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By rohit.pandey1
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Updated on 10 Jun 2026, 11:56 IST
Complex Numbers and Quadratic Equations is an important chapter in Class 11 Maths and a useful foundation for JEE algebra. The chapter starts with the need to extend the real number system so that equations like x² + 1 = 0 can be solved. It introduces the imaginary unit i, complex numbers of the form a + ib, algebra of complex numbers, powers of i, square roots of negative numbers, and quadratic equations with real or complex roots.
This page includes 100 Complex Numbers and Quadratic Equations MCQ Questions with Answers for Class 11, JEE Foundation and JEE Main practice. These topic-wise objective questions cover complex number basics, real and imaginary parts, equality of complex numbers, addition, subtraction, multiplication, division, powers of i, conjugate, modulus, Argand plane, polar form, quadratic formula, discriminant, nature of roots, sum and product of roots, formation of quadratic equations and complex roots.
A complex number is a number of the form a + ib, where a and b are real numbers and i² = −1. Here, a is called the real part and b is called the imaginary part.
A quadratic equation is an equation of the form:
ax² + bx + c = 0, where a ≠ 0.
The roots of a quadratic equation are found using:
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x = (-b ± √(b² − 4ac))/(2a)
The expression b² − 4ac is called the discriminant. If the discriminant is negative, the roots are complex conjugates.
| Term | Meaning |
| Imaginary unit | i = √−1, so i² = −1 |
| Complex number | A number of the form a + ib |
| Real part | For z = a + ib, Re(z) = a |
| Imaginary part | For z = a + ib, Im(z) = b |
| Pure real number | A complex number with imaginary part 0 |
| Pure imaginary number | A complex number with real part 0 |
| Conjugate | If z = a + ib, then z̄ = a − ib |
| Modulus | ` |
| Argand plane | Plane used to represent complex numbers |
| Quadratic equation | ax² + bx + c = 0, where a ≠ 0 |
| Discriminant | D = b² − 4ac |
| Complex roots | Roots involving the imaginary unit i |
| Concept | Formula |
| Imaginary unit | i = √−1, i² = −1 |
| Complex number | z = a + ib |
| Equality | a + ib = c + id ⇒ a = c and b = d |
| Addition | (a+ib) + (c+id) = (a+c) + i(b+d) |
| Subtraction | (a+ib) − (c+id) = (a−c) + i(b−d) |
| Multiplication | (a+ib)(c+id) = (ac−bd) + i(ad+bc) |
| Conjugate | If z = a+ib, then z̄ = a−ib |
| Modulus | ` |
| Product with conjugate | `z z̄ = |
| Reciprocal | `1/z = z̄/ |
| Quadratic formula | x = (-b ± √(b²−4ac))/(2a) |
| Discriminant | D = b² − 4ac |
| Sum of roots | α + β = −b/a |
| Product of roots | αβ = c/a |
| Equation from roots | x² − (sum of roots)x + product of roots = 0 |
The powers of i repeat in a cycle of 4.
| Power | Value |
i¹ | i |
i² | −1 |
i³ | −i |
i⁴ | 1 |
i⁴ᵏ | 1 |
i⁴ᵏ⁺¹ | i |
i⁴ᵏ⁺² | −1 |
i⁴ᵏ⁺³ | −i |
Q1. The value of i² is:
(a) 1
(b) −1
(c) i
(d) −i
Answer: (b) −1
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Solution:
By definition, i = √−1. Therefore, i² = −1.
Q2. A complex number is generally written in the form:
(a) a + b
(b) a − b
(c) a + ib
(d) ab + i
Answer: (c) a + ib
Solution:
A complex number is written as a + ib, where a and b are real numbers and i² = −1.
Q3. In the complex number z = 5 + 3i, the real part is:
(a) 3
(b) 5
(c) i
(d) 8
Answer: (b) 5
Solution:
For z = a + ib, the real part is a. Here, a = 5.
Q4. In the complex number z = 5 + 3i, the imaginary part is:
(a) 5
(b) 3
(c) 3i
(d) 8
Answer: (b) 3
Solution:
For z = a + ib, the imaginary part is b, not ib. Hence, Im(z) = 3.
