Mathematics - Complex Number Question with Solution | TestHub

${\log }_{\frac{1}{\sqrt{2}}}\left(\frac{\left|z\right|+11}{{\left(\left|z\right|-1\right)}^2}\right)\le 2$

$\Rightarrow \frac{\left|z\right|+11}{{\left(\left|z\right|-1\right)}^2}\ge \frac{1}{2}$, $\left\{\because {\log }_a\left(b\right)\le c\Rightarrow b\ge a^c if 0<a<1\right\}$

$\Rightarrow 2\left|z\right|+22\ge {\left(\left|z\right|-1\right)}^2$, $\left\{\because \left|z\right|\ne 1 \& {\left(\left|z\right|-1\right)}^2>0\right\}$

$\Rightarrow 2\left|z\right|+22\ge {\left|z\right|}^2+1-2\left|z\right|$

$\Rightarrow {\left|z\right|}^2-4\left|z\right|-21\le 0$

$\Rightarrow \left(\left|z\right|-7\right)\left(\left|z\right|+3\right)\le 0$

$\Rightarrow \left|z\right|\le 7$

$\therefore$ Largest value of $\left|z\right|$ is $7$