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