jee-main-complex-numbers-quadratic-equations-100mcq

JEE Main - Complex Numbers & Quadratic Equations Practice Test

Targeted 100 MCQ Practice Module focusing on: Complex Numbers, Argand Plane, Modulus & Argument, Roots of Unity, Quadratic Equations, Nature of Roots, and Formation of Equations (JEE Main Unit 2).

Select your answers and click Submit Assessment Answers at the bottom to calculate your total score and review explanations.

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SECTION I: Powers of i and Algebra of Complex Numbers [Questions 1 to 10]

1. The value of i²⁰²⁶ is:

−i i −1 1

Explanation: i⁴ = 1, so i²⁰²⁶ = i^(4·506 + 2) = i² = −1.

2. The value of i⁵⁷ + 1/i¹²⁵ is:

−2i 2 0 2i

Explanation: i⁵⁷ = i (since 57 = 4·14 + 1). i¹²⁵ = i, so 1/i¹²⁵ = 1/i = −i. Sum = i − i = 0.

3. The value of (1 + i)⁸ is:

8 −16 16i 16

Explanation: (1 + i)² = 2i, so (1 + i)⁸ = (2i)⁴ = 16·i⁴ = 16.

4. The imaginary part of (2 + 3i)/(1 − i) is:

1/2 5/2 5 −1/2

Explanation: (2 + 3i)(1 + i)/2 = (2 + 2i + 3i − 3)/2 = (−1 + 5i)/2. Imaginary part = 5/2.

5. If (x + iy)(2 − 3i) = 4 + i with x, y real, then x equals:

5/13 −3/13 8/13 14/13

Explanation: x + iy = (4 + i)/(2 − 3i) = (4 + i)(2 + 3i)/13 = (5 + 14i)/13, so x = 5/13.

6. The multiplicative inverse of 3 + 4i is:

(3 − 4i)/5 (−3 + 4i)/25 (3 − 4i)/25 (3 + 4i)/25

Explanation: 1/z = z̄/|z|² = (3 − 4i)/(9 + 16) = (3 − 4i)/25.

7. The value of i + i² + i³ + … + i¹⁰⁰ is:

0 i −1 1

Explanation: Every four consecutive powers of i sum to 0 (i + i² + i³ + i⁴ = 0). 100 terms = 25 such groups, so the sum is 0.

8. The square roots of −7 + 24i are:

±(3 + 4i) ±(3 − 4i) ±(4 + 3i) ±(4 − 3i)

Explanation: (3 + 4i)² = 9 − 16 + 24i = −7 + 24i. Hence the square roots are ±(3 + 4i).

9. The value of ((1 + i)/(1 − i))²⁰²⁷ is:

−i i 1 −1

Explanation: (1 + i)/(1 − i) = (1 + i)²/2 = i. Then i²⁰²⁷ = i^(4·506 + 3) = i³ = −i.

10. For real θ (θ not a multiple of 2π), the real part of 1/(1 − cos θ + i sin θ) is:

1/(1 − cos θ) 0 1 1/2

Explanation: Multiply numerator and denominator by the conjugate: denominator = (1 − cos θ)² + sin²θ = 2(1 − cos θ). Real part = (1 − cos θ)/(2(1 − cos θ)) = 1/2.

SECTION II: Conjugate and Modulus [Questions 11 to 20]

11. The conjugate of (2 + i)/(1 − i) is:

(−1 − 3i)/2 (1 + 3i)/2 (3 − i)/2 (1 − 3i)/2

Explanation: (2 + i)/(1 − i) = (2 + i)(1 + i)/2 = (1 + 3i)/2. Its conjugate is (1 − 3i)/2.

12. The modulus of the complex number 3 − 4i is:

1 5 7 √7

Explanation: |3 − 4i| = √(3² + (−4)²) = √25 = 5.

