Question

Find all complex roots of `z^3= -8i`?

Answer 1

The roots are: `2i`, `-sqrt(3)-i` and `sqrt(3)-i`

Explanation:

See https://socratic.org/s/aDTc9kmp for one method.

In trigonometric form:

`-8i = 8(cos (-pi/2) + i sin (-pi/2))`

By de Moivre we have:

`(cos theta + i sin theta)^n = cos n theta + i sin n theta`

Hence one of the cube roots of `-8i` is:

`2(cos (-pi/6) + i sin (-pi/6)) = sqrt(3)-i`

The others can be found by adding multiples of `(2pi)/3`...

`2(cos (-pi/6+(2pi)/3) + i sin (-pi/6+(2pi)/3)) = 2(cos (pi/2) + i sin (pi/2)) = 2i`

`2(cos (-pi/6+(4pi)/3) + i sin (-pi/6+(4pi)/3)) = 2(cos ((7pi)/6) + i sin ((7pi)/6)) = -sqrt(3)-i`

Here are the three roots in the complex plane...

graph{(x^2+(y-2)^2-0.005)((x-sqrt(3))^2+(y+1)^2-0.005)((x+sqrt(3))^2+(y+1)^2-0.005) = 0 [-5, 5, -2.5, 2.5]}