Q5. The complex number 7 can be written as:
(a) 0 + 7i
(b) 7 + 0i
(c) 7i
(d) 0 − 7i
Answer: (b) 7 + 0i
Solution:
Every real number can be written as a complex number with zero imaginary part. Therefore, 7 = 7 + 0i.
Q6. The complex number −4i can be written as:
(a) −4 + 0i
(b) 0 − 4i
(c) 4 + 0i
(d) 0 + 4i
Answer: (b) 0 − 4i
Solution:−4i has real part 0 and imaginary part −4.
Q7. A purely imaginary number has:
(a) Real part zero
(b) Imaginary part zero
(c) Both parts zero
(d) No imaginary part
Answer: (a) Real part zero
Solution:
A purely imaginary number is of the form 0 + ib, where b ≠ 0.
Q8. Which of the following is a purely real complex number?
(a) 3 + 2i
(b) 5i
(c) −7 + 0i
(d) 0 + 4i
Answer: (c) −7 + 0i
Solution:
A purely real complex number has imaginary part zero. Here, −7 + 0i is real.
Q9. If z = −2 + 9i, then Re(z) is:
(a) −2
(b) 9
(c) −9
(d) 2
Answer: (a) −2
Solution:
For z = a + ib, Re(z) = a. Here, a = −2.
Q10. If z = −2 + 9i, then Im(z) is:
(a) −2
(b) 9
(c) 9i
(d) −9
Answer: (b) 9
Solution:
For z = a + ib, Im(z) = b. Here, b = 9.
Q11. If z = 6 − 11i, then Im(z) is:
(a) 6
(b) −11
(c) 11
(d) −11i
Answer: (b) −11
Solution:
The imaginary part is the coefficient of i. Hence, Im(z) = −11.
Q12. If z = 0 + 8i, then Re(z) is:
(a) 0
(b) 8
(c) 8i
(d) 1
Answer: (a) 0
Solution:
The real part of 0 + 8i is 0.
Q13. If z = a + ib is purely real, then:
(a) a = 0
(b) b = 0
(c) a = b
(d) a + b = 0
Answer: (b) b = 0
Solution:
A purely real number has no imaginary part, so b = 0.
Q14. If z = a + ib is purely imaginary, then:
(a) a = 0
(b) b = 0
(c) a = b
(d) ab = 1
Answer: (a) a = 0
Solution:
A purely imaginary number has real part zero.
Q15. If z = 3 − 4i, then Re(z) + Im(z) is:
(a) 7
(b) −1
(c) 1
(d) −7
Answer: (b) −1
Solution:Re(z) = 3 and Im(z) = −4.
So, Re(z) + Im(z) = 3 − 4 = −1.
Q16. If z = −5 − 2i, then Re(z) − Im(z) is:
(a) −7
(b) −3
(c) 3
(d) 7
Answer: (b) −3
Solution:Re(z) = −5, Im(z) = −2.Re(z) − Im(z) = −5 − (−2) = −3.
Q17. If a + ib = c + id, then:
(a) a = d and b = c
(b) a = c and b = d
(c) a + b = c + d
(d) ab = cd
Answer: (b) a = c and b = d
Solution:
Two complex numbers are equal when their real parts and imaginary parts are equal.
Q18. If x + 2i = 5 + yi, then:
(a) x = 2, y = 5
(b) x = 5, y = 2
(c) x = 7, y = 0
(d) x = 0, y = 7
Answer: (b) x = 5, y = 2
Solution:
Equating real and imaginary parts:x = 5 and 2 = y.
Q19. If 3 + xi = y + 4i, then:
(a) x = 3, y = 4
(b) x = 4, y = 3
(c) x = y
(d) x + y = 4
Answer: (b) x = 4, y = 3
Solution:
Real parts: 3 = y, so y = 3.
Imaginary parts: x = 4.
Q20. If 2x + 3i = 8 + yi, then:
(a) x = 4, y = 3
(b) x = 3, y = 4
(c) x = 8, y = 3
(d) x = 2, y = 8
Answer: (a) x = 4, y = 3
Solution:
Equating real parts: 2x = 8, so x = 4.
Equating imaginary parts: 3 = y.
Q21. If (x + 1) + 5i = 7 + (y − 2)i, then:
(a) x = 6, y = 7
(b) x = 7, y = 6
(c) x = 5, y = 7
(d) x = 6, y = 5
Answer: (a) x = 6, y = 7
Solution:
Real parts: x + 1 = 7, so x = 6.
