Answer
The sixth roots of 1 are:
$cos~0^{\circ}+i~sin~0^{\circ}$
$cos~60^{\circ}+i~sin~60^{\circ}$
$cos~120^{\circ}+i~sin~120^{\circ}$
$cos~180^{\circ}+i~sin~180^{\circ}$
$cos~240^{\circ}+i~sin~240^{\circ}$
$cos~300^{\circ}+i~sin~300^{\circ}$
We can graph the sixth roots in the complex plane:
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 square 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 $k \in \{0, 1, 2,...,n-1\}$
When k = 0:
$z^{1/6} = 1^{1/6}~[cos(\frac{0^{\circ}}{6}+\frac{(360^{\circ})(0)}{6})+i~sin(\frac{0^{\circ}}{6}+\frac{(360^{\circ})(0)}{6})]$
$z^{1/6} = 1~[cos~0^{\circ}+i~sin~0^{\circ}]$
$z^{1/6} = cos~0^{\circ}+i~sin~0^{\circ}$
When k = 1:
$z^{1/6} = 1^{1/6}~[cos(\frac{0^{\circ}}{6}+\frac{(360^{\circ})(1)}{6})+i~sin(\frac{0^{\circ}}{6}+\frac{(360^{\circ})(1)}{6})]$
$z^{1/6} = 1~[cos~60^{\circ}+i~sin~60^{\circ}]$
$z^{1/6} = cos~60^{\circ}+i~sin~60^{\circ}$
When k = 2:
$z^{1/6} = 1^{1/6}~[cos(\frac{0^{\circ}}{6}+\frac{(360^{\circ})(2)}{6})+i~sin(\frac{0^{\circ}}{6}+\frac{(360^{\circ})(2)}{6})]$
$z^{1/6} = 1~[cos~120^{\circ}+i~sin~120^{\circ}]$
$z^{1/6} = cos~120^{\circ}+i~sin~120^{\circ}$
When k = 3:
$z^{1/6} = 1^{1/6}~[cos(\frac{0^{\circ}}{6}+\frac{(360^{\circ})(3)}{6})+i~sin(\frac{0^{\circ}}{6}+\frac{(360^{\circ})(3)}{6})]$
$z^{1/6} = 1~[cos~180^{\circ}+i~sin~180^{\circ}]$
$z^{1/6} = cos~180^{\circ}+i~sin~180^{\circ}$
When k = 4:
$z^{1/6} = 1^{1/6}~[cos(\frac{0^{\circ}}{6}+\frac{(360^{\circ})(4)}{6})+i~sin(\frac{0^{\circ}}{6}+\frac{(360^{\circ})(4)}{6})]$
$z^{1/6} = 1~[cos~240^{\circ}+i~sin~240^{\circ}]$
$z^{1/6} = cos~240^{\circ}+i~sin~240^{\circ}$
When k = 5:
$z^{1/6} = 1^{1/6}~[cos(\frac{0^{\circ}}{6}+\frac{(360^{\circ})(5)}{6})+i~sin(\frac{0^{\circ}}{6}+\frac{(360^{\circ})(5)}{6})]$
$z^{1/6} = 1~[cos~300^{\circ}+i~sin~300^{\circ}]$
$z^{1/6} = cos~300^{\circ}+i~sin~300^{\circ}$
We can graph the sixth roots in the complex plane: