If z and w are complex numbers such that bar{ | vedclass

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(B) Given $\bar{z} - i \bar{w} = 0$,we have $\bar{z} = i \bar{w}$.Taking the conjugate on both sides,$z = -i w$,which implies $w = \frac{z}{-i} = iz$.

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$\frac{\pi}{8}$

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(B) Given $\bar{z} - i \bar{w} = 0$,we have $\bar{z} = i \bar{w}$.Taking the conjugate on both sides,$z = -i w$,which implies $w = \frac{z}{-i} = iz$.Now,$\operatorname{Arg}(zw) = \operatorname{Arg}(z(iz)) = \operatorname{Arg}(iz^2) = \frac{3 \pi}{4}$.Using the property $\operatorname{Arg}(z_1 z_2) = \operatorname{Arg}(z_1) + \operatorname{Arg}(z_2)$,we get $\operatorname{Arg}(i) + \operatorname{Arg}(z^2) = \frac{3 \pi}{4}$.Since $\operatorname{Arg}(i) = \frac{\pi}{2}$ and $\operatorname{Arg}(z^2) = 2 \operatorname{Arg}(z)$,we have $\frac{\pi}{2} + 2 \operatorname{Arg}(z) = \frac{3 \pi}{4}$.$2 \operatorname{Arg}(z) = \frac{3 \pi}{4} - \frac{\pi}{2} = \frac{\pi}{4}$.Therefore,$\operatorname{Arg}(z) = \frac{\pi}{8}$.

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

If $z$ and $w$ are complex numbers such that $\bar{z} - i \bar{w} = 0$ and $\operatorname{Arg}(zw) = \frac{3 \pi}{4}$,then $\operatorname{Arg} z =$

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