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How to find the third root of z^3=1?

  1. May 13, 2016 #1
    1. The problem statement, all variables and given/known data
    in a given activity: solve for z in C the equation: z^3=1
    2. Relevant equations
    prove that the roots are 1, i, and i^2

    3. The attempt at a solution
    using z^3-1=0 <=> Z^3-1^3 == a^3-b^3=(a-b)(a^2+2ab+b^2)
    it's clear the solution are 1 and i^2=-1 but i didn't find "i" as a solution by using this method
    i need your help please
     
  2. jcsd
  3. May 13, 2016 #2

    jedishrfu

    Staff: Mentor

    Are you sure the equation is right or the roots are right?

    1 is a root but i and i^2 aren't

    When you plug them in to z^3 = 1 you should get 1 each time but i gives -i = 1 and i^2 gives -1 = 1 right?
     
  4. May 13, 2016 #3

    SammyS

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    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Hello hamad12a. Welcome to PF!

    You will see the following request in many threads here at PF, especially in those started by relative new-comers.

    Please state the entire problem as it was given to you.
     
  5. May 13, 2016 #4
    the mistake was in applying the formulae a^3-b^3 which equals to :

    • a3 – b3 = (ab)(a2 + ab + b2)
    i wrote 2ab instead of ab only without myltiplying it by 2
    so, again the roots are: -1/2+sqrt(3)/2*i and -1/2-sqrt(3)/2

    by using the D=b^2-4ac for the quadratic one and a=b for the first term which is between brackets
     
  6. May 13, 2016 #5

    jedishrfu

    Staff: Mentor

    Another way of finding these roots in this case is DeMoivre's formula:

    https://en.wikipedia.org/wiki/De_Moivre's_formula

    and polar symmetry of them being spaced 120 degrees apart with the same modulus of 1.
     
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