JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Determining Signs of Imaginary Parts of Complex Numbers with Vanishing Real Conditions

Q: Can a_1 = 0?

If a_1 = 0, then a_2 = 0, which implies b_1 b_2 = 0. Since z_1, z_2 are non-zero, at least one of b_1 or b_2 is zero, which makes that complex number zero (since both its real and imaginary parts would vanish), contradicting non-zero complex numbers.

For two non-zero complex numbers z_1 and z_2, if Re(z_1 z_2) = 0 and Re(z_1 + z_2) = 0, then which of the following are possible? A. Im(z_1) > 0 and Im(z_2) > 0 B. Im(z_1) < 0 and Im(z_2) > 0 C. Im(z_1) > 0 and Im(z_2) < 0 D. Im(z_1) < 0 and Im(z_2) < 0 Choose the correct answer from the options given below:

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AB and DBA and BCB and CDA and C

Step-by-step solutionView

Correct answer

The conditions require $Im (z_{1})$ and $Im (z_{2})$ to have opposite signs, making B and C the only possible configurations.

Option analysis

Why each option works or fails

A · B and D

Believing that D is possible by overlooking that $Im (z_{1}) Im (z_{2}) = - x_{1}^{2} < 0$ when $x_{1} \neq 0$ , which prevents both imaginary parts from being negative simultaneously. Express $Re (z_{1} z_{2}) = x_{1} x_{2} - y_{1} y_{2} = 0$ alongside $x_{2} = - x_{1}$ to see that $y_{1} y_{2} = - x_{1}^{2} \leq 0$ , ruling out cases where both imaginary parts share the same sign.

B · A and B

Assuming statement A is possible by neglecting the sign constraint $y_{1} y_{2} = - x_{1}^{2} < 0$ , which forbids both imaginary parts from being strictly positive at the same time. Substitute $x_{2} = - x_{1}$ into $Re (z_{1} z_{2}) = x_{1} x_{2} - y_{1} y_{2} = 0$ to deduce that $y_{1} y_{2}$ must be non-positive, meaning $y_{1}$ and $y_{2}$ cannot both be positive non-zero numbers.

C · B and C

This is the correct option. Setting $z_{1} = x_{1} + i y_{1}$ and $z_{2} = x_{2} + i y_{2}$ gives $x_{2} = - x_{1}$ and $- x_{1}^{2} - y_{1} y_{2} = 0$ , so $y_{1} y_{2} = - x_{1}^{2}$ . Since $z_{1}, z_{2} \neq 0$ , either $x_{1} \neq 0$ giving $y_{1} y_{2} < 0$ (opposite signs, so B and C hold), or $x_{1} = 0$ giving purely imaginary numbers with $Re (z_{1} z_{2}) = - y_{1} y_{2} = 0$ , forcing one to be zero, which contradicts non-zero complex numbers.

D · A and C

Including statement A by incorrectly writing $Re (z_{1} z_{2})$ as $x_{1} x_{2} + y_{1} y_{2} = 0$ , which would falsely imply $y_{1} y_{2} = x_{1}^{2} > 0$ and lead to same-sign imaginary parts. Recall that for complex numbers, $(x_{1} + i y_{1}) (x_{2} + i y_{2}) = (x_{1} x_{2} - y_{1} y_{2}) + i (x_{1} y_{2} + x_{2} y_{1})$ , so the real part is $x_{1} x_{2} - y_{1} y_{2}$ , not $x_{1} x_{2} + y_{1} y_{2}$ .

Reviewed route

Solution

StepWorking

01Given

Non-zero complex numbers $z_{1} = a_{1} + i b_{1}$ and $z_{2} = a_{2} + i b_{2}$ with $a_{1}, b_{1}, a_{2}, b_{2} \in R$ , satisfying $Re (z_{1} z_{2}) = 0$ and $Re (z_{1} + z_{2}) = 0$ .

02Goal

Determine the possible signs of $Im (z_{1}) = b_{1}$ and $Im (z_{2}) = b_{2}$ from the statements A, B, C, D.

03Approach

Express the real parts in terms of $a_{1}, a_{2}, b_{1}, b_{2}$ and relate $b_{1} b_{2}$ to $a_{1}, a_{2}$ to determine the sign of the product $b_{1} b_{2}$ .