13. The modulus of (1 + 2i)(2 − i)/(1 + i) is:

5√2 25/√2 5 5/√2

Explanation: |1 + 2i| = √5, |2 − i| = √5, |1 + i| = √2. Modulus = (√5·√5)/√2 = 5/√2.

14. If |z₁| = 2, |z₂| = 3 and |z₁ + z₂| = 4, then |z₁ − z₂|² equals:

6 13 9 10

Explanation: Parallelogram law: |z₁ + z₂|² + |z₁ − z₂|² = 2(|z₁|² + |z₂|²) → 16 + |z₁ − z₂|² = 26 → |z₁ − z₂|² = 10.

15. The number of complex numbers z satisfying z̄ = z² is:

3 4 2 1

Explanation: Taking modulus: |z| = |z|², so |z| = 0 or 1. If |z| = 1, then z³ = z·z̄ = 1, giving 3 roots. Together with z = 0, total = 4.

16. A non-zero complex number z is purely imaginary if and only if:

z + z̄ = 0 z = z̄ z·z̄ = 0 z − z̄ = 0

Explanation: z = iy (y ≠ 0) gives z̄ = −iy, so z + z̄ = 0. The condition z + z̄ = 0 means the real part is 0.

17. The locus of z satisfying |z − 1| = |z + i| is:

the straight line y = −x a circle the straight line y = x the line x = 1

Explanation: (x − 1)² + y² = x² + (y + 1)² → −2x = 2y → x + y = 0, i.e. y = −x (perpendicular bisector of the points 1 and −i).

18. If |z| = 1, the maximum value of |z − 3 + 4i| is:

7 4 5 6

Explanation: |z − (3 − 4i)| ≤ |z| + |3 − 4i| = 1 + 5 = 6, attained when z is opposite in direction to 3 − 4i.

19. If |z| = 1 and z ≠ ±1, then (z − 1)/(z + 1) is:

purely imaginary of modulus 1 equal to zero purely real

Explanation: Let z = cos θ + i sin θ. Then (z − 1)/(z + 1) = i tan(θ/2), which is purely imaginary (non-zero for z ≠ ±1).

20. If |z + 4| ≤ 3, then the maximum value of |z + 1| is:

3 6 4 5

Explanation: |z + 1| = |(z + 4) − 3| ≤ |z + 4| + 3 ≤ 3 + 3 = 6.

SECTION III: Argument and Polar Form [Questions 21 to 30]

21. The principal argument of −1 − i is:

3π/4 5π/4 −π/4 −3π/4

Explanation: The point (−1, −1) lies in the third quadrant, so the principal argument (in (−π, π]) is −π + π/4 = −3π/4.

22. The principal argument of 1 + i√3 is:

π/2 π/6 π/3 2π/3

Explanation: tan θ = √3/1 = √3 and the point lies in the first quadrant, so θ = π/3.

23. The polar form of −√3 + i is:

2(cos 5π/6 + i sin 5π/6) √2(cos 5π/6 + i sin 5π/6) 2(cos π/6 + i sin π/6) 2(cos 2π/3 + i sin 2π/3)

Explanation: Modulus = √(3 + 1) = 2. cos θ = −√3/2 and sin θ = 1/2 gives θ = 5π/6.

24. If arg z = θ with 0 < θ < π, then the principal argument of −z̄ is:

−θ θ − π π − θ π + θ

Explanation: arg z̄ = −θ. Multiplying by −1 adds π: −θ + π = π − θ, which lies in (0, π).

25. For 0 < θ < π, the principal argument of 1 + cos θ + i sin θ is:

2θ θ π − θ/2 θ/2

Explanation: 1 + cos θ + i sin θ = 2cos(θ/2)[cos(θ/2) + i sin(θ/2)] with 2cos(θ/2) > 0, so the argument is θ/2.

