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Math Help - Cube root of a complex number

  1. #1
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    Cube root of a complex number

    Just trying to confirm something if possible:

    a) Complex numbers have 3 cube roots.

    b) The usual method for finding them is to write in polar form, then add 2 pi and 4 pi to the angle and use De Moivre's theorem on the resulting three expressions.

    Cheers
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  2. #2
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    Quote Originally Posted by DangerousDave View Post
    Just trying to confirm something if possible:

    a) Complex numbers have 3 cube roots.

    b) The usual method for finding them is to write in polar form, then add 2 pi and 4 pi to the angle and use De Moivre's theorem on the resulting three expressions.

    Cheers
    You are on the right track. They do have cube roots, find them using

    z^{\frac{1}{n}} = r^{\frac{1}{n}}\times \text{cis}\left( \frac{\theta+2k\pi}{n}\right), k= 0,1,\dots,n-1
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