Hey!

Please find the answer to your query here:

First of all, we will write 8i in polar form.

8i= (2^3) (cos pi/2 + i.sin pi/2)

The cube roots are: 2(cos pi/6+ i.sin Pi/6), 2(cos (pi/6+ 2pi/3)+ i.sin (pi/6+ 2pi/3)) and 2(cos (pi/6+ 4pi/3)+ i.sin (pi/6+ 4pi/3)).

Therefore, these roots are [sqrt(3)+i], [-sqrt(3)+i], [-2i]

Hope this answer helps!