26. If |z₁ + z₂| = |z₁ − z₂| (z₁, z₂ ≠ 0), then arg(z₁/z₂) equals:

±π/4 ±π/2 0 π

Explanation: Equal diagonals of the parallelogram means it is a rectangle, so z₁ and z₂ are perpendicular: Re(z₁ z̄₂) = 0, hence arg(z₁/z₂) = ±π/2.

27. The principal argument of (1 + i√3)³ is:

π −π π/3 0

Explanation: (1 + i√3)³ = [2 e^(iπ/3)]³ = 8 e^(iπ) = −8, whose principal argument is π (range (−π, π]).

28. The value of (cos 30° + i sin 30°)⁶ is:

−1 1 −i i

Explanation: By De Moivre's theorem: cos 180° + i sin 180° = −1.

29. If z = 2(cos π/3 + i sin π/3), then 1/z in polar form is:

2[cos(−π/3) + i sin(−π/3)] (1/2)[cos(−π/3) + i sin(−π/3)] (1/2)(cos π/3 + i sin π/3) (1/2)[cos(2π/3) + i sin(2π/3)]

Explanation: 1/z = (1/r)[cos(−θ) + i sin(−θ)] = (1/2)[cos(−π/3) + i sin(−π/3)].

30. If z = (√3 + i)/2, then z⁶⁹ equals:

−i 1 i −1

Explanation: z = cos(π/6) + i sin(π/6), so z⁶⁹ = cos(69π/6) + i sin(69π/6) = cos(23π/2) + i sin(23π/2) = cos(3π/2) + i sin(3π/2) = −i.

SECTION IV: Argand Plane, Geometry and Roots of Unity [Questions 31 to 40]

31. The area of the triangle whose vertices are represented by 0, 4 and 3i in the Argand plane is:

5 sq. units 12 sq. units 6 sq. units 7 sq. units

Explanation: Vertices (0,0), (4,0), (0,3) form a right triangle: area = ½ × 4 × 3 = 6.

32. The distance between the points 5 − 2i and −7 + 3i in the Argand plane is:

13 17 12 √119

Explanation: Distance = √((5 + 7)² + (−2 − 3)²) = √(144 + 25) = 13.

33. The locus of z satisfying |z − 1| + |z + 1| = 4 is:

a hyperbola a circle an ellipse a parabola

Explanation: The sum of distances from two fixed points (±1) is a constant 4 greater than the distance between them (2), which defines an ellipse.

34. If ω is an imaginary cube root of unity, then (1 + ω − ω²)³ equals:

−8 −8ω 0 8

Explanation: 1 + ω = −ω², so 1 + ω − ω² = −2ω². Hence (−2ω²)³ = −8ω⁶ = −8.

35. The value of (1 + ω)(1 + ω²)(1 + ω⁴)(1 + ω⁸), where ω is a complex cube root of unity, is:

0 1 2 −1

Explanation: (1 + ω)(1 + ω²) = 1 + ω + ω² + ω³ = 0 + 1 = 1. Also 1 + ω⁴ = 1 + ω and 1 + ω⁸ = 1 + ω², so the product is 1·1 = 1.

36. If ω is a complex cube root of unity, then ω¹⁰⁰ + ω²⁰⁰ + ω³⁰⁰ equals:

3 1 −1 0

Explanation: ω¹⁰⁰ = ω, ω²⁰⁰ = ω², ω³⁰⁰ = 1. Sum = 1 + ω + ω² = 0.

37. The number of non-real roots of the equation z⁶ = 1 is:

6 4 2 3

Explanation: z⁶ = 1 has six roots, of which only z = 1 and z = −1 are real. So 4 are non-real.

38. The product of all the fourth roots of unity is:

−1 0 1 i

Explanation: Roots are 1, −1, i, −i. Product = 1·(−1)·(i)(−i) = (−1)(1) = −1.

39. If z = cos(2π/7) + i sin(2π/7), then z + z² + z³ + z⁴ + z⁵ + z⁶ equals:

7 1 −1 0

Explanation: z is a primitive 7th root of unity, so 1 + z + … + z⁶ = 0, giving z + … + z⁶ = −1.

