Solution
$\left|\mathrm{z}_{1}-2 \mathrm{z}_{2}\right|=2\left|\frac{1}{2}-\mathrm{z}_{1} \overline{\mathrm{z}}_{2}\right|=\left|1-2 \mathrm{z}_{1} \overline{\mathrm{z}}_{2}\right| \Rightarrow\left|\mathrm{z}_{1}-2 \mathrm{z}_{2}\right|^{2}=\left|1-2 \mathrm{z}_{1} \overline{\mathrm{z}}_{2}\right|^{2}\left(\because|\mathrm{z}|^{2}=\mathrm{z} \overline{\mathrm{z}}\right)$
\[
\begin{aligned}
& \Rightarrow\left(z_{1}-2 z_{2}\right)\left(\bar{z}_{1}-2 \bar{z}_{2}\right)=\left(1-2 z_{1} \bar{z}_{2}\right)\left(1-2 \bar{z}_{1} z_{2}\right) \Rightarrow z_{1} \bar{z}_{1}-2 z_{2} \bar{z}_{1}-2 z_{1} \bar{z}_{2}+4 z_{2} \bar{z}_{2}=1-2 z_{1} \bar{z}_{2}-2 \bar{z}_{1} z_{2}+4 z_{1} \bar{z}_{1} z_{2} \bar{z}_{2} \\
& \Rightarrow 4\left|z_{1}\right|^{2}\left|z_{2}\right|^{2}-\left|z_{1}\right|^{2}-4\left|z_{2}\right|^{2}+1=0 \Rightarrow\left(\left|z_{1}\right|^{2}-1\right)\left(4\left|z_{2}\right|^{2}-1\right)=0 \Rightarrow\left|z_{1}\right|=1 \text { or }\left|z_{2}\right|=\frac{1}{2}
\end{aligned}
\]
.
