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

If $x+iy = \frac{a+ib}{a-ib}$,prove that $x^{2}+y^{2}=1$.

$\left| {(1 + i)\frac{{(2 + i)}}{{(3 + i)}}} \right| = $

If $z=x+iy$ satisfies the equation $z^2+az+a^2=0$,where $a \in R$,then:

If ${z_1}$ and ${z_2}$ are any two complex numbers,then $|{z_1} + {z_2}|^2 + |{z_1} - {z_2}|^2$ is equal to:

The real part of $(1 - \cos \theta + 2i\sin \theta )^{-1}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module