JEE Mains · Maths · STD 11 - 4.1 complex nubers
Let \(C\) be the set of all complex numbers. Let \(\mathrm{S}_{1}=\{\mathrm{z} \in \mathrm{C}:|\mathrm{z}-2| \leq 1\} \text { and }\) \(\mathrm{S}_{2}=\{\mathrm{z} \in \mathrm{C}: \mathrm{z}(1+\mathrm{i})+\overline{\mathrm{z}}(1-\mathrm{i}) \geq 4\}\) Then, the maximum value of \(\left|z-\frac{5}{2}\right|^{2}\) for \(z \in \mathrm{S}_{1} \cap \mathrm{S}_{2}\) is equal to:
- A \(\frac{3+2 \sqrt{2}}{4}\)
- B \(\frac{5+2 \sqrt{2}}{2}\)
- C \(\frac{3+2 \sqrt{2}}{2}\)
- D \(\frac{5+2 \sqrt{2}}{4}\)
Answer & Solution
Correct Answer
(D) \(\frac{5+2 \sqrt{2}}{4}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{S}_{1}:|\mathrm{z}-2| \leq 1\) \(\mathrm{~S}_{2}: \mathrm{x}(1+\mathrm{i})+\overline{\mathrm{z}}(1-\mathrm{i}) \geq 4\) \(z=x+i y\) \((x+i y)(1+i)+(x-i y)(1-i) \geq 4\) \(x+i(x+y)-y+x-i(x+y)-y \geq 4\) \(2 x-2 y \geq 4\) \(\Rightarrow y \leq x-2\) For…
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