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JEE Main 202622 JanuaryEvening Shift

Exam: JEE Main
Year: 2026
Date: 22 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Complex Number

JEE Main 2026 · 22 January · Evening Shift · Mathematics · Complex Number

Let S=\ z ∈ ℂ: 4 z^ 2 + z =0\ . Then ∑_ z ∈ S |z|^ 2 is equal to:

Let

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

. Then

\sum_{z \in S}|z|^{2}

is equal to:

Options

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

Correct answer

C. $\frac{3}{16}$

Step-by-step solution

Let

z=x+iy

. From

4z^2+\bar{z}=0

, separating real and imaginary parts:
Real:

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

, Imaginary:

y(8x-1)=0

.
Case 1:

y=0

\Rightarrow x(4x+1)=0

\Rightarrow z=0

or

z=-\frac{1}{4}

.
Case 2:

x=\frac{1}{8}

\Rightarrow \frac{1}{16}-4y^2+\frac{1}{8}=0

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

.
So

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

.

\sum|z|^2=0+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}=\frac{3}{16}

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