If a 0 and z = x + iy,then log_{cos^2 theta | vedclass

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(A) Given the inequality: $\log_{\cos^2 	heta} |z - a| > \log_{\cos^2 	heta} |z - ai|$.Since $\cos^2 	heta$ is the base of the logarithm,we must ha

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$x > y$

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(A) Given the inequality: $\log_{\cos^2 	heta} |z - a| > \log_{\cos^2 	heta} |z - ai|$.Since $\cos^2 	heta$ is the base of the logarithm,we must have $\cos^2 	heta \in (0, 1)$.Because the base is between $0$ and $1$,the inequality reverses when removing the logarithm:$|z - a| Substituting $z = x + iy$:$|x + iy - a| $|(x - a) + iy| Squaring both sides:$(x - a)^2 + y^2 $x^2 - 2ax + a^2 + y^2 Subtracting $x^2 + y^2 + a^2$ from both sides:$-2ax Since $a > 0$,dividing by $-2a$ reverses the inequality:$x > y$.

If $a > 0$ and $z = x + iy$,then $\log_{\cos^2 \theta} |z - a| > \log_{\cos^2 \theta} |z - ai|$ for $\theta \in R$ implies:

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