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    complex nos finding cube root

    If (x+iy)^2=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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    Re: complex nos finding cube root

    complex nos finding cube root-24-jun-13-2.png the other part you can do yourself
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    Re: complex nos finding cube root

    x^2 - y^2 = 0 \implies y = \pm x

    So
    2xy = 8 \implies \pm 2x^2 = 8 \implies x = \pm 2
    so (x, y) = ( \pm 2, \pm 2)

    Neither of the remaining two equations share this point.

    -Dan
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    Re: complex nos finding cube root

    sorry the the question is (x+iy)^3=8i
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    Forum Admin topsquark's Avatar
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    Re: complex nos finding cube root

    Quote Originally Posted by spacemenon View Post
    sorry the the question is (x+iy)^3=8i
    The process is the same. Expand the (x + iy)^3 and match coefficients. Give it a try and tell us how it works out.

    -Dan
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