JEE MainMathematics

Reviewed by official_key

+4 marks−1 if incorrectSingle correctpyq

Complex Numbers and Quadratic Equations: Mathematics | JEE Main

Q: Can ?

If , then , which implies . Since are non-zero, at least one of or 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

\text{Re}(z_1 z_2) = 0

and

\text{Re}(z_1 + z_2) = 0

, then which of the following are possible? A.

\text{Im}(z_1) > 0

and

\text{Im}(z_2) > 0

\text{Im}(z_1) < 0

and

\text{Im}(z_2) > 0

\text{Im}(z_1) > 0

and

\text{Im}(z_2) < 0

\text{Im}(z_1) < 0

and

\text{Im}(z_2) < 0

Choose the correct answer from the options given below:

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.

AB and DBA and BCB and CDA and C

Step-by-step solutionView

Correct answer

The conditions require $\operatorname{Im}(z_1)$ and $\operatorname{Im}(z_2)$ 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 $\operatorname{Im}(z_1)\operatorname{Im}(z_2) = -x_1^2 < 0$ when $x_1 \neq 0$ , which prevents both imaginary parts from being negative simultaneously. Express $\operatorname{Re}(z_1 z_2) = x_1 x_2 - y_1 y_2 = 0$ alongside $x_2 = -x_1$ to see that $y_1 y_2 = -x_1^2 \le 0$ , 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 $\operatorname{Re}(z_1 z_2) = x_1 x_2 - y_1 y_2 = 0$ 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 $\operatorname{Re}(z_1 z_2) = -y_1 y_2 = 0$ , forcing one to be zero, which contradicts non-zero complex numbers. N/A

D · A and C

Including statement A by incorrectly writing $\operatorname{Re}(z_1 z_2)$ 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 $a_1, b_1, a_2, b_2 \in \mathbb{R}$ , satisfying $\text{Re}(z_1 z_2) = 0$ and $\text{Re}(z_1 + z_2) = 0$ .

02goal

Determine the possible signs of $\text{Im}(z_1) = b_1$ and $\text{Im}(z_2) = b_2$ 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 $\text{Re}(z_1 + z_2) = 0$ , we have $a_1 + a_2 = 0 \implies a_2 = -a_1$ . Since $z_1 z_2 = (a_1 a_2 - b_1 b_2) + i(a_1 b_2 + a_2 b_1)$ , the condition $\text{Re}(z_1 z_2) = 0$ yields $a_1 a_2 - b_1 b_2 = 0 \implies b_1 b_2 = a_1 a_2 = -a_1^2$ . 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 $z_1 + z_2 = 2i \implies \text{Re}(z_1 + z_2) = 0$ . 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 $z_1 z_2 = (1+i)(-(1+i)) = -2i$ , so $\text{Re}(z_1 z_2) = 0$ and $\text{Re}(z_1 + z_2) = 0$ . Here $\text{Im}(z_1) = 1 > 0$ and $\text{Im}(z_2) = -1 < 0$ (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 $x_1, x_2, y_1, y_2 \in \mathbb{R}$ . What algebraic equations correspond to $\operatorname{Re}(z_1 + z_2) = 0$ and $\operatorname{Re}(z_1 z_2) = 0$ ?
$x_1 + x_2 = 0$ 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 $\operatorname{Im}(z_1)$ and $\operatorname{Im}(z_2)$ ?
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
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
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 $\operatorname{Im}(z_1)$ and $\operatorname{Im}(z_2)$ 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 $\operatorname{Im}(z_1)\operatorname{Im}(z_2) = -x_1^2 < 0$ when $x_1 \neq 0$ , which prevents both imaginary parts from being negative simultaneously. Express $\operatorname{Re}(z_1 z_2) = x_1 x_2 - y_1 y_2 = 0$ alongside $x_2 = -x_1$ to see that $y_1 y_2 = -x_1^2 \le 0$ , 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 $\text{Im}(z_2) < 0$ into $\operatorname{Re}(z_1 z_2) = x_1 x_2 - y_1 y_2 = 0$ 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 $\operatorname{Re}(z_1 z_2) = -y_1 y_2 = 0$ , forcing one to be zero, which contradicts non-zero complex numbers. N/A
D: A and C - Including statement A by incorrectly writing $\operatorname{Re}(z_1 z_2)$ 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 $x_1 x_2 + y_1 y_2$ .

