Answer:
Option C
Explanation:
Central Idea:
if z is unimodular, then |z|=1, Also , use property of modulus i.e, $z\overline{z}=|z|^{2}$
Given, $z_{2}$ is not unimodular i.e, |$z_{2}$|≠ 1
and $\frac{z_{1}-2z_{2}}{2-z_{1}\overline{z_{2}}}$ is unimodular.
$\Rightarrow$ $\mid\frac{z_{1}-2z_{2}}{2-z_{1}\overline{z_{2}}}\mid$ =1
$\Rightarrow$ $\mid z_{1}-2z_{2}\mid ^{2}=\mid2-z_{1}\overline{z_{2}}\mid ^{2}$
$\Rightarrow$ $(z_{1}-2z_{2})(\overline{z_{1}}-2\overline{z_{2}})=(2-z_{1}\overline{z_{2}})(2-\overline{z_{1}}z_{2})$ $(\because z\overline{z}= |z|^{2})$
$\Rightarrow$ $|z_{1}|^{2}+4|z_{2}|^{2}-2\overline{z_{1}}z_{2}-2z_{1}\overline{z_{2}}$
$\Rightarrow$ $(|z_{2}|^{2}-1)(|z_{2}|^{2}-4)=0$
$\because$ $|z_{2}|\neq 1$
$\because$ $|z_{1}|= 2$
Let $z_{1}=x+iy\Rightarrow x^{2}+y^{2}=(2)^{2}$
$\therefore$ point z1 lies on a circle of radius 2.