A Question on Complex Numbers?

I'm too lazy to do the first suggested proof,

but you can verify that ( - 1 - i )^3 = 2 - 2i.

I got it by knowing that 2-2i is on a ray at 315 degrees,

and that 225 degrees times 3 = 675 degrees,

which has the same terminal side as 315 degrees.

Also noticing that |2 - 2i| = sqrt(8), so the cube root

of 2-2i must have a magnitude sqrt(2).

To find the quadratic equation satisfied by the other two roots,

divide x^3 - (2-2i) by (x + 1 + i).

One way to start it is, first multiply both the numerator and denominator by

(x + 1 - i), so that the denominator becomes (x^2 + 2x + 2);

then ordinary long division should work.

The numerator becomes x^4 + x^3 - ix^3 - 2x + 2ix + 4i.

The first term of the quotient is x^2,

remainder is -x^3 - ix^3 -2x^2 - 2x + 2ix + 4i.

The next term of the quotient is -x,

remainder is -ix^3 + 2ix + 4i.

The next term of the quotient is -ix,

remainder is 2ix^2 + 4ix + 4i.

The final term of the quotient is 2i,

no remainder.

So the quadratic equation is

x^2 - x - ix + 2i = 0.

Solving this gives you

x = [ (1+i) plus or minus sqrt (2i - 8i) ] /2

Well, are you supposed to know that sqrt(-6i) is (1-i)sqrt(3) ?

Anyway, throw that in, and you get the other two cube roots, I think.