Question

If $z_1,z_2,z_3\text{and}{z}_4$ are any four complex numbers such that $|z_1|<\frac{1}{\pi },|z_2|=1,|z_3|\le 1$ and $z_3=\frac{z_2(z_1-z_4)}{\overline{z_1}{z}_4-1}.$ Then possible value of $|z_4|$ is /are

• A. $2{\text{log}}_{3}2$
• B. ${\text{log}}_{3}4$
• C. ${\text{log}}_e\pi$
• D. ${\text{log}}_{\pi }e$

Answer 1

The correct option is D ${\text{log}}_{\pi }e$

$\because \frac{z_3}{z_2}=\frac{z_1-z_4}{z_1{z}_4-1}$
and ${\left|\frac{z_3}{z_2}\right|}^2\le 1$

$\Rightarrow \left(\frac{z_1-z_4}{z_1{z}_4-1}\right)\left(\frac{\overline{z_1}-\overline{z_4}}{z_1\overline{z_4}-1}\right)\le 1$

$\Rightarrow z_1\overline{z_1}+z_4\overline{z_4}-\overline{z_1}{z}_4-\overline{z_4}{z}_1\le$

$z_1\overline{z_1}{z}_4\overline{z_4}+1-z_1\overline{z_4}-\overline{z_1}{z}_4$

$\Rightarrow |z_1{|}^2|z_4{|}^2-|z_1{|}^2-|z_4{|}^2+1\ge 0$

$\Rightarrow (|z_1{|}^2-1)(|z_4{|}^2-1)\ge 0$

$\because |z_1{|}^2<1$

So, $|z_4{|}^2\le 1$