Answer
The sixth roots of -64 are:
$2~(cos~30^{\circ}+i~sin~30^{\circ})$
$2~(cos~90^{\circ}+i~sin~90^{\circ})$
$2~(cos~150^{\circ}+i~sin~150^{\circ})$
$2~(cos~210^{\circ}+i~sin~210^{\circ})$
$2~(cos~270^{\circ}+i~sin~270^{\circ})$
$2~(cos~330^{\circ}+i~sin~330^{\circ})$
None of these roots are real.
Work Step by Step
$z = -64 + 0~i$
$z = 64(cos~180^{\circ}+i~sin~180^{\circ})$
$r = 64$ and $\theta = 180^{\circ}$
We can use this equation to find the sixth 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:
$z^{1/6} = 64^{1/6}~[cos(\frac{180^{\circ}}{6}+\frac{(360^{\circ})(0)}{6})+i~sin(\frac{180^{\circ}}{6}+\frac{(360^{\circ})(0)}{6})]$
$z^{1/6} = 2~(cos~30^{\circ}+i~sin~30^{\circ})$
When k = 1:
$z^{1/6} = 64^{1/6}~[cos(\frac{180^{\circ}}{6}+\frac{(360^{\circ})(1)}{6})+i~sin(\frac{180^{\circ}}{6}+\frac{(360^{\circ})(1)}{6})]$
$z^{1/6} = 2~(cos~90^{\circ}+i~sin~90^{\circ})$
When k = 2:
$z^{1/6} = 64^{1/6}~[cos(\frac{180^{\circ}}{6}+\frac{(360^{\circ})(2)}{6})+i~sin(\frac{180^{\circ}}{6}+\frac{(360^{\circ})(2)}{6})]$
$z^{1/6} = 2~(cos~150^{\circ}+i~sin~150^{\circ})$
When k = 3:
$z^{1/6} = 64^{1/6}~[cos(\frac{180^{\circ}}{6}+\frac{(360^{\circ})(3)}{6})+i~sin(\frac{180^{\circ}}{6}+\frac{(360^{\circ})(3)}{6})]$
$z^{1/6} = 2~(cos~210^{\circ}+i~sin~210^{\circ})$
When k = 4:
$z^{1/6} = 64^{1/6}~[cos(\frac{180^{\circ}}{6}+\frac{(360^{\circ})(4)}{6})+i~sin(\frac{180^{\circ}}{6}+\frac{(360^{\circ})(4)}{6})]$
$z^{1/6} = 2~(cos~270^{\circ}+i~sin~270^{\circ})$
When k = 5:
$z^{1/6} = 64^{1/6}~[cos(\frac{180^{\circ}}{6}+\frac{(360^{\circ})(5)}{6})+i~sin(\frac{180^{\circ}}{6}+\frac{(360^{\circ})(5)}{6})]$
$z^{1/6} = 2~(cos~330^{\circ}+i~sin~330^{\circ})$
The imaginary part of the root is zero only when the angle is $0^{\circ}$ or $180^{\circ}$. Since none of the solutions have these angles, the imaginary part is always non-zero. Therefore, there are no real roots.
