If z and $\omega$ are two complex numbers such that $\left| {z\omega } \right| = 1$ and $\arg (z) - \arg (\omega ) = {{3\pi } \over 2}$, then $$\arg \left( {{{1 - 2\overline z \omega } \over {1 + 3\overline z \omega }}} \right)$$ is :

(Here arg(z) denotes the principal argument of complex number z)

As $\left| {z\omega } \right| = 1$<br><br>$\Rightarrow$ If $\left| z \right| = r$, then $\left| \omega \right| = {1 \over r}$<br><br>Let $\arg (z) = \theta$<br><br>$\therefore$ $\arg (\omega ) = \left( {\theta - {{3\pi } \over 2}} \right)$<br><br>So, $z = r{e^{i\theta }}$<br><br>$\Rightarrow \overline z = r{e^{i\theta }}$<br><br>$\omega = {1 \over r}{e^{i\left( {\theta - {{3\pi } \over 2}} \right)}}$<br><br>Now, consider<br><br>$${{1 - 2\overline z \omega } \over {1 + 3\overline z \omega }} = {{1 - 2{e^{i\left( { - {{3\pi } \over 2}} \right)}}} \over {1 + 3{e^{i\left( { - {{3\pi } \over 2}} \right)}}}} = \left( {{{1 - 2i} \over {1 + 3i}}} \right)$$<br><br>$= {{(1 - 2i)(1 - 3i)} \over {(1 + 3i)(1 - 3i)}} = - {1 \over 2}(1 + i)$<br><br>$\therefore$ $$prin\arg \left( {{{1 - 2\overline z \omega } \over {1 + 3\overline z \omega }}} \right)$$<br><br>$$ = prin\arg \left( {{{1 - 2\overline z \omega } \over {1 + 3\overline z \omega }}} \right)$$<br><br>$= \left( { - {1 \over 2}(1 + i)} \right)$<br><br>$= - \left( {\pi - {\pi \over 4}} \right) = {{ - 3\pi } \over 4}$<br><br>So, option (2) is correct.