Question

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

Answer 1

$z=r{e}^{i\theta }$ $\therefore \omega =\frac{1}{r}{e}^{i(\theta -3\pi /2)}$
$\frac{1-2\overline{z}\omega }{1+3\overline{z}\omega }=\frac{1-2e^{-i\theta }\cdot e^{i(-3\pi /2+\theta )}}{1+3e^{-i\theta }\cdot e^{i(-3\pi /2+\theta )}}=\frac{1-2i}{1+3i}$

$\therefore \arg \left(\frac{1-2i}{1+3i}\right)=\arg (-\frac{5}{10}-\frac{5}{10}i)=-\frac{3\pi }{4}$