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  4. Let S1= z1 ∈ C:|z1-3|=(1/2) and S 2= z 2 ∈ C :| z 2-| z 2+1||=| z 2+| z 2-1|| . Then, for z1 ∈ S1 and z2 ∈ S2, the least value of |z2-z1| is :

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Q. Let $S_1=\left\{z_1 \in C:\left|z_1-3\right|=\frac{1}{2}\right\}$ and $S _2=\left\{ z _2 \in C :\left| z _2-\right| z _2+1||=\left| z _2+\right| z _2-1||\right\}$. Then, for $z_1 \in S_1$ and $z_2 \in S_2$, the least value of $\left|z_2-z_1\right|$ is :

JEE MainJEE Main 2022Complex Numbers and Quadratic Equations

Solution:

$ \left| z _2+\right| z _2-\left.1\right|^2=\left| z _2-\right| z _2+\left.1\right|^2 $
$ \Rightarrow \left| z _2+\right| z _2-1||\left(\overline{ z }_2+\left| z _2-1\right|\right)=\left( z _2-\left| z _2+1\right|\right)\left(\overline{ z }_2-\left( z _2+1\right)\right)$
$\Rightarrow z _2\left|\overline{ z }_2+12_2-1\right|-\left(\overline{ z }_2-\left| z _2+1\right|\right)+\overline{ z }_2\left(\left| z _2-1\right|+\left| z _2+1\right|\right)$
$=\left| z _2+1\right|^2=\left| z _2-1\right|^2$
$ \Rightarrow\left[ z _2+\overline{ z }_2\right)\left(\left| z _2-1\right|\right)+\left( z _2+1 \mid\right)=2\left( z _2+\overline{ z }_2\right) $
$\Rightarrow\left( z _2+\overline{ z }_2\right)\left(\left| z _2-1\right|+\left| z _2+1\right|-2\right)=0 $
$ \therefore z _2+\overline{ 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 $\frac{1}{2}$.
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Clearly $\left|z_1-z_2\right| \min =\frac{5}{2}-1=\frac{3}{2}$