Answer
See the explanation below.
Work Step by Step
Apply the division operation on the provided two complex numbers.
$\frac{{{z}_{1}}}{{{z}_{2}}}=\frac{{{r}_{1}}\left( \cos {{\theta }_{1}}+i\sin {{\theta }_{1}} \right)}{{{r}_{2}}\left( \cos {{\theta }_{2}}+i\sin {{\theta }_{2}} \right)}$
Multiply the numerator and denominator of the above equation with the conjugate of the denominator’s second factor.
$\frac{{{z}_{1}}}{{{z}_{2}}}=\frac{{{r}_{1}}\left( \cos {{\theta }_{1}}+i\sin {{\theta }_{1}} \right)}{{{r}_{2}}\left( \cos {{\theta }_{2}}+i\sin {{\theta }_{2}} \right)}\cdot \frac{\left( \cos {{\theta }_{2}}-i\sin {{\theta }_{2}} \right)}{\left( \cos {{\theta }_{2}}-i\sin {{\theta }_{2}} \right)}$ …… (1)
Apply the multiplication operation in equation (1).
$\frac{{{z}_{1}}}{{{z}_{2}}}=\frac{{{r}_{1}}\left[ \cos {{\theta }_{1}}\left( \cos {{\theta }_{2}}-i\sin {{\theta }_{2}} \right)+i\sin {{\theta }_{1}}\left( \cos {{\theta }_{2}}-i\sin {{\theta }_{2}} \right) \right]}{{{r}_{2}}\left[ \cos {{\theta }_{2}}\left( \cos {{\theta }_{2}}-i\sin {{\theta }_{2}} \right)+i\sin {{\theta }_{2}}\left( \cos {{\theta }_{2}}-i\sin {{\theta }_{2}} \right) \right]}$
$=\frac{{{r}_{1}}\left[ \left( \cos {{\theta }_{1}}\cdot \cos {{\theta }_{2}} \right)-i\left( \cos {{\theta }_{1}}\cdot \sin {{\theta }_{2}} \right)+i\left( \sin {{\theta }_{1}}\cdot \cos {{\theta }_{2}} \right)-{{i}^{2}}\left( \sin {{\theta }_{1}}\cdot \sin {{\theta }_{2}} \right) \right]}{{{r}_{2}}\left[ \left( \cos {{\theta }_{2}}\cdot \cos {{\theta }_{2}} \right)-i\left( \cos {{\theta }_{2}}\cdot \sin {{\theta }_{2}} \right)+i\left( \sin {{\theta }_{2}}\cdot \cos {{\theta }_{2}} \right)-{{i}^{2}}\left( \sin {{\theta }_{2}}\cdot \sin {{\theta }_{2}} \right) \right]}$ …… (2)
Substitute the value of the iota in equation (2) and then rearrange the equation.
$\frac{{{z}_{1}}}{{{z}_{2}}}=\frac{{{r}_{1}}\left[ \left( \cos {{\theta }_{1}}\cdot \cos {{\theta }_{2}} \right)+\left( \sin {{\theta }_{1}}\cdot \sin {{\theta }_{2}} \right)+i\left\{ \left( \sin {{\theta }_{1}}\cdot \cos {{\theta }_{2}} \right)-\left( \cos {{\theta }_{1}}\cdot \sin {{\theta }_{2}} \right) \right\} \right]}{{{r}_{2}}\left[ \left( \cos {{\theta }_{2}}\cdot \cos {{\theta }_{2}} \right)+\left( \sin {{\theta }_{2}}\cdot \sin {{\theta }_{2}} \right)+i\left\{ \left( \sin {{\theta }_{2}}\cdot \cos {{\theta }_{2}} \right)-\left( \cos {{\theta }_{2}}\cdot \sin {{\theta }_{2}} \right) \right\} \right]}$
Further solve it to get
$=\frac{{{r}_{1}}\left[ \left( \cos {{\theta }_{1}}\cdot \cos {{\theta }_{2}} \right)+\left( \sin {{\theta }_{1}}\cdot \sin {{\theta }_{2}} \right)+i\left\{ \left( \sin {{\theta }_{1}}\cdot \cos {{\theta }_{2}} \right)-\left( \cos {{\theta }_{1}}\cdot \sin {{\theta }_{2}} \right) \right\} \right]}{{{r}_{2}}\left[ \left( {{\cos }^{2}}{{\theta }_{2}}+{{\sin }^{2}}{{\theta }_{2}} \right)+i\left( 0 \right) \right]}$ …… (3)
Apply the identity of trigonometry ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ in the denominator of equation (3).
$\frac{{{z}_{1}}}{{{z}_{2}}}=\frac{{{r}_{1}}\left[ \left( \cos {{\theta }_{1}}\cdot \cos {{\theta }_{2}} \right)+\left( \sin {{\theta }_{1}}\cdot \sin {{\theta }_{2}} \right)+i\left\{ \left( \sin {{\theta }_{1}}\cdot \cos {{\theta }_{2}} \right)-\left( \cos {{\theta }_{1}}\cdot \sin {{\theta }_{2}} \right) \right\} \right]}{{{r}_{2}}\left[ 1+i\left( 0 \right) \right]}$
Further solve it to get
$\frac{{{z}_{1}}}{{{z}_{2}}}=\frac{{{r}_{1}}}{{{r}_{2}}}\cdot \left[ \left( \cos {{\theta }_{1}}\cdot \cos {{\theta }_{2}} \right)+\left( \sin {{\theta }_{1}}\cdot \sin {{\theta }_{2}} \right)+i\left\{ \left( \sin {{\theta }_{1}}\cdot \cos {{\theta }_{2}} \right)-\left( \cos {{\theta }_{1}}\cdot \sin {{\theta }_{2}} \right) \right\} \right]$ …… (4)
Apply the difference formula of cosine and sine in equation (4).
$\frac{{{z}_{1}}}{{{z}_{2}}}=\frac{{{r}_{1}}}{{{r}_{2}}}\cdot \left[ \cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)+i\sin \left( {{\theta }_{1}}-{{\theta }_{2}} \right) \right]$
Hence, the quotient of the provided two complex numbers is
$\frac{{{r}_{1}}\left( \cos {{\theta }_{1}}+i\sin {{\theta }_{1}} \right)}{{{r}_{2}}\left( \cos {{\theta }_{2}}+i\sin {{\theta }_{2}} \right)}=\frac{{{r}_{1}}}{{{r}_{2}}}\cdot \left[ \cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)+i\sin \left( {{\theta }_{1}}-{{\theta }_{2}} \right) \right]$.
