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Thread: Deduce result by solving equation?

  1. #1
    Super Member fardeen_gen's Avatar
    Jun 2008

    Deduce result by solving equation?

    Solve the equation $\displaystyle Z^7 + 1 = 0$ to deduce that $\displaystyle \cos \frac{\pi}{7}\cos\frac{3\pi}{7}\cos\frac{5\pi}{7} = -\frac{1}{8}$
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  2. #2
    Senior Member
    Apr 2009
    Atlanta, GA

    Getting started...

    Well, if $\displaystyle z^7=-1$, then $\displaystyle e^{7\theta i}=e^{\pi i}$, so $\displaystyle 7\theta=\pi$, for $\displaystyle \theta\in[0,2\pi)$.

    So $\displaystyle z=e^{i\theta}$ for $\displaystyle \theta=\frac\pi7,\frac{3\pi}7,\frac{5\pi}7,\pi,\fr ac{9\pi}7,\frac{11\pi}7,\frac{13\pi}7$

    But why would $\displaystyle Re[z_1]Re[z_2]Re[z_3]=-\frac18$?
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  3. #3
    MHF Contributor
    Opalg's Avatar
    Aug 2007
    Leeds, UK
    Writing Media_Man's answer in a slightly different form, the solutions of $\displaystyle z^7+1=0$ are $\displaystyle z=-1,\:e^{\pm \pi i/7},\:e^{\pm3\pi i/7},\:e^{\pm5\pi i/7}.$ Therefore $\displaystyle z^7+1 = (z+1)(z-e^{\pi i/7})(z-e^{-\pi i/7})(z-e^{3\pi i/7}) (z-e^{-3\pi i/7})(z-e^{5\pi i/7})(z-e^{-5\pi i/7}).$ But $\displaystyle (z-e^{\pi i/7})(z-e^{-\pi i/7}) = z^2-2z\cos\tfrac\pi7+1$, and similarly for the other two complex conjugate pairs. Also, $\displaystyle z^7+1 = (z+1)(z^6-z^5+z^4-z^3+z^2-z+1)$. It follows that

    $\displaystyle (z^2-2z\cos\tfrac{\pi}7+1) (z^2-2z\cos\tfrac{3\pi}7+1) (z^2-2z\cos\tfrac{5\pi}7+1) = z^6-z^5+z^4-z^3+z^2-z+1.$

    Compare coefficients of z to see that $\displaystyle -2\bigl(\cos\tfrac{\pi}7 + \cos\tfrac{3\pi}7 + \cos\tfrac{5\pi}7\bigr) = -1.$ Then compare coefficients of $\displaystyle z^3$ to see that $\displaystyle -4\bigl(\cos\tfrac{\pi}7 + \cos\tfrac{3\pi}7 + \cos\tfrac{5\pi}7\bigr) -8\cos\tfrac{\pi}7\cos\tfrac{3\pi}7\cos\tfrac{5\pi} 7 = -1.$ From those last two equations you find that $\displaystyle \cos\tfrac{\pi}7\cos\tfrac{3\pi}7\cos\tfrac{5\pi}7 = -1/8.$
    Last edited by Opalg; May 26th 2009 at 06:39 AM. Reason: loads of typos
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