Answer
The three cube roots are:
$z^{1/3} = 1$
$z^{1/3} = -\frac{1}{2}+i~(\frac{\sqrt{3}}{2})$
$z^{1/3} = -\frac{1}{2}-i~(\frac{\sqrt{3}}{2})$
Work Step by Step
$z = cos~0^{\circ}+i~sin~0^{\circ}$
$r = 1$ and $\theta = 0^{\circ}$
We can use this equation to find the cube roots:
$z^{1/n} = r^{1/n}~[cos(\frac{\theta}{n}+\frac{2\pi~k}{n})+i~sin(\frac{\theta}{n}+\frac{2\pi~k}{n})]$, where $k \in \{0, 1, 2,...,n-1\}$
When k = 0:
$z^{1/3} = 1^{1/3}~[cos(\frac{0^{\circ}}{3}+\frac{(2\pi)(0)}{3})+i~sin(\frac{0^{\circ}}{3}+\frac{(2\pi)(0)}{3})]$
$z^{1/3} = 1~[cos(0)+i~sin(0)]$
$z^{1/3} = 1~[1+i~(0)]$
$z^{1/3} = 1$
When k = 1:
$z^{1/3} = 1^{1/3}~[cos(\frac{0^{\circ}}{3}+\frac{(2\pi)(1)}{3})+i~sin(\frac{0^{\circ}}{3}+\frac{(2\pi)(1)}{3})]$
$z^{1/3} = 1~[cos(\frac{2\pi}{3})+i~sin(\frac{2\pi}{3})]$
$z^{1/3} = 1~[-\frac{1}{2}+i~(\frac{\sqrt{3}}{2})]$
$z^{1/3} = -\frac{1}{2}+i~(\frac{\sqrt{3}}{2})$
When k = 2:
$z^{1/3} = 1^{1/3}~[cos(\frac{0^{\circ}}{3}+\frac{(2\pi)(2)}{3})+i~sin(\frac{0^{\circ}}{3}+\frac{(2\pi)(2)}{3})]$
$z^{1/3} = 1~[cos(\frac{4\pi}{3})+i~sin(\frac{4\pi}{3})]$
$z^{1/3} = 1~[-\frac{1}{2}-i~(\frac{\sqrt{3}}{2})]$
$z^{1/3} = -\frac{1}{2}-i~(\frac{\sqrt{3}}{2})$
