## Precalculus (6th Edition) Blitzer

$\sqrt{3}+i,\text{ }-\sqrt{3}+i,\text{ and }-2i$.
To find the $n$ distinct complex roots for any complex number $z=r\left( \cos \theta +i\sin \theta \right)$, if $z\ne 0$, in radians we can use the formula given below: ${{z}_{k}}=\sqrt[n]{r}\left[ \cos \left( \frac{\theta +2\pi \cdot k}{n} \right)+i\sin \left( \frac{\theta +2\pi \cdot k}{n} \right) \right]$ Where $k=0,1,2,3,....,n-1$. $z=8i$ can be rewritten as $z=8\left( \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} \right)$. So, the cube roots of $z=8\left( \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} \right)$ are as follows : ${{z}_{k}}=\sqrt[3]{8}\left[ \cos \left( \frac{\frac{\pi }{2}+2\pi \cdot k}{3} \right)+i\sin \left( \frac{\frac{\pi }{2}+2\pi \cdot k}{3} \right) \right],k=0,1,2$ Therefore, we can find the three complex cube roots in the following manner: \begin{align} & {{z}_{0}}=\sqrt[3]{8}\left[ \cos \left( \frac{\frac{\pi }{2}+2\pi \cdot 0}{3} \right)+i\sin \left( \frac{\frac{\pi }{2}+2\pi \cdot 0}{3} \right) \right] \\ & =2\left( \cos \frac{\pi }{6}+i\sin \frac{\pi }{6} \right) \\ & =2\left( \frac{\sqrt{3}}{2}+i\frac{1}{2} \right) \\ & =\sqrt{3}+i \end{align} \begin{align} & {{z}_{1}}=\sqrt[3]{8}\left[ \cos \left( \frac{\frac{\pi }{2}+2\pi \cdot 1}{3} \right)+i\sin \left( \frac{\frac{\pi }{2}+2\pi \cdot 1}{3} \right) \right] \\ & =2\left( \cos \frac{5\pi }{6}+i\sin \frac{5\pi }{6} \right) \\ & =2\left( -\frac{\sqrt{3}}{2}+i\frac{1}{2} \right) \\ & =-\sqrt{3}+i \end{align} \begin{align} & {{z}_{2}}=\sqrt[3]{8}\left[ \cos \left( \frac{\frac{\pi }{2}+2\pi \cdot 2}{3} \right)+i\sin \left( \frac{\frac{\pi }{2}+2\pi \cdot 2}{3} \right) \right] \\ & =2\left( \cos \frac{3\pi }{2}+i\sin \frac{3\pi }{2} \right) \\ & =2\left( 0+i\left( -1 \right) \right) \\ & =-2i \end{align} So, the cube roots are $\sqrt{3}+i,\text{ }-\sqrt{3}+i,\text{ and }-2i$.