Let \[S_{1}=\left\{z_{1} \in \mathbf{C}:\left|z_{1}-3\right|=\frac{1}{2}\right\}\] and \[S_{2}=\left\{z_{2} \in \mathbf{C}:\left|z_{2}-\right| z_{2}+1||=\left|z_{2}+\right| z_{2}-1||\right\}\]. Then, for \[z_{1} \in S_{1}\] and \[z_{2} \in S_{2}\], the least value of \[\left|z_{2}-z_{1}\right|\] is :
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Official solution
\[\because\] \[{\left| {{Z_2} + |{Z_2} - 1|} \right|^2} = {\left| {{Z_2} - |{Z_2} + 1|} \right|^2}\]
\[ \Rightarrow \left( {{Z_2} + |{Z_2} - 1|} \right)\left( {{{\overline Z }_2} + |{Z_2} - 1|} \right) = \left( {{Z_2} - |{Z_2} + 1|} \right)\left( {{{\overline Z }_2} - |{Z_2} + 1|} \right)\]
\[ \Rightarrow {Z_2}\left( {|{Z_2} - 1| + |{Z_2} + 1|} \right) + {\overline Z _2}\left( {|{Z_2} - 1| + |{Z_2} + 1|} \right) = |{Z_2} + 1{|^2} - |{Z_2} - 1{|^2}\]
\[ \Rightarrow \left( {{Z_2} + {{\overline Z }_2}} \right)\left( {|{Z_2} + 1| + |{Z_2} - 1|} \right) = 2\left( {{Z_2} + {{\overline Z }_2}} \right)\]
\[\Rightarrow\] Either \[{Z_2} + {\overline Z _2} = 0\] or ∣Z2+1∣+∣Z2−1∣=2
So, Z2 lies on imaginary axis or on real axis within \[[ - 1,1]\]
Also \[|{Z_1} - 3| = {1 \over 2} \Rightarrow {Z_1}\] lies on the circle having center 3 and radius \[{1 \over 2}\].
Clearly \[|{Z_1} - {Z_2}{|_{\min }} = {3 \over 2}\]