If z1=a+ib and z2=c+id are complex numbers such that z1=z2=1 and Rez1z2¯=0, then the pair of complex numbers w1=a+ic and w2=b+idsatisfies

w 1 = 1 , w 2 = 1 , Re w 1 w 2 ¯ = 0

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If $z_{1} = a + i b a n d z_{2} = c + i d$ are complex numbers such that $|z_{1}| = |z_{2}| = 1 a n d Re (z_{1} \bar{z_{2}}) = 0,$ then the pair of complex numbers $w_{1} = a + i c a n d w_{2} = b + i d$ satisfies

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$|w_{1}| = 1$

$|w_{2}| = 1$

$Re (w_{1} \bar{w_{2}}) = 0$

None of these

answer is A, B, C.

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We have $a^{2} + b^{2} = 1$

$c^{2} + d^{2} = 1$

Also, $z_{1} \bar{z_{2}} = (a + i b) (c - i d)$

$= (a c + b d) + i (b c - a d)$

$Re (z_{1} \bar{z_{2}}) = 0$

$\Rightarrow a c + b d = 0 \Rightarrow \frac{a}{b} = - \frac{d}{c} = k (s a y)$

So, from eqn.(i) and (ii), $(1 + k^{2}) b^{2} = 1 a n d (1 + k^{2}) c^{2} = 1$

Thus, $b^{2} = c^{2} a n d a^{2} = d^{2}$

Now, $|w_{1}| = \sqrt{a^{2} + c^{2}} = \sqrt{a^{2} + b^{2}} = 1$

And $|w_{2}| = \sqrt{b^{2} + d^{2}} = \sqrt{b^{2} + a^{2}} = 1$

$w_{1} \bar{w_{2}} = (a + i c) (b - i d)$

$= (a b + c d) + i (b c - a d)$

$Re (w_{1} \bar{w_{2}}) = a b + c d = k b^{2} - k c^{2} = k (b^{2} - c^{2}) = 0$

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