Let a complex number z,|z| neq 1,satisfy log_ | vedclass

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

Question

(B) Given the inequality: $\log_{\frac{1}{\sqrt{2}}} \left( \frac{|z|+11}{(|z|-1)^2} \right) \leq 2$.Since the base $\frac{1}{\sqrt{2}} < 1$,the inequ

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(B) Given the inequality: $\log_{\frac{1}{\sqrt{2}}} \left( \frac{|z|+11}{(|z|-1)^2} ight) \leq 2$.Since the base $\frac{1}{\sqrt{2}} $\frac{|z|+11}{(|z|-1)^2} \geq \left( \frac{1}{\sqrt{2}} ight)^2 = \frac{1}{2}$.Cross-multiplying (since $(|z|-1)^2 > 0$ for $|z| 
eq 1$):$2(|z|+11) \geq (|z|-1)^2$.$2|z| + 22 \geq |z|^2 - 2|z| + 1$.Rearranging the terms:$|z|^2 - 4|z| - 21 \leq 0$.Factoring the quadratic expression:$(|z|-7)(|z|+3) \leq 0$.Since $|z| \geq 0$,we have $|z|+3 > 0$,so $|z|-7 \leq 0$,which implies $|z| \leq 7$.Given $|z| 
eq 1$,the largest value of $|z|$ is $7$.

Let a complex number $z$,$|z| \neq 1$,satisfy $\log_{\frac{1}{\sqrt{2}}} \left( \frac{|z|+11}{(|z|-1)^2} \right) \leq 2$. Then,the largest value of $|z|$ is equal to ............

Similar Questions

If $f(x)$ is a polynomial of degree $n$ with rational coefficients and $1+2i, 2-\sqrt{3}$ and $5$ are three roots of $f(x)=0$,then the least value of $n$ is

Consider the regions for complex number $z$ defined by $A: \frac{1}{\log_2 |z|} - \frac{1}{\log_2 |z| - 1} - 1 < 0$ and $B: \operatorname{Im}(z) = 0$. The range of values of $\operatorname{Re}(z)$ lying in the region $A \cap B$ is

If $2 + 3i$ is one of the roots of the equation $2x^3 - 9x^2 + kx - 13 = 0,$ where $k \in R,$ then the real root of this equation:

The value of $4x^3 - 4x^2 - 7x + 127$ when $x = \frac{4 + 5i}{2}$ is

If $z_1$ and $z_2$ are two unimodular complex numbers that satisfy $z_1^2 + z_2^2 = 5,$ then $(z_1 - \bar{z}_1)^2 + (z_2 - \bar{z}_2)^2$ is equal to -

Vedclass Test Series

Exam Paper Generator

Online Exam Module