Let $S = \{z \in \mathbb{C} : 4z^2 + \overline{z} = 0\}$. Then $\sum_{z \in S}|z|^2$ is equal to :

JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Complex Numbers

Let

S = {z \in C : 4 z^{2} + \overline{z} = 0}

. Then

\sum_{z \in S} ∣ z ∣^{2}

is equal to :

Answer: (A)

Put

z = x + i y

4 z^{2} + \overset{z}{ˉ} = 0

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

. The imaginary part

y (8 x - 1) = 0

splits into two cases. If

y = 0

, then

4 x^{2} + x = 0

, so

x = 0

x = - \frac{1}{4}

, with

∣ z ∣^{2} = 0

and

\frac{1}{16}

. If

x = \frac{1}{8}

, the real part gives

4 y^{2} = \frac{3}{16}

, so

∣ z ∣^{2} = \frac{1}{64} + \frac{3}{64} = \frac{1}{16}

for each of the two roots. Summing over all four solutions,

z \in S \sum ∣ z ∣^{2} = 0 + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} = \frac{3}{16} .

JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Single correct option