Why is |-iz bar| = |z|?

The modulus of a product is the product of moduli: |-iz bar| = |-i| · |z bar|. Since |-i| = 1 and |z bar| = |z|, it simplifies directly to |z|.

The stem states xy ≠ 0, which requires both x ≠ 0 and y ≠ 0.

JEE MainMathematics

Answer verified against the official NTA key

+4 marks−1 if incorrectSingle correctPrevious-year question

Finding the Squared Modulus of a Complex Number from a Conjugate Equation

If z=x+i y, x y ≠ 0, satisfies the equation z^2+i z bar=0, then |z|^2 is equal to:

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

A9B1C4D1/4

Step-by-step solutionView

Correct answer

Taking the modulus on both sides of $z^{2} = - i \overset{ˉ}{z}$ gives $∣ z ∣^{2} = ∣ \overset{ˉ}{z} ∣ = ∣ z ∣$ , which implies $∣ z ∣ = 1$ (since $x y \neq 0 ⟹ z \neq 0$ ), 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 $∣ - i \overset{ˉ}{z} ∣ = ∣ z ∣$ directly to get $∣ z ∣^{2} = ∣ z ∣$ .

B · 1

The student correctly applies modulus properties to deduce $∣ z ∣ = 1$ and $∣ z ∣^{2} = 1$ . Correctly deduced $∣ z ∣^{2} = 1$ using $∣ z^{2} ∣ = ∣ - i \overset{ˉ}{z} ∣$ and $∣ z ∣ \neq 0$ .

C · 4

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

D · 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

$z = x + i y$ with $x y \neq 0$ , and $z^{2} + i \overset{ˉ}{z} = 0$ .

02Goal

Find the value of $∣ z ∣^{2}$ .

03Approach

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

04Execute

Rewriting the equation: $z^{2} = - i \overset{ˉ}{z}$ Taking modulus on both sides: $∣ z^{2} ∣ = ∣ - i \overset{ˉ}{z} ∣$ $∣ z ∣^{2} = ∣ - i ∣ \cdot ∣ \overset{ˉ}{z} ∣ = 1 \cdot ∣ z ∣ = ∣ z ∣$ $∣ z ∣^{2} - ∣ z ∣ = 0 ⟹ ∣ z ∣ (∣ z ∣ - 1) = 0$ Since $x y \neq 0$ , $z \neq 0$ , which means $∣ z ∣ \neq 0$ . Therefore, $∣ z ∣ = 1 ⟹ ∣ z ∣^{2} = 1^{2} = 1$ .

✓Verify

If $∣ z ∣ = 1$ , let $z = e^{i θ}$ . Then $z^{2} = e^{i 2 θ}$ and $- i \overset{ˉ}{z} = e^{- i π / 2} e^{- i θ} = e^{- i (θ + π / 2)}$ . Equating exponents gives solutions like $θ = π / 2, 7 π / 6, 11 π / 6$ . For non-zero $x, y$ , solutions exist (e.g. $θ = 7 π / 6, 11 π / 6$ ), confirming $∣ z ∣^{2} = 1$ .