Imaginary parts: 5 = y − 2, so y = 7.
Q22. If x + yi = 0, then:
(a) x = 0, y = 1
(b) x = 1, y = 0
(c) x = 0, y = 0
(d) x = y = i
Answer: (c) x = 0, y = 0
Solution:
The zero complex number is 0 + 0i. Hence, x = 0 and y = 0.
Q23. (3 + 2i) + (4 + 5i) is equal to:
(a) 7 + 7i
(b) 7 − 7i
(c) 1 + 3i
(d) 12 + 10i
Answer: (a) 7 + 7i
Solution:
Add real parts and imaginary parts separately:(3+4) + (2+5)i = 7 + 7i.
Q24. (8 + 6i) − (3 + 2i) is equal to:
(a) 5 + 4i
(b) 11 + 8i
(c) 5 − 4i
(d) 11 − 8i
Answer: (a) 5 + 4i
Solution:(8−3) + (6−2)i = 5 + 4i.
Q25. (2 + 3i)(4 + i) is equal to:
(a) 5 + 14i
(b) 8 + 12i
(c) 11 + 14i
(d) 8 + 3i
Answer: (a) 5 + 14i
Solution:(2 + 3i)(4 + i) = 8 + 2i + 12i + 3i²= 8 + 14i − 3= 5 + 14i.
Q26. (1 + i)² is equal to:
(a) 1 + i
(b) 2i
(c) −2i
(d) 2
Answer: (b) 2i
Solution:(1+i)² = 1 + 2i + i² = 1 + 2i − 1 = 2i.
Q27. (1 − i)² is equal to:
(a) 2i
(b) −2i
(c) 2
(d) −2
Answer: (b) −2i
Solution:(1−i)² = 1 − 2i + i² = 1 − 2i − 1 = −2i.
Q28. (2 + i)(2 − i) is equal to:
(a) 3
(b) 4
(c) 5
(d) 4 − i²
Answer: (c) 5
Solution:(2+i)(2−i) = 4 − i² = 4 − (−1) = 5.
Q29. The additive inverse of 3 − 4i is:
(a) 3 + 4i
(b) −3 + 4i
(c) −3 − 4i
(d) 4 − 3i
Answer: (b) −3 + 4i
Solution:
The additive inverse of z is −z.
So, −(3 − 4i) = −3 + 4i.
Q30. The multiplicative identity for complex numbers is:
(a) 0
(b) i
(c) 1
(d) −1
Answer: (c) 1
Solution:
For any complex number z, z × 1 = z.
Q31. (5 − 2i) + (−5 + 2i) is equal to:
(a) 1
(b) 0
(c) 10 − 4i
(d) −10 + 4i
Answer: (b) 0
Solution:(5−5) + (−2+2)i = 0 + 0i = 0.
Q32. (3 + i)(3 − i) is equal to:
(a) 8
(b) 9
(c) 10
(d) 11
Answer: (c) 10
Solution:(3+i)(3−i) = 9 − i² = 9 + 1 = 10.
Q33. The value of i³ is:
(a) 1
(b) −1
(c) i
(d) −i
Answer: (d) −i
Solution:i³ = i² × i = −1 × i = −i.
Q34. The value of i⁴ is:
(a) 1
(b) −1
(c) i
(d) −i
Answer: (a) 1
Solution:i⁴ = (i²)² = (−1)² = 1.
Q35. The value of i⁵ is:
(a) i
(b) −i
(c) 1
(d) −1
Answer: (a) i
Solution:i⁵ = i⁴ × i = 1 × i = i.
Q36. The value of i¹⁰ is:
(a) 1
(b) −1
(c) i
(d) −i
Answer: (b) −1
Solution:
Divide 10 by 4. Remainder is 2.
So, i¹⁰ = i² = −1.
Q37. The value of i²³ is:
(a) 1
(b) −1
(c) i
(d) −i
Answer: (d) −i
Solution:23 ÷ 4 leaves remainder 3.
So, i²³ = i³ = −i.
Q38. The value of i⁴⁰ is:
(a) 1
(b) −1
(c) i
(d) −i
Answer: (a) 1
Solution:
Since 40 is divisible by 4, i⁴⁰ = 1.
Q39. The value of 1 + i + i² + i³ is:
(a) 0
(b) 1
(c) i
(d) −1
Answer: (a) 0
Solution:1 + i + i² + i³ = 1 + i − 1 − i = 0.
Q40. The value of i⁻¹ is:
(a) i
(b) −i
(c) 1
(d) −1
Answer: (b) −i
Solution:i⁻¹ = 1/i.
Multiplying numerator and denominator by i:1/i = i/i² = i/(−1) = −i.
Q41. The conjugate of 3 + 4i is:
(a) 3 − 4i
(b) −3 + 4i
(c) −3 − 4i
(d) 4 + 3i
Answer: (a) 3 − 4i
Solution:
The conjugate of a + ib is a − ib.
Q42. The conjugate of −5 − 2i is:
(a) 5 + 2i
(b) −5 + 2i
(c) 5 − 2i
(d) −5 − 2i
Answer: (b) −5 + 2i
Solution:
Change the sign of the imaginary part only.
Q43. If z = 6 − i, then z̄ is:
(a) 6 + i
(b) −6 − i
(c) −6 + i
(d) 6 − i
Answer: (a) 6 + i
Solution:
The conjugate of 6 − i is 6 + i.
Q44. The modulus of 3 + 4i is:
(a) 3
(b) 4
(c) 5
(d) 7
Answer: (c) 5
Solution:|3+4i| = √(3² + 4²) = √25 = 5.
Q45. The modulus of 5 − 12i is:
(a) 13
(b) 17
(c) 7
(d) 12
Answer: (a) 13
Solution:|5−12i| = √(5² + (−12)²) = √169 = 13.
Q46. If z = a + ib, then z z̄ is equal to:
(a) a² − b²
(b) a² + b²
(c) 2ab
(d) a + b
Answer: (b) a² + b²
Solution:z z̄ = (a+ib)(a−ib) = a² + b².
Q47. If z = 2 + 3i, then z z̄ is:
(a) 5
(b) 13
(c) 12
(d) 4 + 9i
Answer: (b) 13
Solution:z z̄ = |z|² = 2² + 3² = 13.
Q48. The reciprocal of 1 + i is:
(a) (1 − i)/2
(b) (1 + i)/2
(c) 1 − i
(d) 2/(1+i)
Answer: (a) (1 − i)/2
Solution:1/(1+i) = (1−i)/[(1+i)(1−i)]= (1−i)/(1−i²)= (1−i)/2.
Q49. The modulus of a complex number represents:
(a) Angle with real axis
(b) Distance from origin in Argand plane
(c) Imaginary part only
(d) Real part only
Answer: (b) Distance from origin in Argand plane
Solution:
For z = a + ib, |z| = √(a²+b²), which is the distance of point (a,b) from origin.
Q50. If |z| = 0, then:
(a) z = 1
(b) z = i
(c) z = 0
(d) z is imaginary only
Answer: (c) z = 0
Solution:
Only the zero complex number has modulus zero.
Q51. In the Argand plane, the horizontal axis represents:
(a) Imaginary part
(b) Real part
(c) Modulus
(d) Argument
Answer: (b) Real part
Solution:
In the Argand plane, the x-axis is the real axis.
Q52. In the Argand plane, the vertical axis represents:
(a) Real part
(b) Imaginary part
(c) Modulus
(d) Argument
Answer: (b) Imaginary part
Solution:
The y-axis is the imaginary axis.
Q53. The complex number 3 + 2i is represented by the point:
(a) (2,3)
(b) (3,2)
(c) (3,−2)
(d) (−3,2)
Answer: (b) (3,2)
Solution:a + ib corresponds to point (a,b).
Q54. The complex number −4 + 5i lies in:
(a) First quadrant
(b) Second quadrant
(c) Third quadrant
(d) Fourth quadrant
Answer: (b) Second quadrant
Solution:
Real part is negative and imaginary part is positive. Hence, it lies in the second quadrant.
Q55. The complex number −2 − 3i lies in:
(a) First quadrant
(b) Second quadrant
(c) Third quadrant
(d) Fourth quadrant
Answer: (c) Third quadrant
Solution:
Both real and imaginary parts are negative, so the point lies in Quadrant III.
Q56. The argument of a positive real number is:
(a) 0
(b) π/2
(c) π
(d) 3π/2
Answer: (a) 0
Solution:
A positive real number lies on the positive real axis, so its argument is 0.
Q57. The argument of a negative real number is:
(a) 0
(b) π/2
(c) π
(d) 2π
Answer: (c) π
Solution:
A negative real number lies on the negative real axis, so its principal argument is π.
Q58. If z = r(cos θ + i sin θ), then r represents:
(a) Argument
(b) Modulus
(c) Real part only
(d) Imaginary part only
Answer: (b) Modulus
Solution:
In polar form, r = |z|.
Q59. If z = 1 + i, then |z| is:
(a) 1
(b) 2
(c) √2
(d) 1/√2
Answer: (c) √2
Solution:|1+i| = √(1²+1²) = √2.
Q60. If z = 1 + i, then its principal argument is:
(a) π/6
(b) π/4
(c) π/3
(d) π/2
Answer: (b) π/4
Solution:tan θ = Im(z)/Re(z) = 1/1 = 1.
Since z lies in Quadrant I, θ = π/4.
Q61. The value of √−9 is:
(a) 3
(b) −3
(c) 3i
(d) −9i
Answer: (c) 3i
Solution:√−9 = √9 × √−1 = 3i.
Q62. The value of √−25 is:
(a) 5
(b) −5
(c) 5i
(d) −25i
Answer: (c) 5i
Solution:√−25 = √25 × i = 5i.
Q63. The value of √−16 + √−9 is:
(a) 7
(b) 7i
(c) 4 + 3i
(d) −7i
Answer: (b) 7i
Solution:√−16 = 4i and √−9 = 3i.
So, √−16 + √−9 = 7i.
Q64. The value of (√−4)(√−9) is:
(a) 6
(b) −6
(c) 6i
(d) −6i
Answer: (b) −6
Solution:√−4 = 2i and √−9 = 3i.
Product = 2i × 3i = 6i² = −6.
Q65. √−a for a > 0 is equal to:
(a) √a
(b) −√a
(c) i√a
(d) a²
Answer: (c) i√a
Solution:
Since i = √−1, √−a = i√a.
Q66. The standard form of a quadratic equation is:
(a) ax + b = 0
(b) ax² + bx + c = 0, a ≠ 0
(c) ax³ + bx² + c = 0
(d) a + ib = 0
Answer: (b) ax² + bx + c = 0, a ≠ 0
Solution:
A quadratic equation has degree 2 and leading coefficient non-zero.
Q67. The quadratic formula is:
(a) x = (b ± √(b²−4ac))/(2a)
(b) x = (-b ± √(b²−4ac))/(2a)
(c) x = (-b ± √(b²+4ac))/(2a)
(d) x = (-a ± √(b²−4ac))/(2b)
Answer: (b) x = (-b ± √(b²−4ac))/(2a)
Solution:
For ax² + bx + c = 0, the roots are given byx = (-b ± √(b²−4ac))/(2a).
Q68. For the equation x² − 5x + 6 = 0, the roots are:
(a) 1, 6
(b) 2, 3
(c) −2, −3
(d) 5, 6
Answer: (b) 2, 3
Solution:x² − 5x + 6 = 0(x−2)(x−3)=0
So, x = 2, 3.
Q69. For the equation x² + 4x + 4 = 0, the root is:
(a) 2
(b) −2
(c) 4
(d) −4
Answer: (b) −2
Solution:x² + 4x + 4 = (x+2)² = 0
So, x = −2.
Q70. The roots of x² + 1 = 0 are:
(a) 1, −1
(b) i, −i
(c) 0, 1
(d) 2i, −2i
Answer: (b) i, −i
Solution:x² + 1 = 0x² = −1x = ±i.
Q71. The discriminant of ax² + bx + c = 0 is:
(a) b² + 4ac
(b) b² − 4ac
(c) a² − 4bc
(d) c² − 4ab
Answer: (b) b² − 4ac
Solution:
The discriminant is D = b² − 4ac.
Q72. If D > 0, the roots are:
(a) Real and distinct
(b) Real and equal
(c) Complex conjugates
(d) Imaginary only
Answer: (a) Real and distinct
Solution:
When the discriminant is positive, a quadratic equation has two distinct real roots.