40. The value of (1 + i)¹⁰ is:

32(1 + i) −32i 32i 32

Explanation: (1 + i)² = 2i, so (1 + i)¹⁰ = (2i)⁵ = 32 i⁵ = 32i.

SECTION V: Solving Quadratic Equations [Questions 41 to 50]

41. The roots of 2x² − 7x + 3 = 0 are:

1 and 3/2 3 and 1/2 −3 and −1/2 3 and −1/2

Explanation: 2x² − 7x + 3 = (2x − 1)(x − 3) = 0, so x = 1/2 or x = 3.

42. The value of k for which 3x² − 4x + k = 0 has equal roots is:

3/4 −4/3 4/3 16/3

Explanation: D = 16 − 12k = 0 → k = 4/3.

43. The equation x² + kx + 4 = 0 has real roots for:

k ≤ −2 or k ≥ 2 k ≤ −4 or k ≥ 4 −4 < k < 4 −4 ≤ k ≤ 4

Explanation: D = k² − 16 ≥ 0 → k² ≥ 16 → k ≤ −4 or k ≥ 4.

44. The roots of x² − 2x + 5 = 0 are:

1 ± 4i 1 ± 2i 2 ± i −1 ± 2i

Explanation: x = [2 ± √(4 − 20)]/2 = [2 ± 4i]/2 = 1 ± 2i.

45. The roots of x² + x + 1 = 0 are:

(1 ± i√3)/2 (−1 ± √3)/2 (−1 ± i)/2 (−1 ± i√3)/2

Explanation: x = [−1 ± √(1 − 4)]/2 = (−1 ± i√3)/2, i.e. the complex cube roots of unity ω and ω².

46. The roots of z² − (3 + i)z + (2 + i) = 0 are:

2 and 1 + i i and 3 1 and 2 − i 1 and 2 + i

Explanation: Sum of roots = 1 + (2 + i) = 3 + i and product = 2 + i, matching the coefficients. (Also z = 1 satisfies: 1 − 3 − i + 2 + i = 0.)

47. The number of real solutions of |x|² − 3|x| + 2 = 0 is:

4 0 3 2

Explanation: Put |x| = t: t² − 3t + 2 = 0 → t = 1 or 2. So x = ±1, ±2, giving 4 real solutions.

48. The sum of all real values of x satisfying (x² − 5x + 5)^(x² + 4x − 60) = 1 is:

5 4 3 −3

Explanation: Base = 1: x² − 5x + 4 = 0 → x = 1, 4. Exponent = 0: x² + 4x − 60 = 0 → x = −10, 6. Base = −1 with even exponent: x² − 5x + 6 = 0 → x = 2, 3; only x = 2 gives an even exponent (−48). Sum = 1 + 4 − 10 + 6 + 2 = 3.

49. The solution set of √(x + 5) = x − 1 is:

{4, −1} empty set {−1} {4}

Explanation: Squaring: x + 5 = x² − 2x + 1 → x² − 3x − 4 = 0 → x = 4 or −1. For x = −1, RHS = −2 < 0 (extraneous). Only x = 4 works.

50. The number of real solutions of |x² − x − 6| = x + 2 is:

2 3 1 4

Explanation: x² − x − 6 = (x − 3)(x + 2). Case +: x² − 2x − 8 = 0 → x = 4, −2. Case −: 4 − x² = 0 → x = ±2. All satisfy x + 2 ≥ 0, giving distinct solutions −2, 2, 4, i.e. 3 solutions.

SECTION VI: Relations Between Roots and Coefficients [Questions 51 to 60]

51. If α and β are the roots of x² − 7x + 12 = 0, then α² + β² equals:

13 37 25 49

Explanation: α + β = 7, αβ = 12 → α² + β² = 49 − 24 = 25.

