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Thread: complex numbers

  1. #1
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    complex numbers

    if w is a non real cube root of unity prove that (2+5w+2w^2)^6 =729

    i am not really sure how to do this
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  2. #2
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    Re: complex numbers

    $w$ is a cube root of unity so we know it's magnitude is $1$

    Thus we can write $w = e^{i \theta}$

    let $f(w) = 2 + 5w + 2w^2$

    a little work shows that

    $f(e^{i\theta}) = (5+4\cos(\theta))e^{i \theta}$

    and thus

    $\left(f(w)\right)^6 = (5+4\cos(\theta))^6 e^{6i \theta}= 729$

    and as $ (5+4\cos(\theta))^6 \in \mathbb{R}$ it must be that

    $(5+4\cos(\theta))^6 = 729$

    $5+4\cos(\theta) = 3$

    $\cos(\theta) = -\dfrac 1 2$

    $\theta = \dfrac{2\pi}{3},~\dfrac{4\pi}{3}$

    $w = e^{i2\pi/3},~e^{i 4\pi/3}$

    and these are exactly the two non-real cube roots of unity.

    Actually you'll need to reverse these steps as you are given $w$ is a cube root of unity and must show that $f(w)^6 = 729$

    It pretty much just amounts to plugging in the two non-real cube roots of unity and grinding through the algebra using the one shortcut I noted.
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    Re: complex numbers

    I am confused by tour terminology. could you do it in terms of cos(a) and isin(a)
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    Re: complex numbers

    Quote Originally Posted by edwardkiely View Post
    if w is a non real cube root of unity prove that (2+5w+2w^2)^6 =729
    Quote Originally Posted by edwardkiely View Post
    I am confused by tour terminology. could you do it in terms of cos(a) and isin(a)
    There are two non-real cube roots of unity: $\omega=-\frac{1}{2}+\bf{i}\frac{\sqrt3}{2}$ is one of them.
    Now $\omega^2=-\frac{1}{2}-\bf{i}\frac{\sqrt3}{2}$ Can you show that?

    You calculate $2+5\omega +2\omega^2$. If that is $3$ you are done because $3^6=729$.
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    Re: complex numbers

    \omega ^3-1=(\omega -1)\left(\omega ^2+\omega +1\right)

    and since \omega \neq 1 we get \omega ^2+\omega +1=0

    so that 2\omega ^2+2\omega +2=0 and 2+5\omega +2\omega ^2=3\omega

    \left(2+5\omega +2\omega ^2\right)^6=3^6\omega ^6=729
    Thanks from romsek and topsquark
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  6. #6
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    Re: complex numbers

    how do you know w=-.5 +i(√3/2) there are many complex numbers with a magnitude of 1 . and i can do the rest.
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  7. #7
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    Re: complex numbers

    Quote Originally Posted by edwardkiely View Post
    if w is a non real cube root of unity prove that (2+5w+2w^2)^6 =729
    i am not really sure how to do this
    Quote Originally Posted by edwardkiely View Post
    how do you know w=-.5 +i(√3/2) there are many complex numbers with a magnitude of 1 . and i can do the rest.
    That was given.

    There are exactly three cube roots of unity, two of which are non-real:
    $\large{\exp \left( {\dfrac{{2\pi\bf{i} }}{{\sqrt 3 }}} \right) = \cos \left( {\dfrac{{2\pi }}{{\sqrt 3 }}} \right) \pm\bf{i} \sin \left( {\dfrac{{2\pi }}{{\sqrt 3 }}} \right) = \dfrac{1}{2} \pm \bf{i}\dfrac{{\sqrt 3 }}{2}}$

    Those are the only two non-real.
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  8. #8
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    Re: complex numbers

    Quote Originally Posted by edwardkiely View Post
    how do you know w=-.5 +i(√3/2) there are many complex numbers with a magnitude of 1 . and i can do the rest.
    Idea's answer is far more elegant and most likely what they wanted you to come up with.
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