1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Sep 6, 2011 #1

    sharks

    User Avatar
    Gold Member

    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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

    [tex]\frac{a}{b}=\alpha_i[/tex]

    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:

    [tex]
    1 - x = \omega \, x, \; \omega^{5} = 1
    [/tex]

    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]?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook