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Solving Complex Numbers

  1. Dec 29, 2008 #1
    I hope, I've posted this question in the right section.

    1. The problem statement, all variables and given/known data
    Solve the fooling equation over C

    z^3+ 8 = 0


    3. The attempt at a solution

    First Attempt
    z^3 = -8
    cube root (2 ^3) = cube root (8 i^2 )
    z = 2i


    Second Attempt
    z^3 = -8
    z ^3 = -2 ^3
    so, z = -2
     
  2. jcsd
  3. Dec 29, 2008 #2
    Or you could convert -8 into polar form, then using De Moivre's theorem get all three cube roots.
     
  4. Dec 29, 2008 #3
    We're learning Polar Form next year and I have never heard of De Moivre's Theorem.

    So, is my second attempt incorrect?
     
  5. Dec 29, 2008 #4

    gabbagabbahey

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    Homework Helper
    Gold Member

    Your first attempt is incorrect; [itex](2i)^3=-8i\neq -8[/itex]

    Your second attempt is not incorrect but it is incomplete: [itex]z^3+8=0[/itex] is a 3rd degree polynomial equation; so it must have three roots. You have correctly found one root [itex]z=-2[/itex], but you still need to find the other two.

    One method is to divide your polynomial [itex]z^3+8[/itex] by [itex]z+2[/itex] (Since z=-2 is a root, you know (z+2) must be a factor of the polynomial) which will leave you with a quadratic that you can solve to find your other two roots.
     
  6. Dec 29, 2008 #5
    thanks
    I've got it now.
     
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