If z and ω are two complex numbers such that |zω| = 1 and arg(z) -...

$\left|z\omega\right| = 1$

If \left|z\right| = r$, then \left|\omega\right| = \dfrac{1}{r}$

Let $arg\left(z\right) = \theta$

$∴ arg\left(\omega\right) = \left(\theta - \dfrac{3\pi}{2}\right)$

So,

$z = r{e}^{i\theta}$

$⇒ \overset{¯}{z} = r{e}^{i\left(- \theta\right)}$
$⇒ \omega = \dfrac{1}{r}{e}^{i\left(\theta - \dfrac{3\pi}{2}\right)}$

Now, consider
$\dfrac{1 - 2\overset{¯}{z}\omega}{1 + 3\overset{¯}{z}\omega} = \dfrac{1 - 2{e}^{i\left(- \dfrac{3\pi}{2}\right)}}{1 + 3{e}^{i\left(- \dfrac{3\pi}{2}\right)}} = \left(\dfrac{1 - 2i}{1 + 3i}\right)$

$= \dfrac{\left(1 - 2i\right)\left(1 - 3i\right)}{\left(1 + 3i\right)\left(1 - 3i\right)} = \dfrac{1 - 5i + 6{i}^{2}}{10} = \dfrac{- 5 + 5i}{10} = - \dfrac{1}{2}\left(1 + i\right)$

Then,

$principalarg\left(\dfrac{1 - 2\overset{¯}{z}\omega}{1 + 3\overset{¯}{z}\omega}\right)$

$= principalarg\left(- \dfrac{1}{2}\left(1 + i\right)\right)$

$= - \left(\pi - \dfrac{\pi}{4}\right) = \dfrac{- 3\pi}{4}$