Answer
The four complex fourth roots of $-16~i$ are:
$2~(cos~67.5^{\circ}+i~sin~67.5^{\circ})$
$2~(cos~157.5^{\circ}+i~sin~157.5^{\circ})$
$2~(cos~247.5^{\circ}+i~sin~247.5^{\circ})$
$2~(cos~337.5^{\circ}+i~sin~337.5^{\circ})$
Work Step by Step
$z = -16~i$
$z = 16~(0 - i)$
$z = 16(cos~270^{\circ}+i~sin~270^{\circ})$
$r = 16$ and $\theta = 270^{\circ}$
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:
$z^{1/4} = 16^{1/4}~[cos(\frac{270^{\circ}}{4}+\frac{(360^{\circ})(0)}{4})+i~sin(\frac{270^{\circ}}{4}+\frac{(360^{\circ})(0)}{4})]$
$z^{1/4} = 2~(cos~67.5^{\circ}+i~sin~67.5^{\circ})$
When k = 1:
$z^{1/4} = 16^{1/4}~[cos(\frac{270^{\circ}}{4}+\frac{(360^{\circ})(1)}{4})+i~sin(\frac{270^{\circ}}{4}+\frac{(360^{\circ})(1)}{4})]$
$z^{1/4} = 2~(cos~157.5^{\circ}+i~sin~157.5^{\circ})$
When k = 2:
$z^{1/4} = 16^{1/4}~[cos(\frac{270^{\circ}}{4}+\frac{(360^{\circ})(2)}{4})+i~sin(\frac{270^{\circ}}{4}+\frac{(360^{\circ})(2)}{4})]$
$z^{1/4} = 2~(cos~247.5^{\circ}+i~sin~247.5^{\circ})$
When k = 3:
$z^{1/4} = 16^{1/4}~[cos(\frac{270^{\circ}}{4}+\frac{(360^{\circ})(3)}{4})+i~sin(\frac{270^{\circ}}{4}+\frac{(360^{\circ})(3)}{4})]$
$z^{1/4} = 2~(cos~337.5^{\circ}+i~sin~337.5^{\circ})$
