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Math Help - Solve in Complex Numbers

  1. #1
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    Solve in Complex Numbers

    Will you please help me solve this equation.

    Solve this equation in the complex number system:x^3-8=0
    I have only a real number answer, which is 2. How to solve it in complex numbers?

    Thank you
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  2. #2
    MHF Contributor red_dog's Avatar
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    x^3-8=0\Leftrightarrow(x-2)(x^2+2x+4)=0
    x_1=2
    x^2+2x+4=0\Rightarrow\triangle=-12\Rightarrow x_{2,3}=\displaystyle\frac{-2\pm 2i\sqrt{3}}{2}=-1\pm i\sqrt{3}
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  3. #3
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    Thank you very much
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  4. #4
    Eater of Worlds
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    You could take the more 'complex' route and use:

    w=-2\left(cos(\frac{\pi}{3}+\frac{2k{\pi}}{3})+isin(\  frac{\pi}{3}+\frac{-2k{\pi}}{3})\right), k=0,1,2

    Let k=0 and get w=-1-\sqrt{3}i

    k=1: w=2

    k=2: w=-1+\sqrt{3}i
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