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

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(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

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$7$

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(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 ............

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