What are the cube roots of an imaginary number in trigonometric form?

Answer 1

z = y i where y is a real number

1: z^(1/3) = y^(1/3) (cos(30 deg) + i sin(30 deg))

2: z^(1/3) = y^(1/3) (-sin(60 deg) + i cos(60 deg))

3: z^(1/3) = y^(1/3) (- i)