If $z$ and $\omega$ are two complex numbers such that $|z \omega|=1$ and $\arg (z)-\arg (\omega)=\frac{3 \pi}{2}$, then $\arg \left(\frac{1-2 \bar{z} \omega}{1+3 \bar{z} \omega}\right)$ is :
(Here $\arg (z)$ denotes the principal argument of complex number $z$ )
Correct Option: , 2
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.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.