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    By considering the roots of z^5 = 1 , show that cos(2pi/5) + cos(4pi/5) + cos(6pi/5) + cos(8pi/5) = -1

    I found the solutions of z^5 = 1 in polar form but what am I supposed to do next?
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    (Original post by thers)
    By considering the roots of z^5 = 1 , show that cos(2pi/5) + cos(4pi/5) + cos(6pi/5) + cos(8pi/5) = -1

    I found the solutions of z^5 = 1 in polar form but what am I supposed to do next?
    The sum of the 5th roots of unity is zero. What is the real part of this sum?
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    Ok thanks I've done it now. I'm also stuck on:

    Solve 1 - 2z + 4z^2 + 8z^3 = 0 I don't even know how to start.
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    (Original post by thers)
    Ok thanks I've done it now. I'm also stuck on:

    Solve 1 - 2z + 4z^2 + 8z^3 = 0 I don't even know how to start.
    Is the last term supposed to be -8z^3 instead of '+'? If not, then I expect the solution will be very nasty (and possibly unattainable without some sort of very advanced numerical method).

    Assuming you have made the above typo; spot the GP.
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    does the question ask for (assuming it is +8) the real root or all 3 roots?
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    I looked at the answers in the back of the textbook and it was 1/2, +1/2i, -1/2i so the textbook must have made a typo when writing the question. Solved it now.
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    yup - those are the factors for -8z^3....
 
 
 
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