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JEE Mains · Maths · STD 11 - 4.1 complex nubers

Let \(S_{1}=\left\{z_{1} \in C:\left|z_{1}-3\right|=\frac{1}{2}\right\}\) and \(S_{2}=\left\{z_{2} \in C:\left|z_{2}-\right| z_{2}+1||=\left|z_{2}+\right| z_{2}-1||\right\} . \quad\) Then, for \(z_{1} \in S_{1}\) and \(z_{2} \in S_{2}\), the least value of \(\left|z_{2}-z_{1}\right|\) is.

  1. A \(0\)
  2. B \(\frac{1}{2}\)
  3. C \(\frac{3}{2}\)
  4. D \(\frac{5}{2}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{3}{2}\)

Step-by-step Solution

Detailed explanation

\(\left|z_{2}+\right| z_{2}-1||^{2}=\left|z_{2}-\right| z_{2}+1||^{2}\) \(\left|z_{2}+\right| z_{2}-1||\left(\bar{z}_{2}+\left|z_{2}-1\right|\right)=\left(z_{2}-\left|z_{2}+1\right|\right)\left(\bar{z}_{2}-\left(z_{2}+1\right)\right)\)…
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