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in Complex number and Quadratic equations by (20 points)
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 : [Main July 28, 2022(1)] (a) 0 (b) \( \frac{1}{2} \) (c) \( \frac{3}{2} \) (d) \( \frac{5}{2} \)

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by (35.0k points)

Answer: (c) 3/2

The value of z2

So z2 lies on imaginary axis or on real axis within
[–1, 1]. Also |z1-3| = 1/2 \(\Rightarrow z_1\) lies on the circle having centre 3 and radius 1/2.

value of z1-z2

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