$L e t S_{1} = \{z_{1} \in C : |z_{1} - 3| = \frac{1}{2}\} a n d S_{2} = \{z_{2} \in C : |z_{2} - |z_{2} + 1|| = |z_{2} + |z_{2} - 1||\} . T h e n, f o r z_{1} \in S_{1} a n d z_{2} \in S_{2}, t h e l e a s t v a l u e o f |z_{2} - z_{1}| i s :$

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$\frac{5}{2}$

$\frac{1}{2}$

$\frac{3}{2}$

answer is C.

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Detailed Solution

${|z_{2} + |z_{2} - 1||}^{2} = {|z_{2} - |z_{} +||}^{2} \Rightarrow |z_{2} + |z_{2} - 1|| (z_{2} + |z_{2} - 1|) = (z_{2} - |z_{2} + 1|) (z_{2} - (z_{2} + 1)) \Rightarrow z_{2} |z_{2} + 12_{2} - 1| - (z_{2} - |z_{2} + 1|) + z_{2} (|z_{2} - 1| + |z_{2} + 1|) = {|z_{2} + 1|}^{2} = {|z_{2} - 1|}^{2} \Rightarrow [z_{2} + z_{2}] (|z_{2} - 1|) + (z_{2} + 1 - 2) = 0 ∴ z_{2} + z_{2} = 0 o r |z_{2} - 1| + |z_{2} + 1| - 2 = 0 ∴ z_{2} l i e o n i m a g i n a r y a x i s . O r o n r e a l a x i s w i t h i n [- 1, 1] A l s o |z_{1} - 3| = \frac{1}{2} l i e o n c i r c l e h a v n g c e n t r e 3 a n d r a d i u s \frac{1}{2} .$ Question Image