If z and omega are two non-zero complex numbe | vedclass

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(D) Given that $|z \omega| = 1$,we have $|z| |\omega| = 1$. Also,$\operatorname{Arg}(z) - \operatorname{Arg}(\omega) = \frac{\pi}{2}$,which implies $\

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$-i$

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(D) Given that $|z \omega| = 1$,we have $|z| |\omega| = 1$. Also,$\operatorname{Arg}(z) - \operatorname{Arg}(\omega) = \frac{\pi}{2}$,which implies $\operatorname{Arg}(\frac{z}{\omega}) = \frac{\pi}{2}$. Let $z = r_1 e^{i 	heta_1}$ and $\omega = r_2 e^{i 	heta_2}$. Then $|z| = r_1$ and $|\omega| = r_2$,so $r_1 r_2 = 1$. $\operatorname{Arg}(z) = 	heta_1$ and $\operatorname{Arg}(\omega) = 	heta_2$,so $	heta_1 - 	heta_2 = \frac{\pi}{2}$. We need to find $\bar{z} \omega$. $\bar{z} = r_1 e^{-i 	heta_1}$. $\bar{z} \omega = (r_1 e^{-i 	heta_1}) (r_2 e^{i 	heta_2}) = (r_1 r_2) e^{i(	heta_2 - 	heta_1)}$. Since $r_1 r_2 = 1$ and $	heta_2 - 	heta_1 = -\frac{\pi}{2}$,we have: $\bar{z} \omega = 1 \cdot e^{-i \frac{\pi}{2}} = \cos(-\frac{\pi}{2}) + i \sin(-\frac{\pi}{2}) = 0 - i = -i$.

If $z$ and $\omega$ are two non-zero complex numbers such that $|z \omega|=1$ and $\operatorname{Arg}(z) - \operatorname{Arg}(\omega) = \frac{\pi}{2}$,then $\bar{z} \omega =$

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