JEE Mains · Maths · STD 11 - 4.1 complex nubers
Let \( S=\{z\in\mathbb{C}:4z^{2}+\overline{z}=0\} \) Then \( \sum_{z\in S}|z|^{2} \) is equal to:
- A \(\frac{3}{16}\)
- B \(\frac{7}{64}\)
- C \(\frac{1}{16}\)
- D \(\frac{5}{64}\)
Answer & Solution
Correct Answer
(A) \(\frac{3}{16}\)
Step-by-step Solution
Detailed explanation
\(4 z^2+\bar{z}=0\) let \(z=x+i y\) \(4(x+i y)^2+x-i y=0\) \(4 x^2-4 y^2+8 x y i+x-i y=0\) \(4 x^2-4 y^2+x=0 \& y(8 x-1)=0\) \(\Rightarrow y =0\) or \(x=\frac{1}{8}\) If \(y=0,4 x^2+x=0\) \(x =0, \frac{-1}{4}\) \(\therefore z _1=0+0 \cdot i \quad\left| z _1\right|^2=0\)…
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