Answer
The fifth roots of -32 are:
$2~(cos~36^{\circ}+i~sin~36^{\circ})$
$2~(cos~108^{\circ}+i~sin~108^{\circ})$
$2~(cos~180^{\circ}+i~sin~180^{\circ})$
$2~(cos~252^{\circ}+i~sin~252^{\circ})$
$2~(cos~324^{\circ}+i~sin~324^{\circ})$
-32 has one real fifth root.
Work Step by Step
$z = -32 + 0~i$
$z = 32(cos~180^{\circ}+i~sin~180^{\circ})$
$r = 32$ and $\theta = 180^{\circ}$
We can use this equation to find the fifth 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:
$z^{1/5} = 32^{1/5}~[cos(\frac{180^{\circ}}{5}+\frac{(360^{\circ})(0)}{5})+i~sin(\frac{180^{\circ}}{5}+\frac{(360^{\circ})(0)}{5})]$
$z^{1/5} = 2~(cos~36^{\circ}+i~sin~36^{\circ})$
When k = 1:
$z^{1/5} = 32^{1/5}~[cos(\frac{180^{\circ}}{5}+\frac{(360^{\circ})(1)}{5})+i~sin(\frac{180^{\circ}}{5}+\frac{(360^{\circ})(1)}{5})]$
$z^{1/5} = 2~(cos~108^{\circ}+i~sin~108^{\circ})$
When k = 2:
$z^{1/5} = 32^{1/5}~[cos(\frac{180^{\circ}}{5}+\frac{(360^{\circ})(2)}{5})+i~sin(\frac{180^{\circ}}{5}+\frac{(360^{\circ})(2)}{5})]$
$z^{1/5} = 2~(cos~180^{\circ}+i~sin~180^{\circ})$
When k = 3:
$z^{1/5} = 32^{1/5}~[cos(\frac{180^{\circ}}{5}+\frac{(360^{\circ})(3)}{5})+i~sin(\frac{180^{\circ}}{5}+\frac{(360^{\circ})(3)}{5})]$
$z^{1/5} = 2~(cos~252^{\circ}+i~sin~252^{\circ})$
When k = 4:
$z^{1/5} = 32^{1/5}~[cos(\frac{180^{\circ}}{5}+\frac{(360^{\circ})(4)}{5})+i~sin(\frac{180^{\circ}}{5}+\frac{(360^{\circ})(4)}{5})]$
$z^{1/5} = 2~(cos~324^{\circ}+i~sin~324^{\circ})$
The imaginary part of the root is zero only when the angle is $0^{\circ}$ or $180^{\circ}$. One of the solutions has an angle of $180^{\circ}$. Therefore, there is one real root.
