# Thread: Cube root of complex number pls help

1. ## 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

2. ## Re: Cube root of complex number pls help

Originally Posted by spacemenon
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.

3. ## Re: Cube root of complex number pls help

Thank you very very much