The modulus of a product is the product of moduli: . Since and , it simplifies directly to .

The stem states , which requires both and .

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Complex Numbers and Quadratic Equations: Mathematics | JEE Main

\mathrm{z}=\mathrm{x}+\mathrm{i} y, \quad x y \neq 0

, satisfies the equation

\mathrm{z}^{2}+\mathrm{i} \overline{\mathrm{z}}=0

, then

|\mathrm{z}|^{2}

is equal to :

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A9B1C4D

\frac{1}{4}

Step-by-step solutionView

Correct answer

Taking the modulus on both sides of $z^2 = -i\bar{z}$ gives $|z|^2 = |\bar{z}| = |z|$ , which implies $|z| = 1$ (since $xy \neq 0 \implies z \neq 0$ ), so $|z|^2 = 1$ .

Option analysis

Why each option works or fails

A · 9

Squaring the modulus relationship incorrectly or mistaking $|z|=3$ as a solution. Apply the modulus identity $|z^2| = |z|^2$ and $|-i\bar{z}| = |z|$ directly to get $|z|^2 = |z|$ .

B · 1

None. The student correctly applies modulus properties to deduce $|z| = 1$ and $|z|^2 = 1$ . Correctly deduced $|z|^2 = 1$ using $|z^2| = |-i\bar{z}|$ and $|z| \neq 0$ .

C · 4

Confusing the values of $|z|$ and $|z|^2$ after an erroneous power operation. Remember that $|-i\bar{z}| = |-i| \cdot |\bar{z}| = 1 \cdot |z| = |z|$ , leading strictly to $|z|(|z|-1)=0$ .

D ·

\frac{1}{4}

Finding a component such as $y = -1/2$ or $x^2 = 3/4$ from component expansion and mistakenly reporting $y^2 = 1/4$ instead of $|z|^2 = x^2 + y^2$ . Ensure you compute $|z|^2 = x^2 + y^2 = 3/4 + 1/4 = 1$ rather than reporting just $y^2$ .

Reviewed route

Solution

StepWorking

01given

$z = x + iy$ with $xy \neq 0$ , and $z^2 + i\bar{z} = 0$ .

02goal

Find the value of $|z|^2$ .

03approach

Take the modulus on both sides of $z^2 = -i\bar{z}$ , using the properties $|z^2| = |z|^2$ , $|-i| = 1$ , and $|\bar{z}| = |z|$ .

04execute

Rewriting the equation: $z^2 = -i\bar{z}$ Taking modulus on both sides: $|z^2| = |-i\bar{z}|$ $|z|^2 = |-i| \cdot |\bar{z}| = 1 \cdot |z| = |z|$ $|z|^2 - |z| = 0 \implies |z|(|z| - 1) = 0$ Since $xy \neq 0$ , $z \neq 0$ , which means $|z| \neq 0$ . Therefore, $|z| = 1 \implies |z|^2 = 1^2 = 1$ .

✓verify

If $|z| = 1$ , let $z = e^{i\theta}$ . Then $z^2 = e^{i2\theta}$ and $-i\bar{z} = e^{-i\pi/2} e^{-i\theta} = e^{-i(\theta + \pi/2)}$ . Equating exponents gives solutions like $\theta = \pi/2, 7\pi/6, 11\pi/6$ . For non-zero $x, y$ , solutions exist (e.g. $\theta = 7\pi/6, 11\pi/6$ ), confirming $|z|^2 = 1$ .

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 8 September 2026

Quick checks

Students also ask

Why is

|-i\bar{z}| = |z|

The modulus of a product is the product of moduli: $|-i\bar{z}| = |-i| \cdot |\bar{z}|$ . Since $|-i| = 1$ and $|\bar{z}| = |z|$ , it simplifies directly to $|z|$ .

Why can't

x=0

The stem states $xy \neq 0$ , which requires both $x \neq 0$ and $y \neq 0$ .

Answer

Taking the modulus on both sides of $z^2 = -i\bar{z}$ gives $|z|^2 = |\bar{z}| = |z|$ , which implies $|z| = 1$ (since $xy \neq 0 \implies z \neq 0$ ), so $|z|^2 = 1$ .

