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Thread: Complex - six roots of unity :S

  1. #1
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    Complex - six roots of unity :S

    I managed to get to the point where there is an infinite range of values $\displaystyle theta + 2n pi $ (Problem & Ans Attached). I am confused how he put it in exp form.
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  2. #2
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    Quote Originally Posted by trojsi View Post
    I managed to get to the point where there is an infinite range of values $\displaystyle theta + 2n pi $ (Problem & Ans Attached). I am confused how he put it in exp form.
    I think you need to be more specific as to where you are at.

    You should have
    $\displaystyle
    z^6 = 1 = \cos (2k\pi) + i \sin (2k\pi)
    $

    By deMoivre's theorem:

    $\displaystyle z = \cos \frac{k\pi}{3} + i \sin \frac{k\pi}{3} $

    You do not have an infinite range, because you are given the conditions that $\displaystyle -\pi < \phi \leq \pi$
    So you want to pick the values k for which the angle lies between $\displaystyle -\pi$ and $\displaystyle \pi $

    These values of k are: $\displaystyle 0, \pm 1, \pm 2 , 3$ (Note: it's 3, not -3 because your range includes $\displaystyle \pi$, but not $\displaystyle -\pi$)

    To put it into exponential form as required, you'll need euler's formula:

    $\displaystyle e^{i\theta} = \cos \theta + i \sin \theta $
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  3. #3
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    Many thanks

    Perfect explanation... i understood it right away. I need to brush up on the theorem .. i bet maths is only abstract when not knowing the proofs and the whole story behind them.
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  4. #4
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    The other way is to remember that all the roots are evenly spaced around a circle. So if you can figure out the angle between them, you can keep adding this angle and find all the roots.

    In your case, the real roots are easy ($\displaystyle \pm 1$)...
    For the complex roots...

    $\displaystyle z^6 = 1$

    $\displaystyle z^6 = \cos{2\pi} + i\sin{2\pi}$

    $\displaystyle z = \cos{\frac{2\pi}{6}} + i\sin{\frac{2\pi}{6}}$

    $\displaystyle z = \cos{\frac{\pi}{3}} + i\sin{\frac{\pi}{3}}$.


    So all roots differ by an angle of $\displaystyle \frac{\pi}{3}$. From here we can say the roots are:

    $\displaystyle \cos{\frac{\pi}{3}} + i\sin{\frac{\pi}{3}} = \frac{1+i\sqrt{3}}{2}$

    $\displaystyle \cos{\frac{2\pi}{3}} + i\sin{\frac{2\pi}{3}} = \frac{-1+i\sqrt{3}}{2}$

    $\displaystyle \cos{\frac{3\pi}{3}} + i\sin{\frac{3\pi}{3}} = -1$

    $\displaystyle \cos{\frac{4\pi}{3}} + i\sin{\frac{4\pi}{3}} = \frac{-1-i\sqrt{3}}{2}$

    $\displaystyle \cos{\frac{5\pi}{3}} + i\sin{\frac{5\pi}{3}} = \frac{1-i\sqrt{3}}{2}$

    $\displaystyle \cos{\frac{6\pi}{3}} + i\sin{\frac{6\pi}{3}} = 1$.
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  5. #5
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    ohoho lovely sir... I really needed that pattern of the root .. things are crystal clear now.

    thanks
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