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Let $S_{1}=\left\{z_{1} \in C:\left|z_{1}-3\right|=\frac{1}{2}\right\}$ and $S_{2}=\left\{z_{2} \in C:\left|z_{2}-\right| z_{2}+1||=\left|z_{2}+\right| z_{2}-1||\right\} . \quad$ Then, for $z_{1} \in S_{1}$ and $z_{2} \in S_{2}$, the least value of $\left|z_{2}-z_{1}\right|$ is.
$0$
$\frac{1}{2}$
$\frac{3}{2}$
$\frac{5}{2}$
Solution
$\left|z_{2}+\right| z_{2}-1||^{2}=\left|z_{2}-\right| z_{2}+1||^{2}$
$\left|z_{2}+\right| z_{2}-1||\left(\bar{z}_{2}+\left|z_{2}-1\right|\right)=\left(z_{2}-\left|z_{2}+1\right|\right)\left(\bar{z}_{2}-\left(z_{2}+1\right)\right)$
$z_{2}\left|\bar{z}_{2}+12_{2}-1\right|-\left(\bar{z}_{2}-\left|z_{2}+1\right|\right)+\bar{z}_{2}\left(\left|z_{2}-1\right|+\left|z_{2}+1\right|\right)$
$=\left|z_{2}+1\right|^{2}=\left|z_{2}-1\right|^{2}$
${\left[z_{2}+\bar{z}_{2}\right)\left(\left|z_{2}-1\right|\right)+\left(z_{2}+1 \mid\right)=2\left(z_{2}+\bar{z}_{2}\right) }$
$\left(z_{2}+\bar{z}_{2}\right)\left(\left|z_{2}-1\right|+\left|z_{2}+1\right|-2\right)=0$
$z_{2}+\bar{z}_{2}=0$ or $\left|z_{2}-1\right|+\left|z_{2}+1\right|-2=0$
$\therefore z_{2}+\bar{z}_{2}=0 \text { or }\left|z_{2}-1\right|+\left|z_{2}+1\right|-2=0$
$\therefore z _{2}$ lie on imaginary axis. Or on real axis with in $[-1,1]$
Also $\left|z_{1}-3\right|=\frac{1}{2}$ lie on circle having centre 3 and radius $\frac{1}{2}$
Clearly $\left|z_{1}-z_{2}\right| \min =\frac{5}{2}-1=\frac{3}{2}$
