JEE Mains · Maths · STD 11 - 4.1 complex nubers
Sum of squares of modulus of all the complex numbers \(z\) satisfying \(\bar{z}=i z^{2}+z^{2}-z\) is equal to
- A \(50\)
- B \(2\)
- C \(29\)
- D \(9\)
Answer & Solution
Correct Answer
(B) \(2\)
Step-by-step Solution
Detailed explanation
\(z +\overline{ z }=i z ^{2}+ z ^{2}\) Consider \(z=x+i y\) \(2 x=(i+1)\left(x^{2}-y^{2}+2 x y i\right)\) \(\Rightarrow 2 x=x^{2}-y^{2}-2 x y \text { and } x^{2}-y^{2}+2 x y=0\) \(\Rightarrow 2 x=-4 x y\) \(\Rightarrow x=0 \text { or } y=\frac{-1}{2}\) Case…
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