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Math Help - Cube root of complex number pls help

  1. #1
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    Cube root of complex number pls help

    If (x+iy)^3=8i,prove that x((x^2)-3(y^2))=y((y^2)-3(x^2))=2,show that these equations have one solution in which x=y.And hence find one of the cube root of 2-2i.Find the quadratic equation satisfied by the other cube roots of 2-2i



    ans -i-i;z^2-(i+i)z+2i=0;0.5(1+sqrt 3)+0.5(1-sqrt 3)i,0.5(1-sqrt 3)+0.5(1+sqrt 3)i
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  2. #2
    MHF Contributor ebaines's Avatar
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    Re: Cube root of complex number pls help

    Quote Originally Posted by spacemenon View Post
    If (x+iy)^3=8i....
    I believe you have a typo - what you meant to write is this: "If (x+iy)^3=2-2i ...", correct?

    Write out the cube of (x+iy) and set it equal to 2-2i, and you get

    (x^3 - 3xy^2) + (3x^2y - y^3)i = 2 - 2i, hence x^3-3xy^2 = y^3-3x^2y = 2.

    Next if x = y, then this becomes x^3-3x^3 = x^3-3x^3p, or -2x^3 = -2x^3, which is certainly true.

    To find one solution solve -2x^3 = 2. Then since x=y one cube root is x + xi.

    To find the remaining quadratic divide the original equation of z^3-(2 - 2i)=0 by the one root you already know: (z-x-xi). You can do this by long division.
    Last edited by ebaines; June 24th 2013 at 01:28 PM.
    Thanks from topsquark and spacemenon
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    Re: Cube root of complex number pls help

    Thank you very very much
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