52. If α, β are the roots of x² − 6x + 4 = 0, then α³ + β³ equals:

216 144 72 108

Explanation: α + β = 6, αβ = 4 → α³ + β³ = (α + β)³ − 3αβ(α + β) = 216 − 72 = 144.

53. If α, β are the roots of x² − 5x + 3 = 0, then |α − β| equals:

5 13 √13 √37

Explanation: (α − β)² = (α + β)² − 4αβ = 25 − 12 = 13, so |α − β| = √13.

54. If the sum of the roots of kx² + 2x + 3k = 0 equals their product, then k is:

2/3 3/2 −2/3 −3/2

Explanation: Sum = −2/k, product = 3k/k = 3. Equating: −2/k = 3 → k = −2/3.

55. If 4 is a root of x² + px + 12 = 0 and x² + px + q = 0 has equal roots, then q equals:

7 4 49/4 12

Explanation: 16 + 4p + 12 = 0 → p = −7. For x² − 7x + q = 0 to have equal roots: 49 − 4q = 0 → q = 49/4.

56. If the roots of x² − 10x + k = 0 are in the ratio 2 : 3, then k equals:

30 20 12 24

Explanation: Let the roots be 2m and 3m: 5m = 10 → m = 2, so roots are 4 and 6 and k = 24.

57. If α, β are the roots of x² − x − 1 = 0, then α⁴ + β⁴ equals:

7 11 5 9

Explanation: α + β = 1, αβ = −1 → α² + β² = 1 + 2 = 3. α⁴ + β⁴ = (α² + β²)² − 2(αβ)² = 9 − 2 = 7.

58. If α, β are the roots of x² − 2x + 3 = 0, then 1/α + 1/β equals:

−2/3 3/2 2/3 6

Explanation: 1/α + 1/β = (α + β)/(αβ) = 2/3.

59. If the roots of ax² + bx + c = 0 are reciprocals of each other, then:

a = −c b² = 4ac a = c b = 0

Explanation: Product of roots = c/a = 1 → c = a.

60. If the difference between the roots of x² + px + 8 = 0 is 2, then p equals:

±4 ±8 ±2 ±6

Explanation: (α − β)² = p² − 32 = 4 → p² = 36 → p = ±6.

SECTION VII: Formation of Quadratic Equations [Questions 61 to 70]

61. The quadratic equation having roots 3 and −5 is:

x² − 8x − 15 = 0 x² + 2x − 15 = 0 x² + 2x + 15 = 0 x² − 2x − 15 = 0

Explanation: Sum = −2, product = −15, so x² − (sum)x + product = x² + 2x − 15 = 0.

62. The quadratic equation with roots 2 + √3 and 2 − √3 is:

x² − 4x − 1 = 0 x² − 2x + 1 = 0 x² + 4x + 1 = 0 x² − 4x + 1 = 0

Explanation: Sum = 4, product = 4 − 3 = 1, so x² − 4x + 1 = 0.

63. The quadratic equation with roots −3 + i and −3 − i is:

x² − 6x + 10 = 0 x² + 6x + 10 = 0 x² + 6x − 10 = 0 x² + 6x + 8 = 0

Explanation: Sum = −6, product = 9 + 1 = 10, so x² + 6x + 10 = 0.

64. If α, β are roots of x² − 3x + 1 = 0, the equation whose roots are 2α and 2β is:

x² − 6x + 4 = 0 x² − 3x + 4 = 0 x² − 6x + 1 = 0 x² − 6x + 2 = 0

Explanation: New sum = 2(α + β) = 6, new product = 4αβ = 4. Equation: x² − 6x + 4 = 0.

65. If α, β are roots of x² − 4x + 1 = 0, the equation whose roots are α² and β² is:

x² − 4x + 1 = 0 x² − 16x + 1 = 0 x² − 14x − 1 = 0 x² − 14x + 1 = 0

Explanation: α² + β² = 16 − 2 = 14, α²β² = 1. Equation: x² − 14x + 1 = 0.