Step-by-step solution

given: Non-zero complex numbers $z_1 = a_1 + i b_1$ and $z_2 = a_2 + i b_2$ with $a_1, b_1, a_2, b_2 \in \mathbb{R}$ , satisfying $\text{Re}(z_1 z_2) = 0$ and $\text{Re}(z_1 + z_2) = 0$ .
goal: Determine the possible signs of $\text{Im}(z_1) = b_1$ and $\text{Im}(z_2) = b_2$ 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 $x_1 = 0$ to $a_1, a_2$ to determine the sign of the product $b_1 b_2$ .
execute: From $\text{Re}(z_1 + z_2) = 0$ , we have $a_1 + a_2 = 0 \implies a_2 = -a_1$ . Since $z_1 z_2 = (a_1 a_2 - b_1 b_2) + i(a_1 b_2 + a_2 b_1)$ , the condition $\text{Re}(z_1 z_2) = 0$ yields $a_1 a_2 - b_1 b_2 = 0 \implies b_1 b_2 = a_1 a_2 = -a_1^2$ . 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 $\text{Im}(z_1) = b_1$ and $z_2 = -1 + i$ . Then $z_1 + z_2 = 2i \implies \text{Re}(z_1 + z_2) = 0$ . 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 $z_1 z_2 = (1+i)(-(1+i)) = -2i$ , so $\text{Re}(z_1 z_2) = 0$ and $\text{Re}(z_1 + z_2) = 0$ . Here $\text{Im}(z_1) = 1 > 0$ and $\text{Im}(z_2) = -1 < 0$ (Case C). Conjugating gives Case B. Thus B and C are valid.

Complex Numbers and Quadratic Equations: Mathematics | JEE Main

What feels right?

The conditions require $\operatorname{Im}(z_1)$ and $\operatorname{Im}(z_2)$ to have opposite signs, making B and C the only possible configurations.

Why each option works or fails

Solution

01given

02goal

03approach

04execute

✓verify

We think you should solve this next

If $z$ is a complex number such that $|z| \ge 1$ , then the minimum value of…

If $\mathrm{z}=\mathrm{x}+\mathrm{i} y, \quad x y \neq 0$ , satisfies the equation…

Let $\alpha$ and $\beta$ be the roots of the equation…

Students also ask

Complex Numbers and Quadratic Equations: Mathematics | JEE Main

The conditions require Im⁡(z1)\operatorname{Im}(z_1)Im(z1​) and Im⁡(z2)\operatorname{Im}(z_2)Im(z2​) to have opposite signs, making B and C the only possible configurations.

01given

02goal

03approach

04execute

✓verify

If zzz is a complex number such that ∣z∣≥1|z| \ge 1∣z∣≥1, then the minimum value of…

If z=x+iy,xy≠0\mathrm{z}=\mathrm{x}+\mathrm{i} y, \quad x y \neq 0z=x+iy,xy=0, satisfies the equation…

Let α\alphaα and β\betaβ be the roots of the equation…

Students also ask

The conditions require $\operatorname{Im}(z_1)$ and $\operatorname{Im}(z_2)$ to have opposite signs, making B and C the only possible configurations.

If $z$ is a complex number such that $|z| \ge 1$ , then the minimum value of…

If $\mathrm{z}=\mathrm{x}+\mathrm{i} y, \quad x y \neq 0$ , satisfies the equation…

Let $\alpha$ and $\beta$ be the roots of the equation…