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Homework Help: Solve (complex) equation

  1. Jan 25, 2010 #1
    1. The problem statement, all variables and given/known data

    Solve: [tex]x^3 + 4\sqrt{1+i} = 0[/tex]

    and express in both cartesian and polar form.

    2. Relevant equations
    [tex]e^{i\theta} = \cos (\theta) + i \sin (\theta)[/tex]

    3. The attempt at a solution

    What I did was move the constant term to the right hand side and squared both sides to get: [tex]x^6 = 16 + 16 i[/tex]

    which implies: [tex]x = (16+16i)^{1/6} = \left[16\sqrt{2}\right]^{1/6} e^{\frac{(8k+1)\pi i}{6}}[/tex]

    Then I simply sub in k = 0, 1, .., 5 for all my roots. But the original equation is a polynomial of degree 3. There should be only 3 factors. Do I have to test them all to see if they work? Or is there an easier way...

  2. jcsd
  3. Jan 25, 2010 #2
    You introduced these extra roots when you squared both sides. To avoid having to substitute them into the equation, try expressing [tex]4 \sqrt{ 1 + i }[/tex] in polar coordinates.
  4. Jan 25, 2010 #3
    I don't really see how that helps as it looks fairly complicated.

    [tex](1+i)^{\frac{1}{2}} = 2^{\frac{1}{4}} e^{\frac{(8k+1)\pi i}{8}}[/tex]

    which gives us two forms:
    [tex]z_0 = 2^{\frac{1}{4}} \left(\cos \tfrac{\pi}{8} + i\sin \tfrac{\pi}{8}\right)[/tex]

    [tex]z_1 = 2^{\frac{1}{4}} \left(\cos \tfrac{9\pi}{8} + i\sin \tfrac{9\pi}{8}\right)[/tex]

    then substituting each of these back, I get two 3rd degree equations which again would give me 6 roots???
  5. Jan 25, 2010 #4


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    Science Advisor
    Homework Helper

    That's like saying solving x^3+sqrt(1)=0 means solving x^3+1=0 and x^3-1=0. Sure you get 6 roots. The 'sqrt' notation in your notation indicates only one of them. You want the 'principal value' of the square root. It's the pi/8 one.
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