Question

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

• A. $5$
• B. $8$
• C. $6$
• D. $7$

Answer 1

The correct option is D $7$

$\frac{|z|+11}{(|z|-1{)}^2}\ge \frac{1}{2}2|z|+22\ge (|z|-1{)}^{2}2|z|+22\ge |z{|}^2-2|z|+1|z{|}^2-4|z|-21\le 0(|z|-7\left)\right(|z|+3)\le 0\Rightarrow |z|\le 7\therefore |z{|}_{max}=7$