# Chapter 8 - Review Exercises - Page 410: 45

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.

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