## Precalculus (6th Edition) Blitzer

$40\left( \cos \pi +i\sin \pi \right)$
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 ${{z}_{1}}=4\left( \cos \frac{3\pi }{7}+i\sin \frac{3\pi }{7} \right)\text{ and }{{z}_{2}}=10\left( \cos \frac{4\pi }{7}+i\sin \frac{4\pi }{7} \right)$ is \begin{align} & {{z}_{1}}\cdot {{z}_{2}}=4\cdot 10\left[ \cos \left( \frac{3\pi }{7}+\frac{4\pi }{7} \right)+i\sin \left( \frac{3\pi }{7}+\frac{4\pi }{7} \right) \right] \\ & =40\left( \cos \pi +i\sin \pi \right) \end{align} Hence, the product of the complex numbers is $40\left( \cos \pi +i\sin \pi \right)$.