## Precalculus (6th Edition) Blitzer

$\sqrt{3}+i,\text{ }-1+i\sqrt{3},\text{ }-\sqrt{3}-i,\text{ and }1-i\sqrt{3}$.
Method for finding complex roots: To find $n$ distinct complex roots for any complex number $z=r\left( \cos \theta +i\sin \theta \right)$, if $z\ne 0$, in radians we 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 $k=0,1,2,3,....,n-1$. So, the fourth roots of $16\left( \cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3} \right)$ are ${{z}_{k}}=\sqrt[4]{16}\left[ \cos \left( \frac{\frac{2\pi }{3}+2\pi \cdot k}{4} \right)+i\sin \left( \frac{\frac{2\pi }{3}+2\pi \cdot k}{4} \right) \right],k=0,1,2,3$ Therefore, we can find the fourth complex square roots in the following manner: \begin{align} & {{z}_{0}}=\sqrt[4]{16}\left[ \cos \left( \frac{\frac{2\pi }{3}+2\pi \cdot 0}{4} \right)+i\sin \left( \frac{\frac{2\pi }{3}+2\pi \cdot 0}{4} \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[4]{16}\left[ \cos \left( \frac{\frac{2\pi }{3}+2\pi \cdot 1}{4} \right)+i\sin \left( \frac{\frac{2\pi }{3}+2\pi \cdot 1}{4} \right) \right] \\ & =2\left( \cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3} \right) \\ & =2\left( -\frac{1}{2}+i\frac{\sqrt{3}}{2} \right) \\ & =-1+i\sqrt{3} \end{align} \begin{align} & {{z}_{2}}=\sqrt[4]{16}\left[ \cos \left( \frac{\frac{2\pi }{3}+2\pi \cdot 2}{4} \right)+i\sin \left( \frac{\frac{2\pi }{3}+2\pi \cdot 2}{4} \right) \right] \\ & =2\left( \cos \frac{7\pi }{6}+i\sin \frac{7\pi }{6} \right) \\ & =2\left( -\frac{\sqrt{3}}{2}+i\left( -\frac{1}{2} \right) \right) \\ & =-\sqrt{3}-i \end{align} \begin{align} & {{z}_{3}}=\sqrt[4]{16}\left[ \cos \left( \frac{\frac{2\pi }{3}+2\pi \cdot 3}{4} \right)+i\sin \left( \frac{\frac{2\pi }{3}+2\pi \cdot 3}{4} \right) \right] \\ & =2\left( \cos \frac{5\pi }{3}+i\sin \frac{5\pi }{3} \right) \\ & =2\left( \frac{1}{2}+i\left( -\frac{\sqrt{3}}{2} \right) \right) \\ & =1-i\sqrt{3} \end{align} So, the fourth roots are $\sqrt{3}+i,\text{ }-1+i\sqrt{3},\text{ }-\sqrt{3}-i\text{, and }1-i\sqrt{3}$.