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Math Help - Complex Roots

  1. #1
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    Complex Roots

    the first part is straightforward. for the roots i get  \alpha^0 , \alpha^1 , \alpha^2 , \alpha^3 , \alpha^4 , \alpha^5 , \alpha^6.

    For the quadratic I guessed got the root as the complex conjugate which is \alpha^{-1} + \alpha^{-2} \alpha^{-4}

    i get that A= - 2 \left( \cos \frac{2 \pi}{7} +  \cos \frac{4 \pi}{7} +  \cos \frac{8 \pi}{7} \right)

    and B is the product of the roots, but I am having trouble showing the required results.
    Attached Thumbnails Attached Thumbnails Complex Roots-picture-1.png  
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  2. #2
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    If \alpha + \alpha^2 + \alpha^4 is a root then \bar \alpha + (\bar \alpha)^2 + (\bar \alpha)^4 , i.e. conjugate is a root. Note, \alpha \bar \alpha = |\alpha|^2 \implies \alpha \bar \alpha = 1, thus, \bar \alpha = \alpha^{-1}. But since \alpha \cdot \alpha^6 = 1 \implies \alpha^{-1} = \alpha^6. Thus, \bar \alpha = \alpha^6. This immediately implies, ( \bar \alpha)^2 = \alpha^5 and ( \bar \alpha)^4 = \alpha^3. Thus, \alpha^3 + \alpha^5 + \alpha^6 is the other root.

    This means (by Viete) that A = - (\alpha + \alpha^2 + \alpha^4 + \alpha^3+\alpha^5+\alpha^6) = 1. Because we are using the fact that the sum of the roots of unity is zero, i.e. 1+\alpha+...+\alpha^6 = 0.

    Now the product is, (\alpha + \alpha^2 + \alpha^4)(\alpha^3 + \alpha^5 + \alpha^6) = (\alpha + \alpha^2 + ... + \alpha^7) + 2\alpha^7 = 2. Here we would the fact that \alpha^8 = \alpha,\alpha^9 = \alpha^2,\alpha^{10} = \alpha^3. Thus, we have that B=2.

    This means that: z_1 = \alpha+\alpha^2+\alpha^4 and z_2 = \alpha^3+\alpha^5+\alpha^6 solve z^2 + z + 2 = 0.

    Now, -\frac{1}{2}+ i \frac{\sqrt{7}}{2}, - \frac{1}{2}-i\frac{\sqrt{7}}{2} are its solutions. Thus, they are actually z_1,z_2 (not necessarily in that order).

    This means, \sin \frac{2\pi}{7}+\sin \frac{4\pi}{7}+\sin\frac{8\pi}{7} = \pm \frac{\sqrt{7}}{2}.
    You need to determine the sign. That is what the problem says.
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  3. #3
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    I followed that really well, Thank you TPH.

    I get the sign as positive. as \sin\frac{8\pi}{7} = - \sin\frac{\pi}{7}

    and \sin\frac{2\pi}{7} > \sin\frac{\pi}{7}

    \Rightarrow \sin\frac{2\pi}{7} - \sin\frac{\pi}{7} >  0

    \sin\frac{4\pi}{7} > 0 as it is second quadrant

    therefore \sin\frac{2\pi}{7} + \sin\frac{4\pi}{7}  - \sin\frac{\pi}{7} >  0

    that good enough?
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  4. #4
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    Quote Originally Posted by bobak View Post
    that good enough?
    Looks good.
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  5. #5
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    I am just modifying this question of prove your current signature TPH.

    we have \sin\frac{2\pi}{7} + \sin\frac{4\pi}{7}  + \sin\frac{8\pi}{7} = \frac{\sqrt{7}}{2}

    \Rightarrow  2 \sin\frac{2\pi}{7} + 2 \sin\frac{4\pi}{7}  + 2 \sin\frac{8\pi}{7} = \sqrt{7}

    we will use \sin\frac{2\pi}{7} = \sin\frac{5\pi}{7}
    \sin\frac{4\pi}{7} = \sin\frac{3\pi}{7}
    and \sin\frac{8\pi}{7} = -\sin\frac{\pi}{7} = -\sin\frac{6\pi}{7}

    using all of that together we have proved your signature

    -\sin \frac{\pi}{7} + \sin \frac{2\pi}{7} + \sin \frac{3\pi}{7} + \sin \frac{4\pi}{7}+\sin \frac{5\pi}{7} - \sin \frac{6\pi}{7} = \sqrt{7}
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  6. #6
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    How about this?
    \sin \frac{\pi}{11} + \sin\frac{2\pi}{11}+\sin \frac{3\pi}{11} -\sin \frac{4\pi}{11}+\sin\frac{5\pi}{11}+\sin\frac{6\pi  }{11}-\sin \frac{7\pi}{11}+\sin\frac{8\pi}{11}+\sin\frac{9\pi  }{11}+\sin\frac{10\pi}{11} =\sqrt{11}.

    How about this?
    \sin \frac{\pi}{11}\sin \frac{2\pi}{11}\sin \frac{3\pi}{11}\sin\frac{4\pi}{11}\sin\frac{5\pi}{ 11}\sin \frac{6\pi}{11}\sin\frac{7\pi}{11}\sin \frac{8\pi}{11}\sin \frac{9\pi}{11}\sin \frac{10\pi}{11} = \frac{11}{2^{10}}.

    By the way. Identity #2 generalizes.
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  7. #7
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    the Gauss Sum stuff your referring to looks like it is beyond me (for now).

    by generalises do you mean \prod_{k=1}^{n-1}{\sin \left( \frac{k\pi }{n} \right)\; =\; \frac{n}{2^{n-1}}}
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  8. #8
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    Quote Originally Posted by bobak View Post
    by generalises do you mean \prod_{k=1}^{n-1}{\sin \left( \frac{k\pi }{n} \right)\; =\; \frac{n}{2^{n-1}}}
    Exactly. It happens to be not hard at all to derive (but still impressive). If you are familar with roots of unity. Here.
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