66. The equation whose roots are the reciprocals of the roots of 2x² − 5x + 3 = 0 is:

2x² − 5x + 3 = 0 3x² + 5x + 2 = 0 2x² + 5x + 3 = 0 3x² − 5x + 2 = 0

Explanation: Replace x by 1/x: 2/x² − 5/x + 3 = 0 → 3x² − 5x + 2 = 0.

67. If α, β are roots of x² − 3x + 1 = 0, the equation whose roots are α + 1 and β + 1 is:

x² − 3x + 5 = 0 x² + 5x + 5 = 0 x² − 5x + 5 = 0 x² − 5x + 1 = 0

Explanation: Sum = 3 + 2 = 5; product = αβ + (α + β) + 1 = 1 + 3 + 1 = 5. Equation: x² − 5x + 5 = 0.

68. If α, β are roots of x² − 4x + 2 = 0, the equation whose roots are α/β and β/α is:

x² − 6x + 2 = 0 x² − 6x + 1 = 0 x² − 4x + 1 = 0 x² + 6x + 1 = 0

Explanation: α/β + β/α = (α² + β²)/αβ = (16 − 4)/2 = 6; product = 1. Equation: x² − 6x + 1 = 0.

69. A quadratic equation with real coefficients has 2 + 3i as one root. The equation is:

x² − 4x + 13 = 0 x² − 4x + 5 = 0 x² + 4x + 13 = 0 x² − 2x + 13 = 0

Explanation: The other root is 2 − 3i. Sum = 4, product = 4 + 9 = 13. Equation: x² − 4x + 13 = 0.

70. If α, β are roots of x² − 3x + 1 = 0, the equation whose roots are α³ and β³ is:

x² − 18x − 1 = 0 x² − 9x + 1 = 0 x² − 27x + 1 = 0 x² − 18x + 1 = 0

Explanation: α³ + β³ = 27 − 3·1·3 = 18, α³β³ = 1. Equation: x² − 18x + 1 = 0.

SECTION VIII: Nature of Roots and Parameters [Questions 71 to 80]

71. The values of k for which (k − 2)x² + 8x + (k + 4) = 0 has real and equal roots are:

k = 2 or k = −4 k = 4 or k = −6 k = −4 or k = 6 k = 4 or k = 6

Explanation: D = 64 − 4(k − 2)(k + 4) = 0 → (k − 2)(k + 4) = 16 → k² + 2k − 24 = 0 → k = 4 or −6.

72. If a, b, c are real, a ≠ 0 and a + b + c = 0, the roots of ax² + bx + c = 0 are:

non-real imaginary always equal real

Explanation: D = b² − 4ac = (a + c)² − 4ac = (a − c)² ≥ 0, so the roots are always real (one root is x = 1).

73. If p, q are rational and one root of x² + px + q = 0 is 3 + √5, the other root is:

3 − √5 √5 − 3 1/(3 + √5) −3 − √5

Explanation: For rational coefficients, irrational roots occur in conjugate pairs, so the other root is 3 − √5.

74. If the roots of x² − bx + c = 0 differ by 1, then:

b² = 4c − 1 b² = 4c + 1 b² = c + 1 c² = 4b + 1

Explanation: (α − β)² = b² − 4c = 1 → b² = 4c + 1.

75. If one root of ax² + bx + c = 0 is double the other, then:

2b² = 9ac 9b² = 2ac b² = 9ac b² = 4ac

Explanation: Let roots be α, 2α. Then 3α = −b/a and 2α² = c/a. Substituting α = −b/(3a): 2b²/(9a²) = c/a → 2b² = 9ac.

76. If the sum of the roots of ax² + bx + c = 0 equals the sum of the squares of their reciprocals, then ab², bc², ca² satisfy:

they are all equal they are in H.P. they are in G.P. they are in A.P.

