## Precalculus (6th Edition) Blitzer

The quotient of the complex numbers in the polar form is $5\left( \cos 60{}^\circ +i\sin 60{}^\circ \right)$.
Consider any complex number given by \begin{align} & {{z}_{1}}={{r}_{1}}\left( \cos {{\theta }_{1}}+i\sin {{\theta }_{1}} \right) \\ & {{z}_{2}}={{r}_{2}}\left( \cos {{\theta }_{2}}+i\sin {{\theta }_{2}} \right) \\ \end{align} For a complex number in polar form, the quotient is calculated as $\frac{{{z}_{1}}}{{{z}_{2}}}=\frac{{{r}_{1}}}{{{r}_{2}}}\left( \cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)+i\sin \left( {{\theta }_{1}}-{{\theta }_{2}} \right) \right)$ (1) The polar form after the division of the complex numbers: \begin{align} & {{z}_{1}}=50\left( \cos 80{}^\circ +i\sin 80{}^\circ \right) \\ & {{z}_{2}}=10\left( \cos 20{}^\circ +i\sin 20{}^\circ \right) \\ \end{align} (2) Divide it using (1) and (2) to get, \begin{align} & \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) \\ & \frac{{{z}_{1}}}{{{z}_{2}}}=\frac{50}{10}\left( \cos \left( 80{}^\circ -20{}^\circ \right)+i\sin \left( 80{}^\circ -20{}^\circ \right) \right) \\ & =5\left( \cos 60{}^\circ +i\sin 60{}^\circ \right) \end{align} The quotient of the complex numbers in the polar form is $5\left( \cos 60{}^\circ +i\sin 60{}^\circ \right).$