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 8 September 2026

Quick checks

Students also ask

Why is

∣ - i \overset{ˉ}{z} ∣ = ∣ z ∣

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

Why can't

x = 0

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

Answer

Taking the modulus on both sides of $z^{2} = - i \overset{ˉ}{z}$ gives $∣ z ∣^{2} = ∣ \overset{ˉ}{z} ∣ = ∣ z ∣$ , which implies $∣ z ∣ = 1$ (since $x y \neq 0 ⟹ z \neq 0$ ), 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 $∣ - i \overset{ˉ}{z} ∣ = ∣ z ∣$ directly to get $∣ z ∣^{2} = ∣ z ∣$ .
B · correct: 1 - The student correctly applies modulus properties to deduce $∣ z ∣ = 1$ and $∣ z ∣^{2} = 1$ . Correctly deduced $∣ z ∣^{2} = 1$ using $∣ z^{2} ∣ = ∣ - i \overset{ˉ}{z} ∣$ and $∣ z ∣ \neq 0$ .
C: 4 - Confusing the values of $∣ z ∣$ and $∣ z ∣^{2}$ after an erroneous power operation. Remember that $∣ - i \overset{ˉ}{z} ∣ = ∣ - i ∣ \cdot ∣ \overset{ˉ}{z} ∣ = 1 \cdot ∣ z ∣ = ∣ z ∣$ , leading strictly to $∣ z ∣ (∣ z ∣ - 1) = 0$ .
D: 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: $z = x + i y$ with $x y \neq 0$ , and $z^{2} + i \overset{ˉ}{z} = 0$ .
Goal: Find the value of $∣ z ∣^{2}$ .
Approach: Take the modulus on both sides of $z^{2} = - i \overset{ˉ}{z}$ , using the properties $∣ z^{2} ∣ = ∣ z ∣^{2}$ , $∣ - i ∣ = 1$ , and $∣ \overset{ˉ}{z} ∣ = ∣ z ∣$ .
Execute: Rewriting the equation: $z^{2} = - i \overset{ˉ}{z}$ Taking modulus on both sides: $∣ z^{2} ∣ = ∣ - i \overset{ˉ}{z} ∣$ $∣ z ∣^{2} = ∣ - i ∣ \cdot ∣ \overset{ˉ}{z} ∣ = 1 \cdot ∣ z ∣ = ∣ z ∣$ $∣ z ∣^{2} - ∣ z ∣ = 0 ⟹ ∣ z ∣ (∣ z ∣ - 1) = 0$ Since $x y \neq 0$ , $z \neq 0$ , which means $∣ z ∣ \neq 0$ . Therefore, $∣ z ∣ = 1 ⟹ ∣ z ∣^{2} = 1^{2} = 1$ .
Verify: If $∣ z ∣ = 1$ , let $z = e^{i θ}$ . Then $z^{2} = e^{i 2 θ}$ and $- i \overset{ˉ}{z} = e^{- i π / 2} e^{- i θ} = e^{- i (θ + π / 2)}$ . Equating exponents gives solutions like $θ = π / 2, 7 π / 6, 11 π / 6$ . For non-zero $x, y$ , solutions exist (e.g. $θ = 7 π / 6, 11 π / 6$ ), confirming $∣ z ∣^{2} = 1$ .

Shortcut:

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

Given: $z = x + i y$ , $x y \neq 0$ , and $(x + i y)^{2} + i (x - i y) = 0$ .

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: $(x^{2} - y^{2} + 2 i x y) + (i x + y) = 0$ $(x^{2} - y^{2} + y) + i (2 x y + x) = 0$ Setting real and imaginary parts to zero: 1) $x (2 y + 1) = 0$ Since $x y \neq 0$ , $x \neq 0$ , so $2 y + 1 = 0 ⟹ y = - \frac{1}{2}$ . 2) $x^{2} - y^{2} + y = 0 ⟹ x^{2} = y^{2} - y = {(- \frac{1}{2})}^{2} - (- \frac{1}{2}) = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}$ . 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 $x y \neq 0$ , and $x^{2} + y^{2} = 1$ .

Finding the Squared Modulus of a Complex Number from a Conjugate Equation

What feels right?

Taking the modulus on both sides of $z^{2} = - i \overset{ˉ}{z}$ gives $∣ z ∣^{2} = ∣ \overset{ˉ}{z} ∣ = ∣ z ∣$ , which implies $∣ z ∣ = 1$ (since $x y \neq 0 ⟹ z \neq 0$ ), so $∣ z ∣^{2} = 1$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

✓Verify

We think you should solve this next.

If the value of real number $a > 0$ for which $x^{2} - 5 a x + 1 = 0$ and…

Let $a \in R$ and let $α, β$ be the roots of the equation…

The area of the region enclosed by the parabola $y = 4 x - x^{2}$ and…

Students also ask

Finding the Squared Modulus of a Complex Number from a Conjugate Equation

Taking the modulus on both sides of z2=−izˉz2=−izˉ gives ∣z∣2=∣zˉ∣=∣z∣∣z∣2=∣zˉ∣=∣z∣, which implies ∣z∣=1∣z∣=1 (since xy≠0 ⟹ z≠0xy=0⟹z=0), so ∣z∣2=1∣z∣2=1.

01Given

02Goal

03Approach

04Execute

✓Verify

If the value of real number a>0a>0 for which x2−5ax+1=0x2−5ax+1=0 and…

Let a∈Ra∈R and let α,βα,β be the roots of the equation…

The area of the region enclosed by the parabola y=4x−x2y=4x−x2 and…

Students also ask

Taking the modulus on both sides of $z^{2} = - i \overset{ˉ}{z}$ gives $∣ z ∣^{2} = ∣ \overset{ˉ}{z} ∣ = ∣ z ∣$ , which implies $∣ z ∣ = 1$ (since $x y \neq 0 ⟹ z \neq 0$ ), so $∣ z ∣^{2} = 1$ .

If the value of real number $a > 0$ for which $x^{2} - 5 a x + 1 = 0$ and…

Let $a \in R$ and let $α, β$ be the roots of the equation…

The area of the region enclosed by the parabola $y = 4 x - x^{2}$ and…