Complex Numbers – Quadratic Equations – JEE Main 22 Jan 2026 Shift 2

Solution:

Let $z = x + iy$. The given equation is $4z^2 + \bar{z} = 0$.

Substitute $z$ into the equation:
$$ 4(x + iy)^2 + (x – iy) = 0 $$
$$ 4(x^2 – y^2 + 2ixy) + x – iy = 0 $$
$$ (4x^2 – 4y^2 + x) + i(8xy – y) = 0 $$

Comparing real and imaginary parts to zero:
Imaginary part:
$$ y(8x – 1) = 0 $$
This gives two cases: $y = 0$ or $x = \frac{1}{8}$.

**Case 1:** If $y = 0$
Substitute into the real part equation:
$$ 4x^2 + x = 0 $$
$$ x(4x + 1) = 0 $$
$$ \Rightarrow x = 0 \text{ or } x = -\frac{1}{4} $$
So, the solutions are $z_1 = 0$ and $z_2 = -\frac{1}{4}$.

**Case 2:** If $x = \frac{1}{8}$
Substitute into the real part equation:
$$ 4\left(\frac{1}{8}\right)^2 – 4y^2 + \frac{1}{8} = 0 $$
$$ 4\left(\frac{1}{64}\right) – 4y^2 + \frac{1}{8} = 0 $$
$$ \frac{1}{16} + \frac{2}{16} = 4y^2 $$
$$ 4y^2 = \frac{3}{16} \Rightarrow y^2 = \frac{3}{64} $$
$$ y = \pm \frac{\sqrt{3}}{8} $$
So, the solutions are $z_3 = \frac{1}{8} + i\frac{\sqrt{3}}{8}$ and $z_4 = \frac{1}{8} – i\frac{\sqrt{3}}{8}$.

Now, we calculate $\sum_{z \in S} |z|^2$:
$$ |z_1|^2 = 0 $$
$$ |z_2|^2 = \left|-\frac{1}{4}\right|^2 = \frac{1}{16} $$
$$ |z_3|^2 = \left(\frac{1}{8}\right)^2 + \left(\frac{\sqrt{3}}{8}\right)^2 = \frac{1}{64} + \frac{3}{64} = \frac{4}{64} = \frac{1}{16} $$
$$ |z_4|^2 = \frac{1}{16} $$

$$ \text{Total Sum} = 0 + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} = \frac{3}{16} $$

Ans. (1)