Explanation: −b/a = (b² − 2ac)/c² → −bc² = ab² − 2a²c → ab² + bc² = 2ca², so ab², ca², bc² are in A.P.

77. For real x, the range of (x² + x + 1)/(x² − x + 1) is:

[−3, 3] (1/3, 3) (−∞, 1/3] ∪ [3, ∞) [1/3, 3]

Explanation: Let y = (x² + x + 1)/(x² − x + 1). Then (y − 1)x² − (y + 1)x + (y − 1) = 0 has real x iff (y + 1)² − 4(y − 1)² ≥ 0 → (3 − y)(3y − 1) ≥ 0 → 1/3 ≤ y ≤ 3.

78. The minimum value of 2x² − 8x + 9 for real x is:

−1 9 1 2

Explanation: 2x² − 8x + 9 = 2(x − 2)² + 1 ≥ 1, with equality at x = 2.

79. The maximum value of −3x² + 12x − 5 for real x is:

12 7 9 −5

Explanation: −3x² + 12x − 5 = −3(x − 2)² + 7 ≤ 7, with equality at x = 2.

80. The equation x² + mx + 1 = 0 has non-real roots when:

−1 < m < 1 −2 ≤ m ≤ 2 m < −2 or m > 2 −2 < m < 2

Explanation: D = m² − 4 < 0 → −2 < m < 2.

SECTION IX: Locus, Inequalities and Advanced Complex Numbers [Questions 81 to 90]

81. The locus of the point z = x + iy satisfying Re(z²) = 0 is:

a pair of straight lines y = ±x a hyperbola a parabola a circle

Explanation: z² = (x² − y²) + 2ixy, so Re(z²) = x² − y² = 0 → y = x or y = −x.

82. The locus of z satisfying |z − 2| = |z + 2| is:

a circle a hyperbola the imaginary axis the real axis

Explanation: z is equidistant from 2 and −2, so it lies on their perpendicular bisector, i.e. x = 0, the imaginary axis.

83. The locus of z satisfying |z − 1| = 2|z − 4| is:

an ellipse a straight line a circle a parabola

Explanation: The ratio of distances from two fixed points is a constant ≠ 1 — an Apollonius circle.

84. For z ≠ 0 and ω a complex cube root of unity, the points z, ωz and ω²z form:

a collinear set an isosceles right triangle an equilateral triangle a scalene triangle

Explanation: The points lie on a circle of radius |z| centred at the origin and are separated by 120° rotations, forming an equilateral triangle.

85. If |z₁ + z₂| = |z₁| + |z₂|, then arg z₁ − arg z₂ equals:

π/2 0 π/4 π

Explanation: Equality in the triangle inequality holds only when z₁ and z₂ have the same direction, i.e. arg z₁ = arg z₂.

86. If |z + 2/z| = 2, then the greatest value of |z| is:

√3 − 1 1 + √3 1 + √2 2

Explanation: |z| = |(z + 2/z) − 2/z| ≤ 2 + 2/|z|. With r = |z|: r² − 2r − 2 ≤ 0 → r ≤ 1 + √3.

87. If |z − 1| = 1 and arg z = π/4, then z equals:

1 + i (1 + i)/√2 √2 i 1 − i

Explanation: z = r e^(iπ/4). |z − 1|² = r² − √2 r + 1 = 1 → r = √2 (r ≠ 0). So z = √2(cos π/4 + i sin π/4) = 1 + i.

88. The number of complex numbers z satisfying z³ = z̄ is:

3 5 6 4

Explanation: |z|³ = |z| → |z| = 0 or 1. For |z| = 1, z³ = 1/z → z⁴ = 1 (4 roots). With z = 0, total = 5.

89. The smallest positive integer n for which ((1 + i)/(1 − i))ⁿ = 1 is:

2 8 4 1

Explanation: (1 + i)/(1 − i) = i, and iⁿ = 1 first holds at n = 4.