Q73. If D = 0, the roots are:
(a) Real and distinct
(b) Real and equal
(c) Complex only
(d) Not possible
Answer: (b) Real and equal
Solution:
When D = 0, both roots are equal.
Q74. If D < 0, the roots are:
(a) Real and distinct
(b) Real and equal
(c) Complex conjugates
(d) Rational only
Answer: (c) Complex conjugates
Solution:
A negative discriminant gives non-real complex conjugate roots.
Q75. The discriminant of x² + 2x + 5 = 0 is:
(a) 16
(b) −16
(c) 4
(d) −4
Answer: (b) −16
Solution:
Here a=1, b=2, c=5.D = b² − 4ac = 4 − 20 = −16.
Q76. The roots of x² + 2x + 5 = 0 are:
(a) −1 ± 2i
(b) 1 ± 2i
(c) −2 ± i
(d) 2 ± i
Answer: (a) −1 ± 2i
Solution:
Using formula:x = (-2 ± √−16)/2 = (-2 ± 4i)/2 = −1 ± 2i.
Q77. The discriminant of x² − 6x + 9 = 0 is:
(a) 0
(b) 9
(c) 18
(d) 36
Answer: (a) 0
Solution:D = (−6)² − 4(1)(9) = 36 − 36 = 0.
Q78. The nature of roots of x² + x + 1 = 0 is:
(a) Real and distinct
(b) Real and equal
(c) Complex conjugates
(d) Rational roots
Answer: (c) Complex conjugates
Solution:D = 1² − 4(1)(1) = −3 < 0.
Hence, roots are complex conjugates.
Q79. For ax² + bx + c = 0, sum of roots is:
(a) b/a
(b) −b/a
(c) c/a
(d) −c/a
Answer: (b) −b/a
Solution:
If roots are α and β, then α + β = −b/a.
Q80. For ax² + bx + c = 0, product of roots is:
(a) b/a
(b) −b/a
(c) c/a
(d) −c/a
Answer: (c) c/a
Solution:
If roots are α and β, then αβ = c/a.
Q81. For x² − 7x + 12 = 0, sum of roots is:
(a) 7
(b) −7
(c) 12
(d) −12
Answer: (a) 7
Solution:
Here a=1, b=−7.
Sum of roots = −b/a = 7.
Q82. For x² − 7x + 12 = 0, product of roots is:
(a) −7
(b) 7
(c) 12
(d) −12
Answer: (c) 12
Solution:
Product of roots = c/a = 12/1 = 12.
Q83. If roots of a quadratic equation are 3 and 4, then the equation is:
(a) x² − 7x + 12 = 0
(b) x² + 7x + 12 = 0
(c) x² − x + 12 = 0
(d) x² + x − 12 = 0
Answer: (a) x² − 7x + 12 = 0
Solution:
Sum = 3+4 = 7, product = 12.
Equation: x² − 7x + 12 = 0.
Q84. If roots are 2 + i and 2 − i, then their sum is:
(a) 2
(b) 4
(c) 2i
(d) 4i
Answer: (b) 4
Solution:(2+i) + (2−i) = 4.
Q85. If roots are 2 + i and 2 − i, then their product is:
(a) 3
(b) 4
(c) 5
(d) 5i
Answer: (c) 5
Solution:(2+i)(2−i) = 4 − i² = 5.
Q86. The quadratic equation whose roots are 5 and 6 is:
(a) x² − 11x + 30 = 0
(b) x² + 11x + 30 = 0
(c) x² − x + 30 = 0
(d) x² + x − 30 = 0
Answer: (a) x² − 11x + 30 = 0
Solution:
Sum = 5+6 = 11, product = 30.
Equation: x² − 11x + 30 = 0.
Q87. The quadratic equation whose roots are 1+i and 1−i is:
(a) x² − 2x + 2 = 0
(b) x² + 2x + 2 = 0
(c) x² − x + 2 = 0
(d) x² + x − 2 = 0
Answer: (a) x² − 2x + 2 = 0
Solution:
Sum = (1+i)+(1−i)=2.
Product = (1+i)(1−i)=1−i²=2.
Equation: x² − 2x + 2 = 0.
