JEE Mains · Maths · STD 11 - 4.1 complex nubers
Let \(S=\left\{z \in C : z^{2}+\bar{z}=0\right\}\). Then \(\sum \limits_{z \in S}(\operatorname{Re}(z)+\operatorname{Im}(z))\) is equal to\(......\)
- A \(1\)
- B \(2\)
- C \(3\)
- D \(0\)
Answer & Solution
Correct Answer
(D) \(0\)
Step-by-step Solution
Detailed explanation
\(S=\left\{z \in C: z^{2}+\bar{z}=0\right\}\) Let \(z = x + iy\) \(z ^{2}= x ^{2}- y ^{2}+2 ixy\) \(\bar{z}=x-i y\) \(z^{2}+\bar{z}=x^{2}-y^{2}+x+i(2 x y-y)=0\) \(x^{2}+x-y^{2}=0 \text { \& } 2 x y-y=0\) \(y=0\) or \(x=\frac{1}{2}\) If \(y =0 ; x =0,-1\) If…
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