A complex number $z_1$ is unimodular. If $\frac{z_1-2z_2}{2-z_1\bar z_2}$ is unimodular, and $z_2$ is not unimodular, then prove that $z_2$ lies on a circle of radius 2.
Since the given expression is unimodular
$$|z_1-2z_2|=|2-z_1\bar z_2|$$
Which is the same as
$$|z_1-2z_2|=|2-\bar z_1z_2|$$
Since conjugates inside a modulus are equal
So $$|z_1-2z_2|=|2-\frac{z_2}{z_1}|$$
$$|z_1-2z_2|=|2z_1-z_2|$$
What should I do next?
On
The acclaimed result is^ ** not true**. Check, if $z_1=1; z_2=2$ but if $$w=\frac{z_1-2z_2}{2-z_1 \bar z_2}\implies |w|=\infty.$$ OP's approach is correct then further to his last eq. we get $$|z_1-2z_2|^2=|2z_1-z_2|^2 \implies (z_1 -2 z_2)(\bar z_1-2\bar z_2)=(2z_1-z_2)(2\bar z_1-\bar z_2)\implies -3|z_1|^2+3 |z_2|^2=0 \implies |z_1|=|z_2|$$ and since $|z_1|=1$ hence $$|z_2|=1$$
There seems to be a contradiction. From the given $|z_1-2z_2|=|2-z_1\bar z_2|$, we have
$$(z_1-2z_2)(\bar z_1-2\bar z_2 ) =(2-z_1\bar z_2)(2-\bar z_1 z_2) $$
which leads to
$$(|z_1|^2-4)(|z_2|^2-1)=0$$
So, if $z_2$ is not unimodular, $|z_1|^2=4$, which means $z_1$ lies on a circle of radius 2.