If z_1 = a + ib and z_2 = c + id are complex | vedclass

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(D) Given $|z_1| = |z_2| = 1$,we can write $z_1 = \cos 	heta_1 + i \sin 	heta_1$ and $z_2 = \cos 	heta_2 + i \sin 	heta_2$,where $	heta_1 = \arg(

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(D) Given $|z_1| = |z_2| = 1$,we can write $z_1 = \cos 	heta_1 + i \sin 	heta_1$ and $z_2 = \cos 	heta_2 + i \sin 	heta_2$,where $	heta_1 = \arg(z_1)$ and $	heta_2 = \arg(z_2)$.Since $z_1 = a + ib$ and $z_2 = c + id$,we have $a = \cos 	heta_1, b = \sin 	heta_1, c = \cos 	heta_2, d = \sin 	heta_2$.Given $R(z_1\overline{z_2}) = 0$,we have $R[(\cos 	heta_1 + i \sin 	heta_1)(\cos 	heta_2 - i \sin 	heta_2)] = 0$.This simplifies to $R[\cos(	heta_1 - 	heta_2) + i \sin(	heta_1 - 	heta_2)] = \cos(	heta_1 - 	heta_2) = 0$.Thus,$	heta_1 - 	heta_2 = \pm \frac{\pi}{2}$,which implies $	heta_1 = 	heta_2 \pm \frac{\pi}{2}$.Now,$w_1 = a + ic = \cos 	heta_1 + i \cos 	heta_2$. Since $\cos 	heta_2 = \sin 	heta_1$ (or $-\sin 	heta_1$),we have $|w_1| = \sqrt{\cos^2 	heta_1 + \sin^2 	heta_1} = 1$.Similarly,$w_2 = b + id = \sin 	heta_1 + i \sin 	heta_2$. Since $\sin 	heta_2 = \cos 	heta_1$ (or $-\cos 	heta_1$),we have $|w_2| = \sqrt{\sin^2 	heta_1 + \cos^2 	heta_1} = 1$.Finally,$w_1\overline{w_2} = (a + ic)(b - id) = (ab + cd) + i(bc - ad)$.$R(w_1\overline{w_2}) = ab + cd = \cos 	heta_1 \sin 	heta_1 + \cos 	heta_2 \sin 	heta_2 = \frac{1}{2}(\sin 2	heta_1 + \sin 2	heta_2)$.Using $	heta_1 = 	heta_2 + \frac{\pi}{2}$,we get $\sin 2	heta_1 = \sin(2	heta_2 + \pi) = -\sin 2	heta_2$,so $ab + cd = 0$.Thus,all conditions are satisfied.

If $z_1 = a + ib$ and $z_2 = c + id$ are complex numbers such that $|z_1| = |z_2| = 1$ and $R(z_1\overline{z_2}) = 0$,then the pair of complex numbers $w_1 = a + ic$ and $w_2 = b + id$ satisfies

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