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Find roots of complex equation (1-x)^5 = x^5

  1. Sep 6, 2011 #1


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    1. The problem statement, all variables and given/known data
    Find roots of complex equation (1-x)^5 = x^5

    2. Relevant equations
    Probably Euler and/or De Moivre.

    3. The attempt at a solution
    I know i need to have z^5 on one side and all the rest on the other side. But i need 5 roots, and i'm only getting one.

    (1-x)^5 = x^5
    Use fifth root on both sides.
    (1-x) = x
    2x= 1
    x = 1/2 (only one solution, which i know is wrong, but no idea how to proceed).
  2. jcsd
  3. Sep 6, 2011 #2


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    You have the right idea here. But (like you thought), it isn't true that [itex]a^5=b^5[/itex] implies a=b. However, we can fix this:

    If [itex]a^5=b^5[/itex] (and they are nonzero), then [itex]\left(\frac{a}{b}\right)^5=1[/itex]. So a/b are fifth roots of unity. There are 5 fifth roots of unity. So let [itex]\alpha_i[/itex] be one of the fifth roots of unity, then we have


    And thus

    [tex]a=\alpha_i b[/tex]

    So, if [itex]a^5=b^5[/itex], then [itex]a=\alpha_i b[/itex] where we take [itex]\alpha_i[/itex] the fifth roots of unity. This gives us 5 equations.

    Try that on your equation.
  4. Sep 6, 2011 #3
    Take the complex root, to get:

    1 - x = \omega \, x, \; \omega^{5} = 1

    You get a linear equation in [itex]x[/itex] parametrized by the number [itex]\omega[/itex]. Solve it for [itex]x[/itex]. What are the possible values for [itex]\omega[/itex]?
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