complex nos finding cube root

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

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)

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Re: complex nos finding cube root

Attachment 28666 the other part you can do yourself

Re: complex nos finding cube root

$\displaystyle x^2 - y^2 = 0 \implies y = \pm x$

So

$\displaystyle 2xy = 8 \implies \pm 2x^2 = 8 \implies x = \pm 2$

so $\displaystyle (x, y) = ( \pm 2, \pm 2)$

Neither of the remaining two equations share this point.

-Dan

Re: complex nos finding cube root

sorry the the question is (x+iy)^3=8i

Re: complex nos finding cube root

Quote:

Originally Posted by

**spacemenon** sorry the the question is (x+iy)^3=8i

The process is the same. Expand the (x + iy)^3 and match coefficients. Give it a try and tell us how it works out.

-Dan