If a > 0 and z = x + i y , then log cos 2 θ | z - a | > log cos 2 θ | z - a i | , ( θ…

${\log }_{{\cos }^2\theta }|z-a|>{\log }_{{\cos }^2\theta }|z-ai|$

${\cos }^2\theta \ne \left\{0, 1\right\}$, since it is base of the logarithm

So, ${\cos }^2\theta \in \left(0,1\right)$

$\Rightarrow |z-a|<|z-ai|$

$\Rightarrow |x+yi-a|<|x+yi-ai|$

$\Rightarrow \left|\right(x-a)+yi|<|x+(y-a\left)i\right|$

$\Rightarrow x^2+y^2+a^2-2ax<x^2+y^2+a^2-2ay$

$\Rightarrow -2ax<-2ay$

$\Rightarrow x>y$