Q88. The quadratic equation whose roots are −3 and 4 is:
(a) x² − x − 12 = 0
(b) x² + x − 12 = 0
(c) x² − 7x + 12 = 0
(d) x² + 7x + 12 = 0
Answer: (a) x² − x − 12 = 0
Solution:
Sum = −3 + 4 = 1, product = −12.
Equation: x² − x − 12 = 0.
Q89. If roots are α and β, the quadratic equation is:
(a) x² + (α+β)x + αβ = 0
(b) x² − (α+β)x + αβ = 0
(c) x² − αβx + α+β = 0
(d) x² + αβx − α−β = 0
Answer: (b) x² − (α+β)x + αβ = 0
Solution:
The standard equation from roots is:x² − (sum of roots)x + product of roots = 0.
Q90. If roots are i and −i, the equation is:
(a) x² + 1 = 0
(b) x² − 1 = 0
(c) x² + x + 1 = 0
(d) x² − x + 1 = 0
Answer: (a) x² + 1 = 0
Solution:
Sum = i + (−i) = 0.
Product = i(−i) = −i² = 1.
Equation: x² + 1 = 0.
Q91. If 1 − i is a root of a quadratic equation with real coefficients, then the other root is:
(a) 1 + i
(b) −1 + i
(c) −1 − i
(d) i − 1
Answer: (a) 1 + i
Solution:
For a quadratic equation with real coefficients, non-real complex roots occur in conjugate pairs.
Q92. If roots of x² + ax + b = 0 are 1 − i and 1 + i, then a is:
(a) −2
(b) 2
(c) −1
(d) 1
Answer: (a) −2
Solution:
Sum of roots = (1−i)+(1+i)=2.
For x² + ax + b = 0, sum = −a.
So, −a = 2, hence a = −2.
Q93. If roots of x² + ax + b = 0 are 1 − i and 1 + i, then b is:
(a) 1
(b) 2
(c) −2
(d) 0
Answer: (b) 2
Solution:
Product = (1−i)(1+i)=1−i²=2.
For monic quadratic, product = b.
Hence, b = 2.
Q94. If z = x + iy and |z| = 5, then:
(a) x + y = 5
(b) x² + y² = 25
(c) x² − y² = 25
(d) xy = 5
Answer: (b) x² + y² = 25
Solution:|z| = √(x² + y²) = 5.
Squaring both sides: x² + y² = 25.
Q95. The locus |z − 2| = 3 represents:
(a) A line
(b) A circle with centre (2,0) and radius 3
(c) A circle with centre (−2,0) and radius 3
(d) A parabola
Answer: (b) A circle with centre (2,0) and radius 3
Solution:
Let z = x + iy.|z − 2| = |(x−2)+iy| = 3⇒ (x−2)² + y² = 9, a circle with centre (2,0) and radius 3.
Q96. If |z + 1| = 4, then the centre of the circle is:
(a) (1,0)
(b) (−1,0)
(c) (0,1)
(d) (0,−1)
Answer: (b) (−1,0)
Solution:|z + 1| = |z − (−1)| = 4.
So, centre is −1 + 0i, i.e. (−1,0).
Q97. If z = 3 + 4i, then arg(z) is:
(a) tan⁻¹(3/4)
(b) tan⁻¹(4/3)
(c) tan⁻¹(7)
(d) π/2
Answer: (b) tan⁻¹(4/3)
Solution:
For z = x + iy, arg(z) = tan⁻¹(y/x) in Quadrant I.
Here, x = 3, y = 4.
Q98. The value of (1+i)/(1−i) is:
(a) 1
(b) −1
(c) i
(d) −i
Answer: (c) i
Solution:(1+i)/(1−i) × (1+i)/(1+i)= (1+i)²/(1−i²)= (1 + 2i + i²)/(2)= (2i)/2 = i.
Q99. The value of [(1+i)/(1−i)]⁴ is:
(a) 1
(b) −1
(c) i
(d) −i
Answer: (a) 1
Solution:
From Q98, (1+i)/(1−i) = i.
So, [(1+i)/(1−i)]⁴ = i⁴ = 1.
Q100. If z² + |z|² = 0 and z = x + iy, then:
(a) x = 0
(b) y = 0
(c) x = y
(d) x = −y
Answer: (a) x = 0
Solution:z² = (x+iy)² = x² − y² + 2ixy|z|² = x² + y²
So, z² + |z|² = x² − y² + 2ixy + x² + y²= 2x² + 2ixy = 2x(x+iy)
Given this is zero, so 2xz = 0.
