JEE Main 202228 Jul, Morning shiftMathematicsHardComplex Numbers and Quadratic Equations264× since 2002Let S1={z1∈C:∣z1−3∣=12}S_{1}=\left\{z_{1} \in \mathbf{C}:\left|z_{1}-3\right|=\frac{1}{2}\right\}S1={z1∈C:∣z1−3∣=21} and S2={z2∈C:∣z2−∣z2+1∣∣=∣z2+∣z2−1∣∣}S_{2}=\left\{z_{2} \in \mathbf{C}:\left|z_{2}-\right| z_{2}+1||=\left|z_{2}+\right| z_{2}-1||\right\}S2={z2∈C:∣z2−∣z2+1∣∣=∣z2+∣z2−1∣∣}. Then, for z1∈S1z_{1} \in S_{1}z1∈S1 and z2∈S2z_{2} \in S_{2}z2∈S2, the least value of ∣z2−z1∣\left|z_{2}-z_{1}\right|∣z2−z1∣ is :AOption A. 0BOption B. 12\frac{1}{2}21COption C. 32\frac{3}{2}23DOption D. 52\frac{5}{2}25CheckSkip · show solution
