If z and

Solution:

As $|z \omega|=1$

$\Rightarrow \operatorname{If}|z|=r$, then $|\omega|=\frac{1}{r}$

Let $\arg (\mathrm{z})=\theta$

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

So, z $=\mathrm{re}^{\mathrm{i} \theta}$

$\Rightarrow \overline{\mathrm{Z}}=\mathrm{re}^{\mathrm{i}(-\theta)}$

$\omega=\frac{1}{\mathrm{r}} \mathrm{e}^{\mathrm{i}\left(\theta-\frac{3 \pi}{2}\right)}$

Now, consider

$\frac{1-2 \bar{z} \omega}{1+3 \bar{z} \omega}=\frac{1-2 e^{i\left(-\frac{3 \pi}{2}\right)}}{1+3 e^{i\left(\frac{3 \pi}{2}\right)}}=\left(\frac{1-2 i}{1+3 i}\right)$

$=\frac{(1-2 i)(1-3 i)}{(1+3 i)(1-3 i)}=-\frac{1}{2}(1+i)$

$\therefore$ prin arg $\left(\frac{1-2 \bar{z} \omega}{1+3 \bar{z} \omega}\right)$

$=\operatorname{prin} \arg \left(\frac{1-2 \bar{z} \omega}{1+3 \bar{z} \omega}\right)$

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

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

So, option (2) is correct.