Question

If $\left|\frac{z_1-2z_2}{2-z_1\overline{z_2}}\right|=1,then|z_1{|}^2+m|z_2{|}^2=m+|z_1{|}^2|z_2{|}^2.Findm$

Answer 1

$\to \left|\frac{z_1-2z_2}{2-z_1\overline{z_2}}\right|=1$

$\to |z_1-2z_2|=|2-z_1\overline{z_2}|$

On squaring,

$\to {\left|z_1\right|}^2+4{\left|z_2\right|}^2-2Re\left(2z_1\overline{z_2}\right)=2^2+{\left|z_1\overline{z_2}\right|}^2-Re\left(2z_1\overline{z_2}\right)$

$\to {\left|z_1\right|}^2+4{\left|z_2\right|}^2=4+{\left|z_1\overline{z_2}\right|}^2$

On comparing with

${\left|z_1\right|}^2+m{\left|z_2\right|}^2=m+{\left|z_1\overline{z_2}\right|}^2$

$m=4$