Given, $\frac{{\text{z}}_1-\mathrm{2z}{}_2}{2-{\text{z}}_1{\overline{\text{z}}}_2}$  is unimodular

$\Rightarrow \left|\frac{{\text{z}}_1-2 {\text{z}}_2}{2-{\text{z}}_1{\overline{\text{z}}}_2}\right|=1$

$\Rightarrow \left|{\text{z}}_1-2{\text{z}}_2\right|=\left|2- {\text{z}}_1{\overline{\text{z}}}_2\right|$

Squaring both the sides, we get,

${\left|{\text{z}}_1-\mathrm{2z}{}_2\right|}^2={\left|2- {\text{z}}_1{\overline{\text{z}}}_2\right|}^2$

$\Rightarrow \left({\text{z}}_1-2{\text{z}}_2\right)\left({\overline{\text{z}}}_1-2{\overline{\text{z}}}_2\right)=\left(2-{\text{z}}_1{\overline{\text{z}}}_2\right)\left(2-{\overline{\text{z}}}_1{\text{z}}_2\right)$

$\left(\because {\left|\text{z}\right|}^2=\text{z }\overline{\text{z}}\right)$

$\Rightarrow {\text{z}}_1{\overline{\text{z}}}_1-2{\text{z}}_{\mathrm{1 }}{\overline{\text{z}}}_2-2{\overline{\text{z}}}_{\mathrm{1 }}{\text{z}}_2+4{\text{z}}_{\mathrm{2 }}{\overline{\text{z}}}_2$

$=4-2{\overline{\text{z}}}_{\mathrm{1 }}{\text{z}}_2-2{\text{z}}_{\mathrm{1 }}{\overline{\text{z}}}_2+{\text{z}}_{\mathrm{1 }}{\overline{\text{z}}}_{\mathrm{1 }}{\text{z}}_{\mathrm{2 }}{\overline{\text{z}}}_2$

$\Rightarrow {\left|{\text{z}}_1\right|}^2+4{\left|{\text{z}}_2\right|}^2=4+{\left|{\text{z}}_1\right|}^{\mathrm{2 }}{\left|{\text{z}}_2\right|}^2$

$\Rightarrow {\left|{\text{z}}_1\right|}^2-4+4{\left|{\text{z}}_2\right|}^2-{\left|{\text{z}}_1\right|}^{\mathrm{2 }}{\left|{\text{z}}_2\right|}^2=0$

$\Rightarrow \left({\left|{\text{z}}_1\right|}^2-4\right)\left(1-{\left|{\text{z}}_2\right|}^2\right)=0$

$\Rightarrow \left|{\text{z}}_1\right|=\mathrm{2 }\text{or }\left|{\text{z}}_2\right|=1$

 Given, ${\mathrm{z}}_2$ is not unimodular

$\therefore \left|{\text{z}}_1\right|=2$

$\therefore$  Point ${\mathrm{z}}_1$ lies on a circle of radius $2$.