1. Not finding help here? Sign up for a free 30min 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!

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...

    Thanks.
     
  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

    Dick

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




Similar Discussions: Solve (complex) equation
Loading...