If z and w are two complex numbers such that | vedclass

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(D) Let $z = r_1 e^{i	heta_1}$ and $w = r_2 e^{i	heta_2}$.Given $|zw| = |z||w| = r_1 r_2 = 1$,so $r_2 = \frac{1}{r_1}$.Given $\arg(z) - \arg(w) =

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$\bar{z}w = -i$

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(D) Let $z = r_1 e^{i	heta_1}$ and $w = r_2 e^{i	heta_2}$.Given $|zw| = |z||w| = r_1 r_2 = 1$,so $r_2 = \frac{1}{r_1}$.Given $\arg(z) - \arg(w) = 	heta_1 - 	heta_2 = \frac{\pi}{2}$,so $	heta_1 = 	heta_2 + \frac{\pi}{2}$.Now,consider $\bar{z}w = (r_1 e^{-i	heta_1})(r_2 e^{i	heta_2}) = (r_1 r_2) e^{i(	heta_2 - 	heta_1)}$.Substituting the values,$\bar{z}w = (1) e^{i(-\pi/2)} = \cos(-\pi/2) + i\sin(-\pi/2) = -i$.Thus,$\bar{z}w = -i$.

If $z$ and $w$ are two complex numbers such that $|zw| = 1$ and $\arg(z) - \arg(w) = \frac{\pi}{2}$,then

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