As $|z \omega|=1$

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

Let $\arg (z)=q$

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

$\text { So, } z=r e^{1 \theta}$

$\Rightarrow \bar{z}=r e^{i(-\theta)}$

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

Now, consider

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

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

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

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

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