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If $a > 0$ and $z = x + i y$ then $\log_{\cos^{2} θ}^{| z - a |} > \log_{\cos^{2} θ}^{| z - a i |} (θ \in R)$ implies

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$x > y$

$x < y$

$x + y = \cos θ$

$x + y < 0$

answer is A.

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Detailed Solution

a > 0 z = x+iy
0 < cos² $θ$ < 1
$| z - a | < | z - a i | | (x - a) + i y | < | x + i (x - a) | {(x - a)}^{2} + y^{2} < x^{2} + {(y - a)}^{2} - 2 a x < - 2 a y x > y$

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