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જો $z_1 , z_2$ અને $z_3, z_4$ એ $2$ અનુબધ્ધ સંકર સંખ્યાની જોડ હોય તો , $\arg \left( {\frac{{{z_1}}}{{{z_4}}}} \right) + \arg \left( {\frac{{{z_2}}}{{{z_3}}}} \right)$ = .......
$0$
$\frac{\pi}{2}$
$\frac{3\pi}{2}$
$\pi $
Solution
Consider $\arg \left(\frac{z_{1}}{z_{4}}\right)+$ $\arg \left(\frac{z_{2}}{z_{3}}\right)$
$=\arg \left(z_{1}\right)-\arg \left(z_{4}\right)$ $+\arg \left(z_{2}\right)-\arg \left(z_{3}\right)$
$=\left(\arg \left(z_{1}\right)+\arg \left(z_{2}\right)\right)$ $-\left(\arg \left(z_{3}\right)+\arg \left(z_{4}\right)\right)$
given $\left( {{z_2} = {{\bar z}_1} and {z_4} = {{\bar z}_3}} \right)$
$ = \left( {\arg \left( {{z_1}} \right) + \arg \left( {{{\bar z}_1}} \right)} \right) – $ $\left( {\arg \left( {{z_3}} \right) + \arg \left( {{{\bar z}_3}} \right)} \right)$
$\left\{ \begin{gathered}
\operatorname{also} (\arg \left( {{{\bar z}_1}} \right) = – \arg \left( {{z_1}} \right) \hfill \\
\arg \left( {{{\bar z}_3}} \right) = – \arg \left( {{z_3}} \right) \hfill \\
\end{gathered} \right\}$
$=\left(\arg \left(z_{1}\right)-\arg \left(z_{1}\right)\right)-\left(\arg \left(z_{3}\right)-\arg \left(z_{3}\right)\right)$
$=0-0=0$
