(d)Since $|{z_1}| = |{z_2}| = 1$, we have

${z_1} = \cos {\theta _1} + i\sin {\theta _1},{z_2} = \cos {\theta _2} + i\sin {\theta _2}$

where ${\theta _1} = arg({z_1})$ and ${\theta _2} = arg({z_2})$

Also, ${z_1} = a + ib$ and ${z_2} = c + id.$

Therefore$a = \cos {\theta _1}$,$b = \sin {\theta _1},c = \cos {\theta _2}$ and $d = \sin {\theta _2}$

Also, $R({z_1}{\overline z _2}) = 0$

==> $R[(\cos {\theta _1} + i\sin {\theta _1})(\cos {\theta _2} – i\sin {\theta _2})] = 0$

==> $R[(\cos ({\theta _1} – {\theta _2}) + i\sin ({\theta _1} – {\theta _2})] = 0$

==> $\cos ({\theta _1} – {\theta _2}) = 0$

==> ${\theta _1} – {\theta _2} = \frac{\pi }{2}$

==> ${\theta _1} = {\theta _2} + \frac{\pi }{2}$

Now, ${w_1} = a + ic = \cos {\theta _1} + i\cos {\theta _2}$$ = \cos {\theta _1} + i\sin {\theta _1}$

==> $|{w_1}| = 1$

Similarly, $|{w_2}| = 1$

Next ${w_1}{\overline w _2} = (\cos {\theta _1} + i\sin {\theta _1})(\cos {\theta _2} – i\sin {\theta _2})$

$ = \cos ({\theta _1} – {\theta _2}) + i\sin ({\theta _1} – {\theta _2})$

==> $|{w_1}{\overline w _2}| = 1$

Finally, $R({\overline w _1}{w_2}) = R({w_2}{\overline w _1})$

$ = R[(\cos {\theta _2} + i\sin {\theta _2})(\cos {\theta _1} – i\sin {\theta _1})]$

$ = R[\cos ({\theta _2} – {\theta _1}) + i\sin ({\theta _2} – {\theta _1})]$

$ = \cos ({\theta _2} – {\theta _1}) = \cos \left( {\frac{{ – \pi }}{2}} \right) = 0$