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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$x > y$

Vedclass · Answer

(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:

Similar Questions

If $\cosh ^{-1} x = 2 \log _e(\sqrt{2}+1)$,then $x=$

The real part of the principal value of $2^{-i}$ is

$\sqrt{12-\sqrt{68+48 \sqrt{2}}}$ is equal to :

The real part of $(1 - i)^{-i}$ is

$\sqrt[4]{17 + 12\sqrt{2}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module