positive argument) of the equation z^5 - 1 = 0

    show that \alpha^4 = \alpha^*. hence or otherwise obtain z^5 - 1 = 0 as a product of real linear and quadratic facto giving the coefficients in terms of integers and cosines.
    I am good with all of this expect the last part about the quadratic factors. A pointer would be good.

    Many Thanks, Bobak.
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  2. #2
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    Quote Originally Posted by bobak View Post
    I am good with all of this expect the last part about the quadratic factors. A pointer would be good.

    Many Thanks, Bobak.
    since 1 is a zero of the equation and we know that there are five roots that are evenly spaced around the unit cirlce by

    \frac{2 \pi}{5}

    so the five roots are

    1 , e^{\frac{2 \pi}{5}i}, e^{\frac{4 \pi}{5}i} , e^{\frac{6 \pi}{5}i}, and e^{\frac{8 \pi}{5}i}.


    so \alpha =e^{\frac{2 \pi}{5}i} and \alpha^* =e^{\frac{-2 \pi}{5}i}=e^{\frac{8 \pi}{5}i}

    so one of the quadraric factors is
    (z-e^{\frac{2 \pi}{5}i})(z-e^{\frac{8 \pi}{5}i})=z^2-(e^{\frac{8 \pi}{5}i}+e^{\frac{2 \pi}{5}i})z+1

    Now finally use Eulers formula to write the z coeffient is terms of sine and cosine.

    I hope this helps

    Brett
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  3. #3
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    The imaginary parts cancel so all coefficients are real, for the other root I used (z-\alpha^2)(z-\alpha^3) use the conjugate relation and simplified. (and the linear root is just obvious)

    Thanks for the help Brett

    Bobak
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