If $z_1, z_2$ are two distinct complex number such that $\left|\frac{z_1-2 z_2}{\frac{1}{2}-z_1 \bar{z}_2}\right|=2$, then

<p>$$\begin{aligned} & \left|\frac{z_1-2 z_2}{\frac{1}{2}-z_1 \bar{z}_2}\right|=2 \\ & \left|z_1-2 z_2\right|=\left|1-2 z_1 \bar{z}_2\right| \\ & \Rightarrow\left(z_1-2 z_2\right)\left(\bar{z}_1-2 \bar{z}_2\right)=\left(1-2 z_1 \bar{z}_2\right)\left(1-2 \bar{z}_1 z_2\right) \\ & \Rightarrow\left|z_1\right|^2+4\left|z_2\right|^2-2 \bar{z}_1 z_2-2 \bar{z}_2 z_1 \\ & \quad=1+4\left|z_1\right|^2\left|z_2\right|^2-2 z_1 \bar{z}_2-2 \bar{z}_1 z_2 \\ & \Rightarrow\left|z_1\right|^2+4\left|z_2\right|^2-4\left|z_1\right|^2\left|z_2\right|^2-1=0 \\ & \Rightarrow\left(\left|z_1\right|^2-1\right)\left(1-4\left|z_2\right|^2\right)=0 \\ & \Rightarrow\left|z_1\right|=1 \text { and }\left|z_2\right|=\frac{1}{2} \end{aligned}$$</p>