For two complex numbers $z_{1} a n d z_{2}$ we have $| \frac{z_{1} - z_{2}}{1 - \bar{z_{1}} z_{2}} | = 1,$ then

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$| z_{1} | = 2 | z_{2} |$

At least one of $z_{1} a n d z_{2}$ lies on the circle $| z | = 1$

$\arg (\frac{z_{1}}{z_{2}}) = \frac{π}{3}$

Both $z_{1} a n d z_{2}$ lie on circle $| z | = 1$

answer is A.

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Detailed Solution

$| \frac{z_{1} - z_{2}}{1 - \bar{z_{1}} z_{2}} | = 1$
$\Rightarrow | z_{1} - z_{2} | = | 1 - \bar{z_{1}} z_{2} | \Rightarrow {| z_{1} - z_{2} |}^{2} = {| 1 - \bar{z_{1}} z_{2} |}^{2}$
$\Rightarrow {| z_{1} |}^{2} {| z_{2} |}^{2} - {| z_{1} |}^{2} - {| z_{2} |}^{2} + 1 = 0$
$\Rightarrow | z_{1} | = 1, | z_{2} | = 1$ Both $z_{1} a n d z_{2}$ lie on the circle $| z | = 1$