## Trigonometry (10th Edition)

The solutions for $x$ are: $2~(cos~15^{\circ}+i~sin~15^{\circ})$ $2~(cos~105^{\circ}+i~sin~105^{\circ})$ $2~(cos~195^{\circ}+i~sin~195^{\circ})$ $2~(cos~285^{\circ}+i~sin~285^{\circ})$
$x^4-(8+8i~\sqrt{3}) = 0$ $x^4 = (8+8i~\sqrt{3})$ $x^4 = 16~(\frac{1}{2}+\frac{i~\sqrt{3}}{2})$ $x = [16~(\frac{1}{2}+\frac{i~\sqrt{3}}{2})]^{1/4}$ 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: $z^{1/4} = 16^{1/4}~[cos(\frac{60^{\circ}}{4}+\frac{(360^{\circ})(3)}{4})+i~sin(\frac{60^{\circ}}{4}+\frac{(360^{\circ})(3)}{4})]$ $z^{1/4} = 2~(cos~285^{\circ}+i~sin~285^{\circ})$