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

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

जहाँ ${\theta _1} = arg({z_1})$ व ${\theta _2} = arg({z_2})$   

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

अत: $a = \cos {\theta _1}$,$b = \sin {\theta _1},c = \cos {\theta _2}$ तथा

$d = \sin {\theta _2}$

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

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

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

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

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

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

 $|{w_1}| = 1$

 इसी प्रकार, $|{w_2}| = 1$

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

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

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

अन्तत: $R({\overline w _1}{w_2}) = R({w_2}{\overline w _1})$

$ = 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$