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Math Help - Complex numbers and trigonometric result?

  1. #1
    Super Member fardeen_gen's Avatar
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    Complex numbers and trigonometric result?

    By using the concepts of complex numbers, evaluate \sum_{k = 1}^{3} \cos \frac{2k\pi}{7}
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  2. #2
    MHF Contributor red_dog's Avatar
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    Let z=\cos\frac{\pi}{7}+i\sin\frac{\pi}{7}\Rightarrow z^7=-1\Rightarrow(z+1)(z^6-z^5+z^4-z^3+z^2-z+1)=0

    We have: \cos\frac{2\pi}{7}=\frac{1}{2}\left(z^2+\frac{1}{z  ^2}\right)=\frac{z^4+1}{2z^2}

    \cos\frac{4\pi}{7}=\frac{z^8+1}{2z^4}=\frac{1-z}{2z^4}

    \cos\frac{6\pi}{7}=\frac{z^{12}+1}{2z^6}=\frac{1-z^5}{2z^6}

    Then \frac{z^4+1}{2z^2}+\frac{1-z}{2z^4}+\frac{1-z^5}{2z^6}=\frac{-z^5+z^4-z^3+z^2-z+1}{2z^6}=\frac{-z^6}{2z^6}=-\frac{1}{2}
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