Question

If $z$ and $\omega$ are two complex numbers such that $|z \omega|=1$ and $\arg (z)-\arg (\omega)=\frac{3 \pi}{2}$, then $\arg \left(\frac{1-2 \bar{z} \omega}{1+3 z \omega}\right)$ is :
(Here $\arg (z)$ denotes the principal argument of complex number $z$ )

• A. $\frac{\pi}{4}$
• B. $-\frac{3 \pi}{4}$
• C. $-\frac{\pi}{4}$
• D. $\frac{3 \pi}{4}$

Answer 1

Answer: (C)

As $|z \omega|=1$
$\Rightarrow| f | z |=r$, then $|\omega|=\frac{1}{r}$
Let $\arg (z)=\theta$
$\therefore \arg (\omega)=\left(\theta-\frac{3 \pi}{2}\right)$
So, $z=r e^{i \theta}$
$\Rightarrow \bar{Z}=r e^{i(-\theta)} $
$\omega=\frac{1}{r} e^{i\left(\theta-\frac{3 \pi}{2}\right)}$
Now, consider
$\frac{1-2 \bar{z} \omega}{1+3 \bar{z} \omega}=\frac{1-2 e^{i\left(-\frac{3 \pi}{2}\right)}}{1+3 e^{\left(-\frac{3 \pi}{2}\right)}}=\left(\frac{1-2 i}{1+3 i}\right)$
$=\frac{(1-2 i)(1-3 i)}{(1+3 i)(1-3 i)}=-\frac{1}{2}(1+i)$
$\therefore $ prin $\arg \left(\frac{1-2 \bar{z} \omega}{1+3 \bar{z} \omega}\right)$
$=$ prin $\arg \left(\frac{1-2 \bar{z} \omega}{1+3 \bar{z} \omega}\right)$
$=\left(-\frac{1}{2}(1+i)\right)$
$=-\left(\pi-\frac{\pi}{4}\right)=\frac{-3 \pi}{4}$