Question

Solve for the roots in the equation below. In your final answer, include each of the necessary steps and calculations.

Answer 1

$x^3-27i=0$
$x^3=27i$
$x^3=27e^{i\pi/2}$
$x=\left(27e^{i\pi/2}\right)^{1/3}$
$x=3e^{i(\pi/2+2\pi k)/3}$
$x=3e^{i\pi(4k+1)/6}$

where $k\in\{0,1,2\}$. This means you have

$x=3e^{i\pi/6}=\dfrac32(\sqrt3+i)$
$x=3e^{i5\pi/6}=\dfrac32(-\sqrt3+i)$
$x=3e^{i9\pi/6}=-3i$

as the solutions to the original equation.