Answer
The solutions for $x$ are:
$2~(cos~15^{\circ}+i~sin~15^{\circ})$
$2~(cos~105^{\circ}+i~sin~105^{\circ})$
$2~(cos~195^{\circ}+i~sin~195^{\circ})$
$2~(cos~285^{\circ}+i~sin~285^{\circ})$
Work Step by Step
$x^4-(8+8i~\sqrt{3}) = 0$
$x^4 = (8+8i~\sqrt{3})$
$x^4 = 16~(\frac{1}{2}+\frac{i~\sqrt{3}}{2})$
$x = [16~(\frac{1}{2}+\frac{i~\sqrt{3}}{2})]^{1/4}$
Let $z = 16~(\frac{1}{2}+\frac{i~\sqrt{3}}{2})$
$z = 16~(cos~60^{\circ}+i~sin~60^{\circ})$
$r = 16$ and $\theta = 60^{\circ}$
We can use this equation to find the fourth 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:
$z^{1/4} = 16^{1/4}~[cos(\frac{60^{\circ}}{4}+\frac{(360^{\circ})(0)}{4})+i~sin(\frac{60^{\circ}}{4}+\frac{(360^{\circ})(0)}{4})]$
$z^{1/4} = 2~(cos~15^{\circ}+i~sin~15^{\circ})$
When k = 1:
$z^{1/4} = 16^{1/4}~[cos(\frac{60^{\circ}}{4}+\frac{(360^{\circ})(1)}{4})+i~sin(\frac{60^{\circ}}{4}+\frac{(360^{\circ})(1)}{4})]$
$z^{1/4} = 2~(cos~105^{\circ}+i~sin~105^{\circ})$
When k = 2:
$z^{1/4} = 16^{1/4}~[cos(\frac{60^{\circ}}{4}+\frac{(360^{\circ})(2)}{4})+i~sin(\frac{60^{\circ}}{4}+\frac{(360^{\circ})(2)}{4})]$
$z^{1/4} = 2~(cos~195^{\circ}+i~sin~195^{\circ})$
When k = 3:
$z^{1/4} = 16^{1/4}~[cos(\frac{60^{\circ}}{4}+\frac{(360^{\circ})(3)}{4})+i~sin(\frac{60^{\circ}}{4}+\frac{(360^{\circ})(3)}{4})]$
$z^{1/4} = 2~(cos~285^{\circ}+i~sin~285^{\circ})$
