Complex Numbers
7.0 Properties of Argument
7.0 Properties of Argument
$$\begin{equation} \begin{aligned} {z_1} = {r_1}(\cos {\theta _1} + i\sin {\theta _1}) = {r_1}{e^{i{\theta _1}}}{\text{ and }} \\ {z_2} = {r_2}(\cos {\theta _2} + i\sin {\theta _2}) = {r_2}{e^{i{\theta _2}}} \\ \Rightarrow {z_1}{z_2} = {r_1}{r_2}{e^{i{\theta _1}}}{e^{i{\theta _2}}} = {r_1}{r_2}{e^{i({\theta _1} + {\theta _2})}}\quad ...{\text{(1) }} \\ \Rightarrow \frac{{{z_1}}}{{{z_2}}}{\text{ = }}\frac{{{r_1}{e^{i{\theta _1}}}}}{{{r_2}{e^{i{\theta _2}}}}}{\text{ = }}\frac{{{r_1}}}{{{r_2}}}{e^{i({\theta _1} - {\theta _2})}}\quad ...(2) \\\end{aligned} \end{equation} $$
1. ${\text{Arg(}}{z_1}{z_2}) = {\theta _1} + {\theta _2}\quad [{\text{from equation (1)]}}$
${\text{Arg(}}{z_1}{z_2}){\text{ = Arg(}}{z_1}) + {\text{Arg(}}{z_2})$
2. ${\text{Arg}}\left( {\frac{{{z_1}}}{{{z_2}}}} \right) = {\theta _1} - {\theta _2}\quad [{\text{from equation (2)]}}$
${\text{Arg}}\left( {\frac{{{z_1}}}{{{z_2}}}} \right){\text{ = Arg(}}{z_1}) - {\text{Arg(}}{z_2})$
3. ${\text{Arg}}(\overline z ) = - {\text{Arg(}}z)$
4. If ${\text{Arg(}}z) = 0 \Rightarrow z{\text{ is real}}$.
5. ${\text{Arg(}}{z^n}) = n{\text{Arg(}}z)$
Question 8. If ${z_1} = \overline {{z_2}} $ and ${z_3} = \overline {{z_4}} $. Find the value of $${\text{Arg}}\left( {\frac{{{z_1}}}{{{z_3}}}} \right) + {\text{Arg}}\left( {\frac{{{z_2}}}{{{z_4}}}} \right)$$
Solution: $$\begin{equation} \begin{aligned} \because {z_1} = \overline {{z_2}} \Rightarrow {\text{Arg(}}{z_1}) = {\text{Arg(}}\overline {{z_2}} ) = - {\text{Arg(}}{z_2}) \\ \because {z_3} = \overline {{z_4}} \Rightarrow {\text{Arg(}}{z_3}) = {\text{Arg(}}\overline {{z_4}} ) = - {\text{Arg(}}{z_4}) \\\end{aligned} \end{equation} $$
Now, $$\begin{equation} \begin{aligned} {\text{Arg}}\left( {\frac{{{z_1}}}{{{z_3}}}} \right) + {\text{Arg}}\left( {\frac{{{z_2}}}{{{z_4}}}} \right) = {\text{Arg}}({z_1}) - {\text{Arg}}({z_3}) + {\text{Arg}}({z_2}) - {\text{Arg}}({z_4}) \\ {\text{ = }} - {\text{Arg}}({z_2}) - ( - {\text{Arg}}({z_4})) + {\text{Arg}}({z_2}) - {\text{Arg}}({z_4}) \\ {\text{ = 0}} \\\end{aligned} \end{equation} $$
