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

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

Quote:

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**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.

implies or

Is that consistent with the rest of the post? Read the post again. The grouping symbols are mismatched.

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

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 nor are equal to 2. Check the problem statement again.

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

-Dan