Why each option works or fails

A: 9 - Squaring the modulus relationship incorrectly or mistaking $|z|=3$ as a solution. Apply the modulus identity $|z^2| = |z|^2$ and $|-i\bar{z}| = |z|$ directly to get $|z|^2 = |z|$ .
B · correct: 1 - None. The student correctly applies modulus properties to deduce $|z| = 1$ and $|z|^2 = 1$ . Correctly deduced $|z|^2 = 1$ using $|z^2| = |-i\bar{z}|$ and $|z| \neq 0$ .
C: 4 - Confusing the values of $|z|$ and $|z|^2$ after an erroneous power operation. Remember that $|-i\bar{z}| = |-i| \cdot |\bar{z}| = 1 \cdot |z| = |z|$ , leading strictly to $|z|(|z|-1)=0$ .
D: $\frac{1}{4}$ - Finding a component such as $y = -1/2$ or $x^2 = 3/4$ from component expansion and mistakenly reporting $y^2 = 1/4$ instead of $|z|^2 = x^2 + y^2$ . Ensure you compute $|z|^2 = x^2 + y^2 = 3/4 + 1/4 = 1$ rather than reporting just $y^2$ .

Step-by-step solution

given: $z = x + iy$ with $xy \neq 0$ , and $z^2 + i\bar{z} = 0$ .
goal: Find the value of $|z|^2$ .
approach: Take the modulus on both sides of $z^2 = -i\bar{z}$ , using the properties $|z^2| = |z|^2$ , $|-i| = 1$ , and $|\bar{z}| = |z|$ .
execute: Rewriting the equation: $z^2 = -i\bar{z}$ Taking modulus on both sides: $|z^2| = |-i\bar{z}|$ $|z|^2 = |-i| \cdot |\bar{z}| = 1 \cdot |z| = |z|$ $|z|^2 - |z| = 0 \implies |z|(|z| - 1) = 0$ Since $xy \neq 0$ , $z \neq 0$ , which means $|z| \neq 0$ . Therefore, $|z| = 1 \implies |z|^2 = 1^2 = 1$ .
verify: If $|z| = 1$ , let $z = e^{i\theta}$ . Then $z^2 = e^{i2\theta}$ and $-i\bar{z} = e^{-i\pi/2} e^{-i\theta} = e^{-i(\theta + \pi/2)}$ . Equating exponents gives solutions like $\theta = \pi/2, 7\pi/6, 11\pi/6$ . For non-zero $x, y$ , solutions exist (e.g. $\theta = 7\pi/6, 11\pi/6$ ), confirming $|z|^2 = 1$ .

Shortcut:

When to use it: When one wants to explicitly find the values of x and y.

given: $z = x + iy$ , $xy \neq 0$ , and $(x + iy)^2 + i(x - iy) = 0$ .

goal: Find $|z|^2 = x^2 + y^2$ .

approach: Expand into real and imaginary parts, set each to zero, and solve for $x^2 + y^2$ .

execute: Expand: $(x^2 - y^2 + 2ixy) + (ix + y) = 0$ $(x^2 - y^2 + y) + i(2xy + x) = 0$ Setting real and imaginary parts to zero: 1) $x(2y + 1) = 0$ Since $xy \neq 0$ , $x \neq 0$ , so $2y + 1 = 0 \implies y = -\frac{1}{2}$ . 2) $x^2 - y^2 + y = 0 \implies x^2 = y^2 - y = \left(-\frac{1}{2}\right)^2 - \left(-\frac{1}{2}\right) = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}$ . Then $|z|^2 = x^2 + y^2 = \frac{3}{4} + \frac{1}{4} = 1$ .

verify: $x = \pm \frac{\sqrt{3}}{2}$ and $y = -\frac{1}{2}$ satisfy $xy \neq 0$ , and $x^2+y^2 = 1$ .

Complex Numbers and Quadratic Equations: Mathematics | JEE Main

What feels right?

Taking the modulus on both sides of $z^2 = -i\bar{z}$ gives $|z|^2 = |\bar{z}| = |z|$ , which implies $|z| = 1$ (since $xy \neq 0 \implies z \neq 0$ ), so $|z|^2 = 1$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

✓verify

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