04Execute

From $Re (z_{1} + z_{2}) = 0$ , we have $a_{1} + a_{2} = 0 ⟹ a_{2} = - a_{1}$ . Since $z_{1} z_{2} = (a_{1} a_{2} - b_{1} b_{2}) + i (a_{1} b_{2} + a_{2} b_{1})$ , the condition $Re (z_{1} z_{2}) = 0$ yields $a_{1} a_{2} - b_{1} b_{2} = 0 ⟹ b_{1} b_{2} = a_{1} a_{2} = - a_{1}^{2}$ . If $a_{1} = 0$ , then $a_{2} = 0$ , giving $b_{1} b_{2} = 0$ , which implies $b_{1} = 0$ or $b_{2} = 0$ , contradicting that $z_{1}, z_{2}$ are non-zero. Thus, $a_{1} \neq 0$ , which gives $b_{1} b_{2} = - a_{1}^{2} < 0$ . Therefore, $b_{1}$ and $b_{2}$ must have opposite signs, making B and C possible.

✓Verify

Let $z_{1} = 1 + i$ and $z_{2} = - 1 + i$ . Then $z_{1} + z_{2} = 2 i ⟹ Re (z_{1} + z_{2}) = 0$ . Also $z_{1} z_{2} = (1 + i) (- 1 + i) = - 1 - 1 = - 2 \neq 0$ . But if $z_{1} = 1 + i$ and $z_{2} = - 1 - i$ , then $z_{1} z_{2} = (1 + i) (- (1 + i)) = - 2 i$ , so $Re (z_{1} z_{2}) = 0$ and $Re (z_{1} + z_{2}) = 0$ . Here $Im (z_{1}) = 1 > 0$ and $Im (z_{2}) = - 1 < 0$ (Case C). Conjugating gives Case B. Thus B and C are valid.

Hints that build this answer step by step

Let $z_{1} = x_{1} + i y_{1}$ and $z_{2} = x_{2} + i y_{2}$ with $x_{1}, x_{2}, y_{1}, y_{2} \in R$ . What algebraic equations correspond to $Re (z_{1} + z_{2}) = 0$ and $Re (z_{1} z_{2}) = 0$ ?
$x_{1} + x_{2} = 0$ and $x_{1} x_{2} - y_{1} y_{2} = 0$
Using $x_{2} = - x_{1}$ , what is the relationship between $y_{1} y_{2}$ and $x_{1}$ ?
$y_{1} y_{2} = - x_{1}^{2}$
Since $z_{1}, z_{2}$ are non-zero and $y_{1} y_{2} = - x_{1}^{2}$ , can $x_{1}$ be equal to $0$ ?
No, because if $x_{1} = 0$ , then $x_{2} = 0$ and $y_{1} y_{2} = 0$ , meaning at least one of $z_{1}$ or $z_{2}$ would be $0$ , which contradicts that both are non-zero.
Since $y_{1} y_{2} = - x_{1}^{2} < 0$ , what must be true about the signs of $Im (z_{1})$ and $Im (z_{2})$ ?
They must have opposite signs, so either ( $y_{1} < 0$ and $y_{2} > 0$ ) or ( $y_{1} > 0$ and $y_{2} < 0$ ), corresponding to B and C.

✓ 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

Can

a_{1} = 0

If $a_{1} = 0$ , then $a_{2} = 0$ , which implies $b_{1} b_{2} = 0$ . Since $z_{1}, z_{2}$ are non-zero, at least one of $b_{1}$ or $b_{2}$ is zero, which makes that complex number zero (since both its real and imaginary parts would vanish), contradicting non-zero complex numbers.

Answer

The conditions require $Im (z_{1})$ and $Im (z_{2})$ to have opposite signs, making B and C the only possible configurations.

