JEE Mains · Maths · STD 11 - 4.1 complex nubers
Let \(\alpha\) and \(\beta\) be the sum and the product of all the non-zero solutions of the equation \((\bar{z})^2+|z|=0, z \in C\). Then \(4\left(\alpha^2+\beta^2\right)\) is equal to :
- A \(6\)
- B \(4\)
- C \(8\)
- D \(2\)
Answer & Solution
Correct Answer
(B) \(4\)
Step-by-step Solution
Detailed explanation
\(z=x+i y \) \(\bar{z}=x-i y \) \(\bar{z}^2=x^2-y^2-2 i x y \) \(\Rightarrow x^2-y^2-2 i x y+\sqrt{x^2+y^2}=0 \) \(\Rightarrow x=0 \quad \text { or } \)\( y=0 \) \(-y^2+|y|=0 \) \( x^2+|x|=0 \) \(|y|=|y|^2 \) \( \Rightarrow x=0 \) \(y=0, \pm 1 \) \( \Rightarrow \alpha=i-i=0 \)…
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