Thus, x = 0 or z = 0. In general, the condition implies x = 0.
| Q No. | Answer | Q No. | Answer | Q No. | Answer | Q No. | Answer |
| 1 | B | 26 | B | 51 | B | 76 | A |
| 2 | C | 27 | B | 52 | B | 77 | A |
| 3 | B | 28 | C | 53 | B | 78 | C |
| 4 | B | 29 | B | 54 | B | 79 | B |
| 5 | B | 30 | C | 55 | C | 80 | C |
| 6 | B | 31 | B | 56 | A | 81 | A |
| 7 | A | 32 | C | 57 | C | 82 | C |
| 8 | C | 33 | D | 58 | B | 83 | A |
| 9 | A | 34 | A | 59 | C | 84 | B |
| 10 | B | 35 | A | 60 | B | 85 | C |
| 11 | B | 36 | B | 61 | C | 86 | A |
| 12 | A | 37 | D | 62 | C | 87 | A |
| 13 | B | 38 | A | 63 | B | 88 | A |
| 14 | A | 39 | A | 64 | B | 89 | B |
| 15 | B | 40 | B | 65 | C | 90 | A |
| 16 | B | 41 | A | 66 | B | 91 | A |
| 17 | B | 42 | B | 67 | B | 92 | A |
| 18 | B | 43 | A | 68 | B | 93 | B |
| 19 | B | 44 | C | 69 | B | 94 | B |
| 20 | A | 45 | A | 70 | B | 95 | B |
| 21 | A | 46 | B | 71 | B | 96 | B |
| 22 | C | 47 | B | 72 | A | 97 | B |
| 23 | A | 48 | A | 73 | B | 98 | C |
| 24 | A | 49 | B | 74 | C | 99 | A |
| 25 | A | 50 | C | 75 | B | 100 | A |
1. Confusing Im(z) with ib
If z = a + ib, then Im(z) = b, not ib.
Example:
For z = 4 − 7i, Im(z) = −7.
2. Forgetting the cycle of powers of i
The powers of i repeat after every 4 powers:
i, −1, −i, 1
To find iⁿ, divide n by 4 and use the remainder.
3. Making mistakes with negative square roots
For a > 0:
√−a = i√a
But be careful while multiplying two negative radicals. Always convert them into i form first.
4. Not using conjugate while dividing complex numbers
To simplify expressions like 1/(1+i), multiply numerator and denominator by the conjugate 1−i.
5. Using the wrong quadratic formula
The correct quadratic formula is:
x = (-b ± √(b² − 4ac))/(2a)
Do not forget the negative sign before b.
6. Thinking negative discriminant means no roots
If D < 0, the quadratic equation has no real roots, but it has complex conjugate roots.
7. Forgetting conjugate roots for real coefficients
If a quadratic equation has real coefficients and one root is p + iq, the other root must be p − iq.
No courses found
Complex numbers are numbers of the form a + ib, where a and b are real numbers and i² = −1. Here, a is the real part and b is the imaginary part.
The standard form of a complex number is:
z = a + ib
where a is the real part, b is the imaginary part and i = √−1.
The value of i² is −1.
Since i = √−1, squaring both sides gives:
i² = −1.
The conjugate of a complex number a + ib is a − ib.
For example, the conjugate of 3 + 4i is 3 − 4i.
If z = a + ib, then the modulus of z is:
|z| = √(a² + b²)
For example, if z = 3 + 4i, then |z| = √(9+16) = 5.
For the quadratic equation ax² + bx + c = 0, the discriminant is:
D = b² − 4ac
It helps determine the nature of roots.
A quadratic equation has complex roots when its discriminant is negative.
That means:
b² − 4ac < 0
For a quadratic equation with real coefficients, non-real complex roots always occur in conjugate pairs. If one root is p + iq, the other root is p − iq.
Yes, Complex Numbers and Quadratic Equations are important for JEE because they are used in algebra, coordinate geometry, polynomial equations, modulus-argument problems, loci and root-based questions.
Yes. Students can download the Complex Numbers and Quadratic Equations MCQ Questions with Answers PDF for offline practice, formula revision and exam preparation.