JEE (Main) · Mathematics, Class 11
Complex Numbers: JEE Main PYQ, 2020–2025
Complex Numbers (Mathematics, Class 11) has been asked 119 times in the 111 JEE (Main) shift papers from 2020–2025: about 1.1 questions a paper, 4% of all Mathematics questions. It is the 15th most-asked of 19 Mathematics chapters. The 2025 shifts had 15 from it in all. Last checked 28 October 2026.
Questions per year, all shifts
Year by year.
| Year | Questions |
|---|---|
| 2020 | 14in 14 papers |
| 2021 | 28in 26 papers |
| 2022 | 29in 24 papers |
| 2023 | 13in 12 papers |
| 2024 | 20in 18 papers |
| 2025 | 15in 17 papers |
JEE (Main) is sat in many shifts a year, each with its own paper; the counts add up every shift. 2021–2024 papers had 30 questions a subject, the others 25.
Inside the chapter
Topics asked most.
- Algebra of complex numbers45
- Modulus and argument43
- Argand diagram31
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10 Complex Numbers questions to practise.
Our own questions, written in the JEE Main pattern and checked, each with an explanation. Pick an option to see the answer.
Q1 · Mathematics · Complex Numbers
The multiplicative inverse of a non-zero complex number (a, b) is (a/(a² + b²), −b/(a² + b²)). What is the multiplicative inverse of (0, 2)?
Answer: (C)
With a = 0 and b = 2, a² + b² = 4. The first component is 0/4 = 0. The second is −2/4 = −1/2. So the inverse is (0, −1/2). Verify with the product rule: (0, 2)(0, −1/2) = (0·0 − 2·(−1/2), 0·(−1/2) + 2·0) = (1, 0), the multiplicative identity.
Q2 · Mathematics · Complex Numbers
With ordered-pair multiplication (a,b)(c,d) = (ac−bd, ad+bc), which of the following ordered pairs (a,b) satisfies (a,b)(a,b) = (−3, 4)?
Answer: (A)
Squaring gives (a,b)(a,b) = (a²−b², 2ab). So we need a²−b² = −3 and 2ab = 4, hence ab = 2. Test the candidates: for (1,2), a²−b² = 1−4 = −3 and 2ab = 4, so both conditions hold. The other options fail: (2,1) gives (3,4), (1,−2) gives (−3,−4) and (2,−1) gives (3,−4). So the answer is (1,2).
Q3 · Mathematics · Complex Numbers
Let i² = −1 and S = 1·i + 2·i² + 3·i³ + … + 102·i¹⁰², that is, S = Σ n·iⁿ for n = 1 to 102. What is the value of S?
Answer: (A)
Powers of i repeat with period 4: i, −1, −i, 1. Group the terms in fours, from n = 4k+1 to 4k+4. The block sum is (4k+1)i − (4k+2) − (4k+3)i + (4k+4) = 2 − 2i. This does not depend on k. Terms 1 to 100 make 25 blocks, so they sum to 25(2 − 2i) = 50 − 50i. For the remaining terms, 101 ≡ 1 (mod 4) gives i¹⁰¹ = i, so the term is 101i. Also 102 ≡ 2 (mod 4) gives i¹⁰² = −1, so the term is −102. Therefore S = 50 − 50i + 101i − 102 = −52 + 51i.
Q4 · Mathematics · Complex Numbers
In the Argand plane, the points A and B represent the complex numbers 2 + 3i and 4 − 7i respectively. Which complex number does the midpoint of the segment AB represent? (i² = −1)
Answer: (A)
The point A is (2, 3) and B is (4, −7). The midpoint of a segment has the averaged coordinates, so it represents (z₁ + z₂)/2. Here z₁ + z₂ = (2 + 4) + (3 − 7)i = 6 − 4i. Halving gives 3 − 2i, the point (3, −2).
Q5 · Mathematics · Complex Numbers
What is the principal argument of the complex number −1 + i√3?
Answer: (A)
The point (−1, √3) is in the second quadrant, and tan α = √3/1 gives the reference angle α = π/3. So the principal argument is π − π/3 = 2π/3.
