## Precalculus (6th Edition) Blitzer

$2\left( \cos \frac{7\pi }{6}+i\sin \frac{7\pi }{6} \right)$
Division of two complex numbers: For diving two complex numbers ${{z}_{1}}={{r}_{1}}\left( \cos {{\theta }_{1}}+i\sin {{\theta }_{1}} \right)$ and ${{z}_{2}}={{r}_{2}}\left( \cos {{\theta }_{2}}+i\sin {{\theta }_{2}} \right)$, we use the formula given below: $\frac{{{z}_{1}}}{{{z}_{2}}}=\frac{{{r}_{1}}}{{{r}_{2}}}\left[ \cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)+i\sin \left( {{\theta }_{1}}-{{\theta }_{2}} \right) \right]$ So, the product of the given complex numbers ${{z}_{1}}=2\left( \cos \frac{5\pi }{3}+i\sin \frac{5\pi }{3} \right)\text{ and }{{z}_{2}}=\left( \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} \right)$ can be given by: \begin{align} & \frac{{{z}_{1}}}{{{z}_{2}}}=\frac{2}{1}\left[ \cos \left( \frac{5\pi }{3}-\frac{\pi }{2} \right)+i\sin \left( \frac{5\pi }{3}-\frac{\pi }{2} \right) \right] \\ & =2\left[ \cos \left( \frac{10\pi -3\pi }{6} \right)+i\sin \left( \frac{10\pi -3\pi }{6} \right) \right] \\ & =2\left( \cos \frac{7\pi }{6}+i\sin \frac{7\pi }{6} \right) \end{align} Hence, the product of the complex numbers is $2\left( \cos \frac{7\pi }{6}+i\sin \frac{7\pi }{6} \right)$.