JEE Main 2026 — Jan 22, Shift 2 — Mathematics solution

JEE Main2026Jan 22, Shift 2Mathematics

Let $S = {z \in C : 4 z^{2} + \overset{z}{ˉ} = 0}$ . Then $z \in S \sum ∣ z ∣^{2}$ is equal to :

A
$\frac{3}{16}$
B
$\frac{7}{64}$
C
$\frac{1}{16}$
D
$\frac{5}{64}$

Solution

Let $z = x + i y$ . Then $4 (x + i y)^{2} + (x - i y) = 0$ , which expands to

$(4 x^{2} - 4 y^{2} + x) + i (8 x y - y) = 0$ .

Imaginary part: $y (8 x - 1) = 0 \Rightarrow y = 0$ or $x = \frac{1}{8}$ .

Case $y = 0$ : $4 x^{2} + x = 0 \Rightarrow x = 0$ or $x = - \frac{1}{4}$ . Gives $∣ z ∣^{2} = 0$ and $\frac{1}{16}$ .

Case $x = \frac{1}{8}$ : $\frac{1}{16} - 4 y^{2} + \frac{1}{8} = 0 \Rightarrow y^{2} = \frac{3}{64}$ . Two solutions, each with $∣ z ∣^{2} = \frac{1}{64} + \frac{3}{64} = \frac{1}{16}$ .

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

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