Q6 · Mathematics · Complex Numbers
Complex numbers are treated as ordered pairs of reals with (a, b)(c, d) = (ac − bd, ad + bc), and (1, 0) is the multiplicative identity. For how many real pairs (x, y) ≠ (0, 0) is the multiplicative inverse of (x, y) equal to (y, x)?
Answer: (D)
The inverse of (x, y) is (y, x) exactly when (x, y)(y, x) = (1, 0). Now (x, y)(y, x) = (xy − yx, x·x + y·y) = (0, x² + y²). For this to equal (1, 0) we need 0 = 1 in the first component, which is impossible. So no pair works. The trap is to notice only x² + y² = 1 and conclude there are infinitely many solutions.
Q7 · Mathematics · Complex Numbers
Complex numbers are treated as ordered pairs of real numbers, two pairs being equal exactly when their corresponding components are equal. If (2x + 1, 3) = (5, y − 2) for real x and y, what is x + y?
Answer: (A)
Two ordered pairs are equal if and only if both their first components and their second components are equal. First components: 2x + 1 = 5, so x = 2. Second components: 3 = y − 2, so y = 5. Hence x + y = 7.
Q8 · Mathematics · Complex Numbers
Let z be a non-zero complex number with |z| = 2, and let ω be a non-real cube root of unity. The points z, ωz and ω²z are the vertices of a triangle in the Argand plane. What is the area of this triangle (in square units)?
Answer: (A)
Multiplying by ω rotates a point by 120° about the origin without changing its modulus. So z, ωz, ω²z lie on |w| = 2 at 120° intervals and form an equilateral triangle with circumradius R = 2. Its side is R√3 = 2√3, so the area is (√3/4)(2√3)² = (√3/4)(12) = 3√3. Equivalently, the triangle is made of three triangles with apex at the origin, each of area ½·2·2·sin120° = √3, which gives 3√3.
Q9 · Mathematics · Complex Numbers
Let z = 1 + cos(6π/5) + i sin(6π/5), where i² = −1. What is the principal argument of z (in radians, taken in the interval (−π, π])?
Answer: (A)
Write 1 + cosθ + i sinθ = 2cos(θ/2)[cos(θ/2) + i sin(θ/2)] with θ = 6π/5, so θ/2 = 3π/5. A polar form needs a positive modulus, but 2cos(3π/5) < 0. So z = 2|cos(3π/5)|·[−cos(3π/5) − i sin(3π/5)] = 2|cos(3π/5)|·[cos(3π/5 − π) + i sin(3π/5 − π)]. The argument is 3π/5 − π = −2π/5. Check: Re z = 1 + cos216° ≈ 0.19 > 0 and Im z = sin216° ≈ −0.59 < 0, so z is in the fourth quadrant and tan⁻¹(−0.59/0.19) ≈ −72° = −2π/5.
Q10 · Mathematics · Complex Numbers
What is the distance from the origin of the point representing z = 6 − 8i in the Argand plane? (i² = −1)
Answer: (C)
The distance of the point (x, y) from the origin is the modulus |z| = √(x² + y²). With x = 6 and y = −8, |z| = √(36 + 64) = √100 = 10.
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All 119 real JEE Main questions from Complex Numbers.
The questions from the actual papers, with the official answer and an explanation for each — chapter by chapter, or as full timed papers.
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About Complex Numbers in JEE Main.
How many questions come from Complex Numbers in JEE Main?
Complex Numbers (Mathematics, Class 11) has been asked 119 times in the 111 JEE (Main) shift papers from 2020–2025: about 1.1 questions a paper, 4% of all Mathematics questions. It is the 15th most-asked of 19 Mathematics chapters. The 2025 shifts had 15 from it in all.
Which topics of Complex Numbers are asked most in JEE Main?
Algebra of complex numbers (45), Modulus and argument (43), Argand diagram (31), counted over the papers from 2020–2025.
Is Complex Numbers important for JEE Main 2027?
It is in the syllabus in force, so it can be asked. By past papers it ranks 15 of 19 in Mathematics; use that to decide how much time it gets, not whether to study it.
Where can I solve Complex Numbers JEE Main previous year questions?
On Nikaalle, free: sign up and practise every past JEE Main question from Complex Numbers with an explanation for each, or sit the full papers timed. The ten questions on this page are our own, written in the same pattern.
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