JEE Mathematics 2026 Question with Solution

Given: $S = \{ z \in \mathbb{C} : 4z^2 + \bar{z} = 0 \}$

Find: $\sum_{z \in S} |z|^2$

Write

$$z = x + iy, \quad \bar{z} = x - iy$$

where $x, y \in \mathbb{R}$.

Substitute into the given equation:

$$4(x + iy)^2 + (x - iy) = 0$$ $$4(x^2 - y^2 + 2ixy) + x - iy = 0$$ $$(4x^2 - 4y^2 + x) + i(8xy - y) = 0$$

Equating real and imaginary parts,

$$4x^2 - 4y^2 + x = 0$$

and

$$8xy - y = 0 \Rightarrow y(8x - 1) = 0$$

Now solve the two cases.

Case 1: $y = 0$

$$4x^2 + x = 0 \Rightarrow x(4x + 1) = 0$$

So,

$$x = 0, \; -\frac{1}{4}$$

Hence,

$$z = 0, \; -\frac{1}{4}$$

Case 2: $8x - 1 = 0 \Rightarrow x = \frac{1}{8}$ Substitute in the real-part equation:

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

Thus,

$$z = \frac{1}{8} \pm i\frac{\sqrt{3}}{8}$$

Now compute $|z|^2$ for each solution:

$$|0|^2 = 0$$ $$\left|-\frac{1}{4}\right|^2 = \frac{1}{16}$$ $$\left|\frac{1}{8} \pm i\frac{\sqrt{3}}{8}\right|^2 = \frac{1}{64} + \frac{3}{64} = \frac{4}{64} = \frac{1}{16}$$

Therefore,

$$\sum_{z \in S} |z|^2 = 0 + \frac{1}{16} + \frac{1}{16} = \frac{3}{16}$$

The correct option is B.

Given: $4z^2 + \bar{z} = 0$

Find: the sum of $|z|^2$ over all solutions.

For equations involving both $z$ and $\bar{z}$, express $z$ in terms of real and imaginary parts and then compare components.

Let $z = x + iy$. Then $\bar{z} = x - iy$.

After substitution,

$$(4x^2 - 4y^2 + x) + i(8xy - y) = 0$$

A complex number is zero only when both its real and imaginary parts are zero. So we get:

$$4x^2 - 4y^2 + x = 0$$ $$y(8x - 1) = 0$$

From $y(8x - 1) = 0$, either $y = 0$ or $x = \frac{1}{8}$.

If $y = 0$, then the real equation becomes

$$4x^2 + x = 0$$

which gives $x = 0$ or $x = -\frac{1}{4}$.

If $x = \frac{1}{8}$, then

$$4\left(\frac{1}{64}\right) - 4y^2 + \frac{1}{8} = 0$$

which simplifies to

$$y^2 = \frac{3}{64}$$

So $y = \pm \frac{\sqrt{3}}{8}$.

Hence the solutions are:

$$0, \quad -\frac{1}{4}, \quad \frac{1}{8} + i\frac{\sqrt{3}}{8}, \quad \frac{1}{8} - i\frac{\sqrt{3}}{8}$$

Now their squared moduli are:

$$0, \quad \frac{1}{16}, \quad \frac{1}{16}, \quad \frac{1}{16}$$

The two non-real roots contribute the same value, so the total is

$$\frac{3}{16}$$

Therefore, the required sum is $\frac{3}{16}$.