Let z_1, z_2 be two complex numbers such that | vedclass

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(C) Given $\bar{z}_1 - i \bar{z}_2 = 0$. Taking the conjugate on both sides,we get $z_1 + i z_2 = 0$,which implies $z_1 = -i z_2$. We know that $-i =

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$\frac{\pi}{8}$

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(C) Given $\bar{z}_1 - i \bar{z}_2 = 0$. Taking the conjugate on both sides,we get $z_1 + i z_2 = 0$,which implies $z_1 = -i z_2$. We know that $-i = e^{-i \pi / 2}$,so $z_1 = z_2 e^{-i \pi / 2}$. Taking the argument on both sides,$\arg(z_1) = \arg(z_2) - \frac{\pi}{2}$,which implies $\arg(z_2) = \arg(z_1) + \frac{\pi}{2}$. Given $\arg(z_1 z_2) = \arg(z_1) + \arg(z_2) = \frac{3 \pi}{4}$. Substituting $\arg(z_2)$,we get $\arg(z_1) + (\arg(z_1) + \frac{\pi}{2}) = \frac{3 \pi}{4}$. $2 \arg(z_1) = \frac{3 \pi}{4} - \frac{\pi}{2} = \frac{\pi}{4}$. Therefore,$\arg(z_1) = \frac{\pi}{8}$.

List-$I$ (Complex number)	List-$II$ (Polar form)
$(i) \sqrt{3}-i$	$(a) 2 \operatorname{cis} \frac{\pi}{6}$
$(ii) \sqrt{3}+i$	$(b) 2 \operatorname{cis} \frac{5 \pi}{6}$
$(iii) -\sqrt{3}+i$	$(c) 2 \operatorname{cis}\left(-\frac{5 \pi}{6}\right)$
$(iv) -\sqrt{3}-i$	$(d) 2 \operatorname{cis}\left(-\frac{\pi}{6}\right)$

Let $z_1, z_2$ be two complex numbers such that $\bar{z}_1 - i \bar{z}_2 = 0$ and $\arg(z_1 z_2) = \frac{3 \pi}{4}$,then $\arg(z_1) =$

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