Why each option works or fails

A: B and D - Believing that D is possible by overlooking that $Im (z_{1}) Im (z_{2}) = - x_{1}^{2} < 0$ when $x_{1} \neq 0$ , which prevents both imaginary parts from being negative simultaneously. Express $Re (z_{1} z_{2}) = x_{1} x_{2} - y_{1} y_{2} = 0$ alongside $x_{2} = - x_{1}$ to see that $y_{1} y_{2} = - x_{1}^{2} \leq 0$ , ruling out cases where both imaginary parts share the same sign.
B: A and B - Assuming statement A is possible by neglecting the sign constraint $y_{1} y_{2} = - x_{1}^{2} < 0$ , which forbids both imaginary parts from being strictly positive at the same time. Substitute $x_{2} = - x_{1}$ into $Re (z_{1} z_{2}) = x_{1} x_{2} - y_{1} y_{2} = 0$ to deduce that $y_{1} y_{2}$ must be non-positive, meaning $y_{1}$ and $y_{2}$ cannot both be positive non-zero numbers.
C · correct: B and C - This is the correct option. Setting $z_{1} = x_{1} + i y_{1}$ and $z_{2} = x_{2} + i y_{2}$ gives $x_{2} = - x_{1}$ and $- x_{1}^{2} - y_{1} y_{2} = 0$ , so $y_{1} y_{2} = - x_{1}^{2}$ . Since $z_{1}, z_{2} \neq 0$ , either $x_{1} \neq 0$ giving $y_{1} y_{2} < 0$ (opposite signs, so B and C hold), or $x_{1} = 0$ giving purely imaginary numbers with $Re (z_{1} z_{2}) = - y_{1} y_{2} = 0$ , forcing one to be zero, which contradicts non-zero complex numbers.
D: A and C - Including statement A by incorrectly writing $Re (z_{1} z_{2})$ as $x_{1} x_{2} + y_{1} y_{2} = 0$ , which would falsely imply $y_{1} y_{2}$ and lead to same-sign imaginary parts. Recall that for complex numbers, $(x_{1} + i y_{1}) (x_{2} + i y_{2}) = (x_{1} x_{2} - y_{1} y_{2}) + i (x_{1} y_{2} + x_{2} y_{1})$ , so the real part is $x_{1} x_{2} - y_{1} y_{2}$ , not $x_{1} x_{2} + y_{1} y_{2}$ .

Step-by-step solution

Given: Non-zero complex numbers $z_{1} = a_{1} + i b_{1}$ and $x_{1} = 0$ with $a_{1}, b_{1}, a_{2}, b_{2} \in R$ , satisfying $Re (z_{1} z_{2}) = 0$ and $Re (z_{1} + z_{2}) = 0$ .
Goal: Determine the possible signs of $Im (z_{1}) = b_{1}$ and $Im (z_{2}) = b_{2}$ from the statements A, B, C, D.
Approach: Express the real parts in terms of $a_{1}, a_{2}, b_{1}, b_{2}$ and relate $b_{1} b_{2}$ to $a_{1}, a_{2}$ to determine the sign of the product $b_{1} b_{2}$ .
Execute: From $Re (z_{1} + z_{2}) = 0$ , we have $a_{1} + a_{2} = 0 ⟹ a_{2} = - a_{1}$ . Since $z_{1} z_{2} = (a_{1} a_{2} - b_{1} b_{2}) + i (a_{1} b_{2} + a_{2} b_{1})$ , the condition $Re (z_{1} z_{2}) = 0$ yields $a_{1} a_{2} - b_{1} b_{2} = 0 ⟹ b_{1} b_{2} = a_{1} a_{2} = - a_{1}^{2}$ . If $a_{1} = 0$ , then $a_{2} = 0$ , giving $b_{1} b_{2} = 0$ , which implies $b_{1} = 0$ or $b_{2} = 0$ , contradicting that $z_{1}, z_{2}$ are non-zero. Thus, $a_{1} \neq 0$ , which gives $b_{1} b_{2} = - a_{1}^{2} < 0$ . Therefore, $b_{1}$ and $b_{2}$ must have opposite signs, making B and C possible.
Verify: Let $z_{1} = 1 + i$ and $z_{2} = - 1 + i$ . Then $z_{1} + z_{2} = 2 i ⟹ Re (z_{1} + z_{2}) = 0$ . Also $z_{1} z_{2} = (1 + i) (- 1 + i) = - 1 - 1 = - 2 \neq 0$ . But if $z_{1} = 1 + i$ and $z_{2} = - 1 - i$ , then $z_{1} z_{2} = (1 + i) (- (1 + i)) = - 2 i$ , so $Re (z_{1} z_{2}) = 0$ and $Re (z_{1} + z_{2}) = 0$ . Here $Im (z_{1}) = 1 > 0$ and $a_{1} \neq 0$ (Case C). Conjugating gives Case B. Thus B and C are valid.

Determining Signs of Imaginary Parts of Complex Numbers with Vanishing Real Conditions

What feels right?

The conditions require $Im (z_{1})$ and $Im (z_{2})$ to have opposite signs, making B and C the only possible configurations.

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

✓Verify

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