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    (Original post by Mihael_Keehl)
    I was talking in general for a polynomial with a high degree, not in this particular case, sorry ifI was vague
    in general, no

    these are not complex conjugates
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    (Original post by TeeEm)
    in general, no

    these are not complex conjugates
    I see thanks.
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    (Original post by B_9710)
    I do have the answers. With the general solution being x=tan(nπ/7 + π/28)
    I was just seeing the different ways that people suggest to go about this question
    So, here's my solution given the hint above:

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    Observe that if t = \tan \theta, then \arg(1+it) = \theta and so \arg(1+it)^7 = 7\theta. Since tan(arg(x+iy)) = y / x we can deduce Im((1+it)^7) / Re(1+it)^7 = \tan(7\theta)
    That is, \dfrac{7t - 35t^3 + 21t^5-t^7}{1-21t^2+35t^4-7t^6} = \tan 7\theta
    Comparing with our original equation we see we need 7t - 35t^3 + 21t^5-t^7 = 1-21t^2+35t^4-7t^6, so \tan 7 \theta = 1 and so t = (arctan(1) + n\pi) / 7 = \pi/28 + n\pi /7 as desired.
 
 
 
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Updated: October 26, 2015
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