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If $z_1 , z_2$ and $z_3, z_4$ are $2$ pairs of complex conjugate numbers, then $\arg \left( {\frac{{{z_1}}}{{{z_4}}}} \right) + \arg \left( {\frac{{{z_2}}}{{{z_3}}}} \right)$ equals
$0$
$\frac{\pi}{2}$
$\frac{3\pi}{2}$
$\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) – $ $\left( {\arg \left( {{z_3}} \right) + \arg \left( {{{\bar z}_3}} \right)} \right)$
$\left\{ \begin{gathered}
\operatorname{also} (\arg \left( {{{\bar z}_1}} \right) = – \arg \left( {{z_1}} \right) \hfill \\
\arg \left( {{{\bar z}_3}} \right) = – \arg \left( {{z_3}} \right) \hfill \\
\end{gathered} \right\}$
$=\left(\arg \left(z_{1}\right)-\arg \left(z_{1}\right)\right)-\left(\arg \left(z_{3}\right)-\arg \left(z_{3}\right)\right)$
$=0-0=0$
Similar Questions
Let $z$ be complex number satisfying $|z|^3+2 z^2+4 z-8=0$, where $\bar{z}$ denotes the complex conjugate of $z$. Let the imaginary part of $z$ be nonzero.
Match each entry in List-$I$ to the correct entries in List-$II$.
| List-$I$ | List-$II$ |
| ($P$) $|z|^2$ is equal to | ($1$) $12$ |
| ($Q$) $|z-\bar{z}|^2$ is equal to | ($2$) $4$ |
| ($R$) $|z|^2+|z+\bar{z}|^2$ is equal to | ($3$) $8$ |
| ($S$) $|z+1|^2$ is equal to | ($4$) $10$ |
| ($5$) $7$ |
The correct option is:
