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Math Help - Roots of an equation

  1. #1
    MHF Contributor
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    Roots of an equation

    \varphi'''-\lambda\varphi=0

    m^3-\lambda=0\Rightarrow m^3=\lambda

    My book then does this:

    \displaystyle m=s, \  m=\left(\frac{-1\pm i\sqrt{3}}{2}\right)s,\ \ s=\lambda^{1/3}

    How were the roots of m determined?
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  2. #2
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    \displaystyle w_k=r^{1/n}\left[\cos\left(\frac{0+2\pi k}{n}\right)+i\sin\left(\frac{0+2\pi k}{n}\right)\right]

    r=\lambda^{1/3}=s

    Solved.
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  3. #3
    A Plied Mathematician
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    Right. Just the third roots of unity.
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