Answer
The roots have this form:
$cos~\theta+i~sin~\theta$
where $\theta = 0, \frac{360^{\circ}}{n}, \frac{(360^{\circ})(2)}{n}, \frac{(360^{\circ})(3)}{n}, \frac{(360^{\circ})(4)}{n},..., \frac{(360^{\circ})(n-1)}{n}$
We can see that solutions of this form will be equally spaced around with an angle of $\frac{360^{\circ}}{n}$ between each successive solution.
Work Step by Step
$z = 1+0~i$
$z = cos~0^{\circ}+i~sin~0^{\circ}$
$r = 1$ and $\theta = 0^{\circ}$
We can use this equation to find the n-th roots:
$z^{1/n} = r^{1/n}~[cos(\frac{\theta}{n}+\frac{360^{\circ}~k}{n})+i~sin(\frac{\theta}{n}+\frac{360^{\circ}~k}{n})]$, where $k \in \{0, 1, 2,...,n-1\}$
The roots have this form:
$z^{1/n} = 1^{1/n}~[cos(\frac{0^{\circ}}{n}+\frac{360^{\circ}~k}{n})+i~sin(\frac{0^{\circ}}{n}+\frac{360^{\circ}~k}{n})]$, where $k \in \{0, 1, 2,...,n-1\}$
$z^{1/n} = 1~[cos(\frac{360^{\circ}~k}{n})+i~sin(\frac{360^{\circ}~k}{n})]$, where $k \in \{0, 1, 2,...,n-1\}$
The roots have this form:
$cos~\theta+i~sin~\theta$
where $\theta = 0, \frac{360^{\circ}}{n}, \frac{(360^{\circ})(2)}{n}, \frac{(360^{\circ})(3)}{n}, \frac{(360^{\circ})(4)}{n},..., \frac{(360^{\circ})(n-1)}{n}$
We can see that solutions of this form will be equally spaced around with an angle of $\frac{360^{\circ}}{n}$ between each successive solution.
