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     z^3 = 1

    You are given that w is a non-real root of the above equation, show that:

    1+w+w^2 =0

    So I go:

    3 roots are 1, e^{\pm \frac{2ki\pi}{3}}

    And

    z^3 - 1=0

    Thus the sum of the roots must equal the coefficient of the z^2 which is 0 and so:
    1+w+w^2 =0


    Is that good enough ? Any other ways ?
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    I assume you mean 1 + w + w^2 = 0?

    The sum of the roots is acceptable I believe, but only if you have proved that 1, w and w^2 are the roots already. (which you did by saying what the 3 roots are, but you might have to do a little more to show that they are in fact the 3 roots of the equation)

    Another way is to factorise z^3-1=0
    into (z-1)(z^2+z+1) = 0 (you can figure this out the first time through polynomial division and then I think you can quote the result, at least you can in my system)
    But w is non-real so w=/=1 so 1 + w + w^2 = 0
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    (Original post by Aeonstorm)
    I assume you mean 1 + w + w^2 = 0?

    The sum of the roots is acceptable I believe, but only if you have proved that 1, w and w^2 are the roots already. (which you did by saying what the 3 roots are, but you might have to do a little more to show that they are in fact the 3 roots of the equation)

    Another way is to factorise z^3-1=0
    into (z-1)(z^2+z+1) = 0 (you can figure this out the first time through polynomial division and then I think you can quote the result, at least you can in my system)
    But w is non-real so w=/=1 so 1 + w + w^2 = 0
    Ah yea, After finding the 3 roots I would let one non-real root be w and then show that w^6 = (w^2)^3 = 1^6 = 1 and so w^2 is also a root and then proceed as I did above.

    I probably wouldn't have spotted your method and it would take too long for me to divide the thing by z-1
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    (Original post by member910132)
    Ah yea, After finding the 3 roots I would let one non-real root be w and then show that w^6 = (w^2)^3 = 1^6 = 1 and so w^2 is also a root and then proceed as I did above.

    I probably wouldn't have spotted your method and it would take too long for me to divide the thing by z-1
    I recommend you ask your teacher whether you are allowed to quote the general result:

    z^n -1 = (z-1)(z^(n-1) + z^(n-2) + z^(n-3) + ... + 1)

    rather than actually deriving it each time. If you are allowed to, it is a slight bit quicker than showing that the roots are 1, w and w^2 and then using the sum of the roots. If not, your method is perfectly valid.
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    (Original post by Aeonstorm)
    I recommend you ask your teacher whether you are allowed to quote the general result:

    z^n -1 = (z-1)(z^(n-1) + z^(n-2) + z^(n-3) + ... + 1)

    rather than actually deriving it each time. If you are allowed to, it is a slight bit quicker than showing that the roots are 1, w and w^2 and then using the sum of the roots. If not, your method is perfectly valid.
    Self teaching FM lol, that's why I post about 3-4 questions in here a day.
  6. Online

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    (Original post by member910132)
    Self teaching FM lol, that's why I post about 3-4 questions in here a day.
    I also used this forum regularly when I self-taught FM. The combination of my textbooks and this forum was all I really needed.

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