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Complex roots of a cubic equation

  1. Dec 3, 2011 #1
    1. The problem statement, all variables and given/known data

    Hi all,

    I was wondering if there is a procedure you can follow to calculate the complex roots of a cubic equation.


    2. Relevant equations

    For example the equation

    x3 - 1 = 0

    has roots of x = 1
    x = -0.5 + √3/2 i
    x = -0.5 - √3/2 i

    Admittedly, I got those solutions off wolfram alpha, but I am wondering how to work it out without wolfram!

    Thanks!


    3. The attempt at a solution
     
  2. jcsd
  3. Dec 3, 2011 #2

    micromass

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    In this case, it's not so hard. If you have a complex number z, then you first need to write it in the form

    [tex]z=R(\cos(\theta)+i\sin(\theta))[/tex]

    It is now a theorem (prove this!!), that the n-th roots are exactly

    [tex]\sqrt[n]{R}(\cos(\frac{\theta+2k\pi}{n})+ i\sin(\frac{\theta+2k\pi}{n}))^n[/tex]

    for [itex]0\leq k<n[/itex]. This is due to De Moivre's identity.

    So, can you use this information to calculate the third roots of 1?
     
  4. Dec 3, 2011 #3
    Haha, great. The formula works!

    Thanks very much micromass!
     
  5. Dec 3, 2011 #4

    HallsofIvy

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    Here's another way to do it (not as "sophisticated"): [itex]x^3- 1= 0[/itex] has the obvious solution x= 1 so x- 1 is a factor. Dividing [itex]x^3- 1[/itex] by x- 1, we find that [itex]x^3- 1= (x- 1)(x^2+ x+ 1)[/itex]. If x is not 0 then [itex]x^2+ x+ 1= 0[/itex]. That's a quadratic equation so use the quadratic formula to solve it.
     
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