## Precalculus (6th Edition) Blitzer

$2\left( \cos 5{}^\circ +i\sin 5{}^\circ \right)$
Division of two complex numbers: The division of two complex numbers ${{z}_{1}}={{r}_{1}}\left( \cos {{\theta }_{1}}+i\sin {{\theta }_{1}} \right)$ and ${{z}_{2}}={{r}_{2}}\left( \cos {{\theta }_{2}}+i\sin {{\theta }_{2}} \right)$, is given by $\frac{{{z}_{1}}}{{{z}_{2}}}=\frac{{{r}_{1}}}{{{r}_{2}}}\left[ \cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)+i\sin \left( {{\theta }_{1}}-{{\theta }_{2}} \right) \right]$ So, the product of the given complex numbers ${{z}_{1}}=10\left( \cos 10{}^\circ +i\sin 10{}^\circ \right)\text{ and }{{z}_{2}}=5\left( \cos 5{}^\circ +i\sin 5{}^\circ \right)$ is \begin{align} & \frac{{{z}_{1}}}{{{z}_{2}}}=\frac{10}{5}\left[ \cos \left( 10{}^\circ -5{}^\circ \right)+i\sin \left( 10{}^\circ -5{}^\circ \right) \right] \\ & =2\left( \cos 5{}^\circ +i\sin 5{}^\circ \right) \end{align} Hence, the product of the complex numbers is $2\left( \cos 5{}^\circ +i\sin 5{}^\circ \right)$.