## Precalculus (6th Edition) Blitzer

$\cos 265{}^\circ +i\sin 265{}^\circ$
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 ${{z}_{1}}=\left( \cos 210{}^\circ +i\sin 210{}^\circ \right)\text{ and }{{z}_{2}}=\left( \cos 55{}^\circ +i\sin 55{}^\circ \right)$ is as follows: \begin{align} & {{z}_{1}}\cdot {{z}_{2}}=1\cdot 1\left[ \cos \left( 210+55 \right)+i\sin \left( 210+55 \right) \right] \\ & =\cos 265{}^\circ +i\sin 265{}^\circ \end{align} Hence, the product of the complex numbers is $\cos 265{}^\circ +i\sin 265{}^\circ$.