# Cube root of complex number pls help

• Jun 24th 2013, 04:20 AM
spacemenon
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(Thinking)(Thinking)(Thinking)(Thinking)(Thinki ng)(Thinking)(Thinking)
• Jun 24th 2013, 01:25 PM
ebaines
Re: Cube root of complex number pls help
Quote:

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

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

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

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

To find the remaining quadratic divide the original equation of \$\displaystyle z^3-(2 - 2i)=0\$ by the one root you already know: (z-x-xi). You can do this by long division.
• Jun 25th 2013, 07:10 AM
spacemenon
Re: Cube root of complex number pls help
Thank you very very much