90. The value of ((1 + i)/√2)²⁰²⁶ is:

i −1 −i 1

Explanation: (1 + i)/√2 = e^(iπ/4). Raised to 2026: e^(i·2026π/4) = e^(i·1013π/2) = e^(i(506π + π/2)) = i.

SECTION X: Mixed JEE-Style Problems [Questions 91 to 100]

91. If α and β are the roots of x² − 6x − 2 = 0 with α > β, and aₙ = αⁿ − βⁿ for n ≥ 1, then (a₁₀ − 2a₈)/(2a₉) equals:

2 4 6 3

Explanation: α² = 6α + 2 → α¹⁰ − 2α⁸ = α⁸(α² − 2) = 6α⁹; similarly for β. So a₁₀ − 2a₈ = 6a₉, and the ratio = 6a₉/(2a₉) = 3.

92. If α and β are the roots of x² + x + 1 = 0, then α²⁰²⁵ + β²⁰²⁵ equals:

−1 2 1 0

Explanation: α = ω, β = ω². ω²⁰²⁵ = (ω³)⁶⁷⁵ = 1 and ω⁴⁰⁵⁰ = (ω³)¹³⁵⁰ = 1. Sum = 2.

93. If x² + ax + b = 0 and x² + bx + a = 0 (a ≠ b) have a common root, then a + b equals:

0 2 1 −1

Explanation: Subtracting: (a − b)x + (b − a) = 0 → x = 1. Substituting: 1 + a + b = 0 → a + b = −1.

94. If x² + 2x + 3 = 0 and ax² + bx + c = 0 (a, b, c real) have a common root, then a : b : c is:

1 : 3 : 2 1 : 2 : 3 3 : 2 : 1 2 : 1 : 3

Explanation: x² + 2x + 3 = 0 has non-real roots; for real coefficients, complex roots come in conjugate pairs, so both roots are common and the equations are proportional: a : b : c = 1 : 2 : 3.

95. The sum of squares of the roots of x² − (sin α − 2)x − (1 + sin α) = 0 is least when α equals:

π/4 π/2 0 π/6

Explanation: Sum of squares = (sin α − 2)² + 2(1 + sin α) = sin²α − 2 sin α + 6 = (sin α − 1)² + 5, minimum when sin α = 1, i.e. α = π/2.

96. If |z| = 1 and z ≠ −1, then (1 + z)/(1 + z̄) equals:

z −z 1 z̄

Explanation: Since z̄ = 1/z: (1 + z)/(1 + 1/z) = z(1 + z)/(z + 1) = z.

97. If (1 + i)z = (1 − i)z̄, where z = x + iy, then z can be written as (x real):

x(1 − i) xi x(1 + i) x

Explanation: Expanding: (x − y) + i(x + y) = (x − y) − i(x + y) → x + y = 0 → y = −x, so z = x − ix = x(1 − i).

98. If α, β are the roots of x² − 2x + 4 = 0, then α⁶ + β⁶ equals:

−128 64 128 0

Explanation: Roots are 1 ± i√3 = 2e^(±iπ/3). α⁶ = 64e^(2πi) = 64 and β⁶ = 64. Sum = 128.

99. If z = x + iy and z^(1/3) = a + ib, then x/a + y/b equals:

2(a² − b²) 4(a² + b²) 4(a² − b²) a² − b²

Explanation: z = (a + ib)³ = (a³ − 3ab²) + i(3a²b − b³). So x/a = a² − 3b² and y/b = 3a² − b². Sum = 4a² − 4b² = 4(a² − b²).

100. The number of complex solutions of z² + |z| = 0 is:

1 4 2 3

Explanation: z = 0 is a solution. For z ≠ 0: |z| = −z² means z² is a negative real, so z = ±ir (r > 0): −r² + r = 0 → r = 1, z = ±i. Total 3 solutions.

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