If {z_1} = a + ib and {z_2} = c + id are comp | vedclass

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(D) Given $|{z_1}| = |{z_2}| = 1$,we have ${z_1} = \cos {	heta _1} + i\sin {	heta _1}$ and ${z_2} = \cos {	heta _2} + i\sin {	heta _2}$,where ${

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All the above

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(D) Given $|{z_1}| = |{z_2}| = 1$,we have ${z_1} = \cos {	heta _1} + i\sin {	heta _1}$ 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 $R[(\cos {	heta _1} + i\sin {	heta _1})(\cos {	heta _2} - i\sin {	heta _2})] = \cos({	heta _1} - {	heta _2}) = 0$.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 $|{w_1}|^2 = \cos^2 {	heta _1} + \cos^2 {	heta _2} = \cos^2 {	heta _1} + \sin^2 {	heta _1} = 1$,so $|{w_1}| = 1$.Similarly,$|{w_2}|^2 = b^2 + d^2 = \sin^2 {	heta _1} + \sin^2 {	heta _2} = \sin^2 {	heta _1} + \cos^2 {	heta _1} = 1$,so $|{w_2}| = 1$.Finally,$R({w_1}\overline {{w_2}} ) = R((a + ic)(b - id)) = R(ab - iad + ibc + cd) = ab + cd = \cos {	heta _1}\sin {	heta _1} + \cos {	heta _2}\sin {	heta _2} = \frac{1}{2}(\sin 2{	heta _1} + \sin 2{	heta _2})$.Since ${	heta _1} = {	heta _2} + \frac{\pi }{2}$,$\sin 2{	heta _1} = \sin(2{	heta _2} + \pi) = -\sin 2{	heta _2}$,so $R({w_1}\overline {{w_2}} ) = 0$.Thus,all the above 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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