If z and are two complex numbers such that |z|=1 and (z)-()=32, then !(1-2z1+3z) is

Question

If $z$ and $\omega$ are two complex numbers such that $|z\omega|=1$ and $\arg(z)-\arg(\omega)=\dfrac{3\pi}2$, then $\arg\!\left(\dfrac{1-2z\omega}{1+3z\omega}\right)$ is

Accepted Answer

With $|z\omega|=1$ and the given argument difference, $z\omega=i$.

$$\frac{1-2i}{1+3i}=\frac{(1-2i)(1-3i)}{10}=\frac{-5-5i}{10}=-\frac12-\frac12i,$$

whose argument is $-\pi+\dfrac\pi4=-\dfrac{3\pi}4$.

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