JEE Main 2026 — 22 January, Evening Shift — Question 4

Let $z = x + iy$. Then $4z^2 + \bar{z} = 0$ becomes $4(x+iy)^2 + (x - iy) = 0$. Expand: $4(x^2 - y^2 + 2ixy) + x - iy = 0$. Separate real and imaginary parts: $4x^2 - 4y^2 + x = 0$ and $8xy - y = 0$. From the imaginary part: $y(8x - 1) = 0$, so $y = 0$ or $x = \frac{1}{8}$. Case 1: $y = 0$.

Then real part: $4x^2 + x = 0 \Rightarrow x(4x+1)=0 \Rightarrow x = 0$ or $x = -\frac{1}{4}$.

Thus $z_1 = 0$, $|z_1|^2 = 0$; $z_2 = -\frac{1}{4}$, $|z_2|^2 = \frac{1}{16}$. Case 2: $x = \frac{1}{8}$.

Then real part: $4\left(\frac{1}{64}\right) - 4y^2 + \frac{1}{8} = 0 \Rightarrow \frac{1}{16} - 4y^2 + \frac{1}{8} = 0 \Rightarrow 4y^2 = \frac{3}{16} \Rightarrow y = \pm \frac{\sqrt{3}}{8}$.

Thus $z_3 = \frac{1}{8} + i\frac{\sqrt{3}}{8}$, $|z_3|^2 = \frac{1}{64} + \frac{3}{64} = \frac{1}{16}$; $z_4 = \frac{1}{8} - i\frac{\sqrt{3}}{8}$, $|z_4|^2 = \frac{1}{16}$. Sum: $0 + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} = \frac{3}{16}$.