## Precalculus (6th Edition) Blitzer

The product of the complex numbers is $15\left( \cos 110{}^\circ +i\sin 110{}^\circ \right)$.
Product of two complex numbers: The product 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 ${{z}_{1}}\cdot {{z}_{2}}={{r}_{1}}\cdot {{r}_{2}}\left[ \cos \left( {{\theta }_{1}}+{{\theta }_{2}} \right)+i\sin \left( {{\theta }_{1}}+{{\theta }_{2}} \right) \right]$ So, the product of the given complex numbers ${{z}_{1}}=3\left( \cos 40{}^\circ +i\sin 40{}^\circ \right),{{z}_{2}}=5\left( \cos 70{}^\circ +i\sin 70{}^\circ \right)$ is \begin{align} & {{z}_{1}}\cdot {{z}_{2}}=3\cdot 5\left[ \cos \left( 40+70 \right)+i\sin \left( 40+70 \right) \right] \\ & =15\left( \cos 110{}^\circ +i\sin 110{}^\circ \right) \end{align} Hence, the product of the complex numbers is $15\left( \cos 110{}^\circ +i\sin 110{}^\circ \right)$.