Complex Numbers: JEE Advanced PYQ, 2007–2026

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).

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.

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.

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.

Answer: (A)

Multiplying out gives x² − y² = −5 and 2xy = 12. Use the identity (x² + y²)² = (x² − y²)² + (2xy)². This gives 25 + 144 = 169, so x² + y² = 13. Solving directly gives (x, y) = (2, 3) or (−2, −3), and both agree.

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.

Answer: (A)

Write z = r(cos θ + i sin θ) with θ = 5π/6. Then z̄ has argument −θ, and arguments add under multiplication and subtract under division. So Arg(z²/z̄) = 2θ − (−θ) = 3θ = 5π/2 (mod 2π). Subtracting 2π gives π/2, which lies in (−π, π]. Check: z²/z̄ = z³/|z|², so it has the same argument as z³, namely 3θ.

Answer: (D)

Let r = |z|. By the reverse triangle inequality, |z − 1/z| ≥ | |z| − |1/z| | = |r − 1/r|. So the given condition forces |r − 1/r| ≤ 2. Considering r − 1/r ≤ 2 with r > 0: r² − 2r − 1 ≤ 0, so r ≤ 1 + √2. Equality needs z and 1/z to point in the same direction, which holds for real z. Take z = 1 + √2: 1/z = √2 − 1, and z − 1/z = 2. So the maximum value of |z| is 1 + √2.

Answer: (A)

The inverse of z is z̄/|z|². First (2 + 3i)² = 4 + 12i − 9 = −5 + 12i. Then |z|² = (−5)² + 12² = 169. So the inverse is (−5 − 12i)/169. Check: (−5 + 12i)(−5 − 12i) = 25 + 144 = 169, so the product with the inverse is 1.

Answer: (A)

The equality |a + b| = |a| + |b| holds only when a and b point in the same direction (b = ta with t > 0, as a and b are non-zero). Apply it with a = z₁ and b = −z₂: |z₁ + (−z₂)| = |z₁| + |−z₂|. So −z₂ = t z₁ with t > 0, i.e. z₂ = −t z₁. Then |z₂| = t|z₁| = 13t = 26, so t = 2 and z₂ = −2(−5 + 12i) = 10 − 24i. Hence Re z₂ + Im z₂ = 10 − 24 = −14.