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Thread: cube roots of unity

  1. #1
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    cube roots of unity

    Hi folks,

    there is a very well worn method of finding the cube roots of unity, but there are always more than one way of doing things, so I tried a few, hoping that all roads lead to the mathematical rome.

    $z^3 = 1$


    $(x + iy)^3 = 1$


    $x^3 - 3xy^{2} + 3x^{2}yi -iy^{3} = 1$


    $(x^3 - 3xy{^2}) + i(3x{^2}y - y^3) = 1$


    so


    $x^3 - 3xy^{2} = 1$ ...... (1)




    $3x^{2}y - y^3 = 0$ .......(2)


    from (2)


    $y ( y^2 - 3x^2) = 0$


    $ y^2 = 3x^2$


    from (1)


    $x ( x^2 - 3y^2) = 1$


    $x = 1$ or $x^2 = 3y^2 + 1$


    $x^2 = 9x^2 + 1$


    $-8x^2 = 1$


    this is not going well, but I can't see the flaw in it. Where am I going wrong?
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  2. #2
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    Re: cube roots of unity

    Quote Originally Posted by s_ingram View Post
    from (1)


    $x ( x^2 - 3y^2) = 1$


    $x = 1$ or $x^2 = 3y^2 + 1$


    $x^2 = 9x^2 + 1$


    $-8x^2 = 1$


    this is not going well, but I can't see the flaw in it. Where am I going wrong?
    $x(x^2-3y^2) = 1$

    From (2), you found $y=0$ or $y^2=3x^2$.

    If $y=0$, then you have $x^3=1$, which means $x=1$ since $x\in \mathbb{R}$.
    If $y^2=3x^2$, then you have:
    $x(x^2-9x^2) = 1$
    $-8x^3 = 1$
    $x^3 = -\dfrac{1}{8}$
    $x = -\dfrac{1}{2}$

    Plugging back in, we have $y^2 = 3x^2 = \dfrac{3}{4}$ so $y = \pm \dfrac{\sqrt{3}}{2}$

    The problem with your method is that $ab=1$ does NOT imply $a=1$ or $b=1$. Example: $2\cdot \dfrac{1}{2} = 1$ and neither factor is 1. You were confusing this with $ab=0$. This implies $a=0$ or $b=0$.
    Last edited by SlipEternal; Mar 30th 2018 at 04:56 AM.
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  3. #3
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    Re: cube roots of unity

    Quote Originally Posted by s_ingram View Post
    Hi folks,

    there is a very well worn method of finding the cube roots of unity, but there are always more than one way of doing things, so I tried a few, hoping that all roads lead to the mathematical rome.

    $z^3 = 1$


    $(x + iy)^3 = 1$


    $x^3 - 3xy^{2} + 3x^{2}yi -iy^{3} = 1$


    $(x^3 - 3xy{^2}) + i(3x{^2}y - y^3) = 1$


    so


    $x^3 - 3xy^{2} = 1$ ...... (1)




    $3x^{2}y - y^3 = 0$ .......(2)


    from (2)


    $y ( y^2 - 3x^2) = 0$


    $ y^2 = 3x^2$
    Actually, either $y^2= 3x^2$ or y= 0. y= 0 reduces (1) to $x^2= 1$ which gives x= 1.


    $from (1)


    $x ( x^2 - 3y^2) = 1$


    $x = 1$ or $x^2 = 3y^2 + 1$
    NO!! The "zero property", that if xy= 0 either x= 0 or y= 0, is special to 0. It is NOT true that if xy= a then either x= a or y= a for any a other than 0.

    $x^2 = 9x^2 + 1$


    $-8x^2 = 1$


    this is not going well, but I can't see the flaw in it. Where am I going wrong?
    Thanks from s_ingram
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  4. #4
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    Re: cube roots of unity

    thanks guys.
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  5. #5
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    Re: cube roots of unity

    Quote Originally Posted by HallsofIvy View Post
    Actually, either $y^2= 3x^2$ or y= 0. y= 0 reduces (1) to $x^2= 1$ which gives x= 1.
    That should be "reduces to $x^3 = 1$."
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  6. #6
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    Re: cube roots of unity

    yes, I assumed it was just a typo. Also, after all the help I got, it seemed kind of rude to be so picky. But you guys all all mathematicians, precision really counts. In future I will be more rigerous!
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  7. #7
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    Re: cube roots of unity

    No, not rude at all, just a helpful correction.

    (Would it be rude of me to point out that "rigorous" is not spelled with a "e"?)
    Thanks from topsquark
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  8. #8
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    Re: cube roots of unity

    Ah... you see! You got me again!
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