If log_{tan 30^{circ}} left( frac{2|z|^2 + 2| | vedclass

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

Question

(C) Given the inequality: $\log_{	an 30^{\circ}} \left( \frac{2|z|^2 + 2|z| - 3}{|z| + 1} ight) < -2$. Since $	an 30^{\circ} = \frac{1}{\sqrt{3}}$

Vedclass · Accepted Answer

$|z| > 2$

Vedclass · Answer

(C) Given the inequality: $\log_{	an 30^{\circ}} \left( \frac{2|z|^2 + 2|z| - 3}{|z| + 1} ight) Since $	an 30^{\circ} = \frac{1}{\sqrt{3}}$,the base is $0 When the base of a logarithm is between $0$ and $1$,the inequality sign reverses when removing the logarithm: $\frac{2|z|^2 + 2|z| - 3}{|z| + 1} > (\frac{1}{\sqrt{3}})^{-2}$. $\frac{2|z|^2 + 2|z| - 3}{|z| + 1} > 3$. $2|z|^2 + 2|z| - 3 > 3(|z| + 1)$. $2|z|^2 + 2|z| - 3 > 3|z| + 3$. $2|z|^2 - |z| - 6 > 0$. Factoring the quadratic expression: $(2|z| + 3)(|z| - 2) > 0$. Since $|z| \geq 0$,$2|z| + 3$ is always positive. Therefore,$|z| - 2 > 0$,which implies $|z| > 2$.

If $\log_{\tan 30^{\circ}} \left( \frac{2|z|^2 + 2|z| - 3}{|z| + 1} \right) < -2$,then:

Similar Questions

Sum of the moduli of the complex roots of the equation $(x^2+\frac{1}{x^2})-5(x+\frac{1}{x})+6=0$ is

Find the conjugate of $\frac{5i}{7+i}$.

If $\alpha_1, \alpha_2, \alpha_3$ respectively denote the moduli of the complex numbers $-i, \frac{1}{3}(1+i)$ and $-1+i$,then their increasing order is

If $a > 0$ and $z = \frac{(1+i)^2}{a-i}$,where $i = \sqrt{-1}$,has a magnitude of $\frac{2}{\sqrt{5}}$,then $\bar{z}$ is

The modulus of the conjugate of $z = \frac{-2+i}{(1-2i)^2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module