Answer
The quotient of the complex numbers in the polar form is $5\left( \cos 60{}^\circ +i\sin 60{}^\circ \right)$.
Work Step by Step
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 $5\left( \cos 60{}^\circ +i\sin 60{}^\circ \right).$
