Find the three cube roots of the complex number 8i. give your answers in the form x + iy

8i is:
 z = 8(cos(90) + i sin(90)) ; now take the 1/3 power of it.

When we take this to a normal integer power of 3 we get:

z^3 = 8^3(cos(90*3) + i sin(90*3)) right? its the same for rational exponents as well.

z^(1/3) - 2(cos(90/3) + i sin(90/3))

90/3 = 30; the cos(30)=sqrt(3)/2 and the sin(30) = 1/2 2(sqrt(3)/2 + i 1/2) = sqrt(3) + 1i should be one of the cube roots.

(sqrt(3) + i)^2 = 2 +2sqrt(3) i
(sqrt(3)+i) (2 +2sqrt(3)i
2sqrt(3) + 6i +2i -2sqrt(3) = 8i