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(C) Given $|z \omega| = 1$ and $\arg(z) - \arg(\omega) = \frac{3 \pi}{2}$.Let $z = r e^{i 	heta_1}$ and $\omega = \frac{1}{r} e^{i 	heta_2}$.Then $\

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$-\frac{3 \pi}{4}$

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(C) Given $|z \omega| = 1$ and $\arg(z) - \arg(\omega) = \frac{3 \pi}{2}$.Let $z = r e^{i 	heta_1}$ and $\omega = \frac{1}{r} e^{i 	heta_2}$.Then $\bar{z} = r e^{-i 	heta_1}$.Thus,$\bar{z} \omega = r e^{-i 	heta_1} \cdot \frac{1}{r} e^{i 	heta_2} = e^{i(	heta_2 - 	heta_1)}$.Since $	heta_1 - 	heta_2 = \frac{3 \pi}{2}$,we have $	heta_2 - 	heta_1 = -\frac{3 \pi}{2} = \frac{\pi}{2} - 2\pi \equiv \frac{\pi}{2} \pmod{2\pi}$.So,$\bar{z} \omega = e^{i \pi/2} = i$.Now,substitute this into the expression:$\frac{1 - 2 \bar{z} \omega}{1 + 3 \bar{z} \omega} = \frac{1 - 2i}{1 + 3i}$.To find the argument,multiply by the conjugate of the denominator:$\frac{1 - 2i}{1 + 3i} 	imes \frac{1 - 3i}{1 - 3i} = \frac{1 - 3i - 2i + 6i^2}{1^2 + 3^2} = \frac{1 - 5i - 6}{10} = \frac{-5 - 5i}{10} = -\frac{1}{2} - \frac{1}{2}i$.This complex number lies in the third quadrant.The argument is $	an^{-1}\left(\frac{-1/2}{-1/2}ight) - \pi = 	an^{-1}(1) - \pi = \frac{\pi}{4} - \pi = -\frac{3 \pi}{4}$.

List-$I$	List-$II$
$(P)$ $\|z\|^2$ is equal to	$(1)$ $12$
$(Q)$ $\|z-\bar{z}\|^2$ is equal to	$(2)$ $4$
$(R)$ $\|z\|^2+\|z+\bar{z}\|^2$ is equal to	$(3)$ $8$
$(S)$ $\|z+1\|^2$ is equal to	$(4)$ $10$
	$(5)$ $7$

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 $z$)

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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 $z$)