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

• Jun 23rd 2013, 06:42 AM
spacemenon
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
• Jun 23rd 2013, 08:22 AM
Plato
Re: Find the quadratic equation satisfied by the other cube roots of 2-2i
Quote:

Originally Posted by spacemenon
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.
• Jun 23rd 2013, 04:45 PM
spacemenon
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
• Jun 23rd 2013, 09:21 PM
topsquark
Re: Find the quadratic equation satisfied by the other cube roots of 2-2i
Quote:

Originally Posted by spacemenon
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