## Precalculus (6th Edition) Blitzer

$\frac{1}{2}+\frac{\sqrt{3}}{2}i,\text{ }-1,\text{ and }\frac{1}{2}-\frac{\sqrt{3}}{2}i$.
For any complex number $z=r\left( \cos \theta +i\sin \theta \right)$, if $z\ne 0$, $n$ distinct complex roots in radians will be found by using the following formula: ${{z}_{k}}=\sqrt[n]{r}\left[ \cos \left( \frac{\theta +2\pi \cdot k}{n} \right)+i\sin \left( \frac{\theta +2\pi \cdot k}{n} \right) \right]$ Where $k=0,1,2,3,....,n-1$. $z=-1$ can be rewritten as $z=1\left( \cos \pi +i\sin \pi \right)$. So, the cube roots of $z=1\left( \cos \pi +i\sin \pi \right)$ are ${{z}_{k}}=\sqrt[3]{1}\left[ \cos \left( \frac{\pi +2\pi \cdot k}{3} \right)+i\sin \left( \frac{\pi +2\pi \cdot k}{3} \right) \right],k=0,1,2$ Therefore, the three complex cube roots are calculated in the following manner: \begin{align} & {{z}_{0}}=\sqrt[3]{1}\left[ \cos \left( \frac{\pi +2\pi \cdot 0}{3} \right)+i\sin \left( \frac{\pi +2\pi \cdot 0}{3} \right) \right] \\ & =\cos \frac{\pi }{3}+i\sin \frac{\pi }{3} \\ & =\frac{1}{2}+i\frac{\sqrt{3}}{2} \end{align} \begin{align} & {{z}_{1}}=\sqrt[3]{1}\left[ \cos \left( \frac{\pi +2\pi \cdot 1}{3} \right)+i\sin \left( \frac{\pi +2\pi \cdot 1}{3} \right) \right] \\ & =\cos \pi +i\sin \pi \\ & =-1+i\left( 0 \right) \\ & =-1 \end{align} \begin{align} & {{z}_{2}}=\sqrt[3]{1}\left[ \cos \left( \frac{\pi +2\pi \cdot 2}{3} \right)+i\sin \left( \frac{\pi +2\pi \cdot 2}{3} \right) \right] \\ & =\cos \frac{5\pi }{3}+i\sin \frac{5\pi }{3} \\ & =\frac{1}{2}+i\left( -\frac{\sqrt{3}}{2} \right) \\ & =\frac{1}{2}-\frac{\sqrt{3}}{2}i \end{align} So, the cube roots are $\frac{1}{2}+\frac{\sqrt{3}}{2}i,\text{ }-1,\text{ and }\frac{1}{2}-\frac{\sqrt{3}}{2}i$.