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Math Help - Roots of unity

  1. #1
    Newbie SuumEorum's Avatar
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    Roots of unity

    z^2-2wz-1=0, where z is a complex number and w the third root of unity. I've tried substituting w=e^(i2pi/3)

    z^2-2(e^(i2pi/3))z-1=0
    z^2+2*[0.5-(i*root(3)/2)]z-1=0

    Using the quadratic equation gives me a square root of i, which I've never encountered before, any help is greatly appreciated, thanks in advance .
    Last edited by SuumEorum; January 13th 2009 at 11:50 AM.
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  2. #2
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    I havn't checked your working so far but to calculate \sqrt{i}

    note that

    \sqrt{i}=\sqrt{e^{i\frac{\pi}{2}+2ki\pi}}=e^{i\fra  c{\pi}{4}+ki\pi}=\pm \left ( \frac{1}{\sqrt 2}+i\frac{1}{\sqrt 2} \right )

    where k=0,1
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  3. #3
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    Quote Originally Posted by SuumEorum View Post
    z^2-2wz-1=0, where z is a complex number and w the third root of unity. I've tried substituting w=e^(i2pi/3)

    z^2-2(e^(i2pi/3))z-1=0
    z^2+2*[0.5-(i*root(3)/2)]z-1=0

    Using the quadratic equation gives me a square root of i, which I've never encountered before, any help is greatly appreciated, thanks in advance .
    From the quadratic formula: z = w \pm \sqrt{w^2 + 1}.

    If w = 1 the solution is simple to express.

    Otherwise note that

    w^3 = 1 \Rightarrow w^3 - 1 = 0

     \Rightarrow (w - 1)(w^2 + w + 1) = 0 \Rightarrow w^2 + w + 1 = 0

     \Rightarrow w^2 + 1 = -w

    and so z = w \pm \sqrt{-w}

    etc.
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