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Math Help - Find the quadratic equation satisfied by the other cube roots of 2-2i

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    Find the quadratic equation satisfied by the other cube roots of 2-2i

    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
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    Re: Find the quadratic equation satisfied by the other cube roots of 2-2i

    Quote Originally Posted by spacemenon View Post
    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
    I find this question very very confused.

    (x+yi)^2=8i implies x^2-y^2=0~\&~2xy=8 or x=y=2

    Is that consistent with the rest of the post? Read the post again. The grouping symbols are mismatched.
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    Re: Find the quadratic equation satisfied by the other cube roots of 2-2i

    Sorry, u r right ,the question was wrong.
    Any idea will be appreciated ,pls do solve the prb without using de moviers theorem


    If (x+iy)^3=8i,where x and y are real nos 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: Find the quadratic equation satisfied by the other cube roots of 2-2i

    Quote Originally Posted by spacemenon View Post
    x((x^2)-3(y^2)=y((y^2)-3(x^2)=2
    I believe Plato was referring to the parenthesis, not to mention the fact that neither x(x^2) - (3y^2) nor x(x^2 - 3y^2) are equal to 2. Check the problem statement again.

    And de Moivre's theorem doesn't play into this. Just expand (x + iy)^2 and match coefficients with 8i. (Which is what Plato did.)

    -Dan
    Last edited by topsquark; June 23rd 2013 at 10:26 PM.
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