JEE Mains · Maths · STD 11 - 4.1 complex nubers
Let \(S=\left\{Z \in C: \bar{z}=i\left(z^2+\operatorname{Re}(\bar{z})\right)\right\}\). Then \(\sum_{z \in S}|z|^2\) is equal to
- A \(\frac{7}{2}\)
- B \(4\)
- C \(\frac{5}{2}\)
- D \(3\)
Answer & Solution
Correct Answer
(B) \(4\)
Step-by-step Solution
Detailed explanation
Let \(Z=x+\) iy, \(x \in R, y \in R\) \(x-i y=i\left(x^2-y^2+(2 x y) i+x\right)\) \(x =- 2 x x\) \(- y =- y ^2+ x ^2+ x\) \(\Rightarrow x=0, y=-\frac{1}{2}(\text { from }(1))\) If \(x \neq 0\), then \(y =0,1\) If \(y =-\frac{1}{2}\), then \(x =\frac{1}{2},-\frac{3}{2}\)…
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