Precalculus (6th Edition) Blitzer

The above statement can be verified with the help of a derivation. Let ${{z}_{1}}={{r}_{1}}\left( \cos {{\theta }_{1}}+i\sin {{\theta }_{1}} \right)$ and ${{z}_{2}}={{r}_{2}}\left( \cos {{\theta }_{2}}+i\sin {{\theta }_{2}} \right)$ be two complex numbers in polar form. The product of these two complex numbers can be found by performing a simple multiplication as shown below: \begin{align} & {{z}_{1}}\cdot {{z}_{2}}={{r}_{1}}\cdot {{r}_{2}}\left[ \left( \cos {{\theta }_{1}}+i\sin {{\theta }_{1}} \right)\left( \cos {{\theta }_{2}}+i\sin {{\theta }_{2}} \right) \right] \\ & ={{r}_{1}}\cdot {{r}_{2}}\left[ \left( \cos {{\theta }_{1}}\cdot \cos {{\theta }_{2}} \right)+\left( \cos {{\theta }_{1}}\cdot i\sin {{\theta }_{2}} \right)+\left( i\sin {{\theta }_{1}}\cdot \cos {{\theta }_{2}} \right)+\left( i\sin {{\theta }_{1}}\cdot i\sin {{\theta }_{2}} \right) \right] \\ & ={{r}_{1}}\cdot {{r}_{2}}\left[ \left\{ \left( \cos {{\theta }_{1}}\cdot \cos {{\theta }_{2}} \right)-\left( \sin {{\theta }_{1}}\cdot \sin {{\theta }_{2}} \right) \right\}+i\left\{ \left( \cos {{\theta }_{1}}\cdot \sin {{\theta }_{2}} \right)+\left( \sin {{\theta }_{1}}\cdot \cos {{\theta }_{2}} \right) \right\} \right] \\ & ={{r}_{1}}\cdot {{r}_{2}}\left[ \cos \left( {{\theta }_{1}}+{{\theta }_{2}} \right)+i\sin \left( {{\theta }_{1}}+{{\theta }_{2}} \right) \right] \end{align} Above, the derivation sum formula for cosines and sines is used. Hence, the provided statement make sense.