Q31 / 75 · JEE Main 2026 (22 January Shift 2) · Mathematics
Let $S=\left\{z \in \mathbb{C}: 4 z^{2}+\bar{z}=0\right\}$. Then $\sum_{z \in S}|z|^{2}$ is equal to:
- A$\frac{5}{64}$
- B$\frac{1}{16}$
- Correct answerC$\frac{3}{16}$
- D$\frac{7}{64}$
Solution
Let $z=x+iy$. From $4z^2+\bar{z}=0$, separating real and imaginary parts:
Real: $4x^2-4y^2+x=0$, Imaginary: $y(8x-1)=0$.
Case 1: $y=0$
$\Rightarrow x(4x+1)=0$
$\Rightarrow z=0$ or $z=-\frac{1}{4}$.
Case 2: $x=\frac{1}{8}$
$\Rightarrow \frac{1}{16}-4y^2+\frac{1}{8}=0$
$\Rightarrow y^2=\frac{3}{64}$.
So $|z|^2=\frac{1}{64}+\frac{3}{64}=\frac{1}{16}$.
$\sum|z|^2=0+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}=\frac{3}{16}$
Real: $4x^2-4y^2+x=0$, Imaginary: $y(8x-1)=0$.
Case 1: $y=0$
$\Rightarrow x(4x+1)=0$
$\Rightarrow z=0$ or $z=-\frac{1}{4}$.
Case 2: $x=\frac{1}{8}$
$\Rightarrow \frac{1}{16}-4y^2+\frac{1}{8}=0$
$\Rightarrow y^2=\frac{3}{64}$.
So $|z|^2=\frac{1}{64}+\frac{3}{64}=\frac{1}{16}$.
$\sum|z|^2=0+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}=\frac{3}{16}$
Hence, the correct answer is option: 'C'.