Question

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

• A. $5$
• B. $8$
• C. $6$
• D. $7$

Answer 1

The correct option is D

$7$

Explanation for correct answer:

Finding the maximum value of $\left|z\right|$:

Given$\left|z\right|\ne 1$ satisfies

If ${\log }_{a}b>m$

$\Rightarrow b>a^m\text{if}a>1\text{or}b<a^m\text{if}a<1$

Let's consider the given equation,

$$\begin{gathered} {\log }_{\frac{1}{\sqrt{2}}}\left(\frac{\left|z\right|+11}{{\left(\left|z\right|-1\right)}^2}\right)\le 2 \\ \Rightarrow \left(\frac{\left|z\right|+11}{{\left(\left|z\right|-1\right)}^2}\right)\ge {\left(\frac{1}{\sqrt{2}}\right)}^2\left[\because \frac{1}{\sqrt{2}}<1\right] \\ \Rightarrow \left(\frac{\left|z\right|+11}{{\left(\left|z\right|-1\right)}^2}\right)\ge \frac{1}{2} \\ \Rightarrow 2\left(\left|z\right|+11\right)\ge 1{\left(\left|z\right|-1\right)}^2 \\ \Rightarrow 2\left|z\right|+22\ge {\left|z\right|}^2-2\left|z\right|+1 \end{gathered}$$

$$\begin{gathered} \Rightarrow {\left|z\right|}^2-4\left|z\right|-21\le 0 \\ \Rightarrow {\left|z\right|}^2-7\left|z\right|+3\left|z\right|-21\le 0 \\ \Rightarrow \left(\left|z\right|-7\right)\left(\left|z\right|+3\right)\le 0 \end{gathered}$$

$\Rightarrow \left|z\right|\le 7\text{or}\left|z\right|\le -3$

$\left|z\right|$ is always positive.

Therefore, the maximum value of ${\left|z\right|}_{max}=7$.

Hence, option (D) is the correct answer.