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Chapter 2 · Class 11 Mathematics

Complex Numbers and Quadratic Equations — Important Questions

33 questions With answers CBSE format

SUMMARY: This chapter introduces complex numbers, their algebraic properties, and their application in solving quadratic equations.
KEY TOPICS: complex numbers, imaginary unit, algebra of complex numbers, polar representation, quadratic equations, roots of quadratic equations, discriminant, modulus and argument, conjugate of a complex number, solving quadratic equations using complex numbers

Previous chapter Binomial Theorem Next chapter Conic Sections
Multiple Choice Questions (MCQs) 5 questions
Q1 1 Mark

The value of i² is:

A1
B−1
Ci
D−i
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Correct answer: Option 2 — −1
Q2 1 Mark

The modulus of the complex number 3 + 4i is:

A5
B7
C12
D1
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Correct answer: Option 1 — 5
Q3 1 Mark

The conjugate of (1 + i) is:

A1 − i
B−1 + i
C−1 − i
D1 + i
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Correct answer: Option 1 — 1 − i
Q4 1 Mark

The roots of x² + 1 = 0 are:

A±1
B±i
C1 ± i
D−1 ± i
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Correct answer: Option 2 — ±i
Q5 1 Mark

For the quadratic ax² + bx + c = 0 the discriminant is:

Ab² + 4ac
Bb² − 4ac
C4ac − b²
D−b² − 4ac
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Correct answer: Option 2 — b² − 4ac
Short Answer Questions 5 questions
Q6 3 Marks

Express (2 + 3i) + (1 − 2i) in standard form.

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(2 + 3i) + (1 − 2i) = (2 + 1) + (3 − 2)i = 3 + i.
Q7 3 Marks

Find the multiplicative inverse of (3 + 4i).

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1/(3 + 4i) = (3 − 4i)/((3+4i)(3−4i)) = (3 − 4i)/(9 + 16) = (3 − 4i)/25 = 3/25 − (4/25) i.
Q8 3 Marks

Solve x² + 4 = 0 over the complex numbers.

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x² = −4 ⇒ x = ±√(−4) = ±2i. So roots are 2i and −2i.
Q9 3 Marks

Find the modulus and argument of −1 + i.

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|−1 + i| = √(1 + 1) = √2. arg(−1 + i): the point lies in the second quadrant; tan θ = 1/−1 = −1 with reference angle π/4 so θ = π − π/4 = 3π/4.
Q10 3 Marks

If z₁ = 2 + 3i and z₂ = 1 − i compute z₁ z₂.

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z₁ z₂ = (2 + 3i)(1 − i) = 2 − 2i + 3i − 3i² = 2 + i + 3 = 5 + i.
Long Answer Questions 6 questions
Q11 6 Marks

Find the roots of 2x² − 3x + 5 = 0 and express them in the form a + bi.

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Discriminant = (−3)² − 4·2·5 = 9 − 40 = −31. Roots = (3 ± √(−31))/(4) = 3/4 ± (√31/4) i. So roots are 3/4 + (√31/4) i and 3/4 − (√31/4) i.
Q12 6 Marks

Express (1 + i)/(1 − i) in standard form a + bi.

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Multiply numerator and denominator by the conjugate (1 + i): ((1+i)(1+i))/((1−i)(1+i)) = (1 + 2i + i²)/(1 − i²) = (1 + 2i − 1)/(1 + 1) = 2i/2 = i. So (1 + i)/(1 − i) = 0 + 1·i.
Q13 6 Marks

Show that for any complex number z |z|² = z · z̄ where z̄ is the conjugate. Verify with z = 3 − 4i.

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Let z = a + bi. Then z̄ = a − bi and z·z̄ = (a + bi)(a − bi) = a² − (bi)² = a² + b² = |z|². For z = 3 − 4i: |z|² = 9 + 16 = 25; z·z̄ = (3 − 4i)(3 + 4i) = 9 + 16 = 25. ✓
Q14 6 Marks

Find the square root of −15 − 8i in the form a + bi.

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Let √(−15 − 8i) = a + bi where a b ∈ R. Then a² − b² = −15 and 2ab = −8 i.e. ab = −4. From a² − b² = −15 and a² + b² = √(15² + 8²) = √(225 + 64) = √289 = 17. So a² = 1 and b² = 16. Combined with ab = −4: (a b) = (1 −4) or (−1 4). Hence square roots are ±(1 − 4i).
Q15 6 Marks

For the quadratic equation x² − 2x + 5 = 0 find the roots and verify the sum and product of roots formulas.

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Discriminant = 4 − 20 = −16. Roots = (2 ± √(−16))/2 = (2 ± 4i)/2 = 1 ± 2i. Sum = (1 + 2i) + (1 − 2i) = 2 = −b/a = 2 ✓. Product = (1 + 2i)(1 − 2i) = 1 + 4 = 5 = c/a = 5 ✓.
Q16 6 Marks

Compare real and imaginary parts of a complex number with the help of a table.

Assertion–Reason Questions 5 questions
Q17 1 Mark

Assertion (A): i² = −1.

Reason (R): The imaginary unit i is defined as √(−1) so its square equals −1.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.
Q18 1 Mark

Assertion (A): The conjugate of (a + bi) is (a − bi).

Reason (R): Conjugating a complex number reverses the sign of the imaginary part while leaving the real part unchanged.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.
Q19 1 Mark

Assertion (A): For complex numbers z₁ z₂: |z₁ z₂| = |z₁| · |z₂|.

Reason (R): The modulus is multiplicative under complex multiplication.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.
Q20 1 Mark

Assertion (A): A quadratic equation with discriminant less than zero has complex (non-real) roots.

Reason (R): The square root of a negative number is imaginary so the quadratic formula gives complex roots when D < 0.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.
Q21 1 Mark

Assertion (A): If z + z̄ is twice the real part of z then it is always real.

Reason (R): Adding a complex number to its conjugate cancels the imaginary parts leaving only the real part doubled.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.
Statement-Based Questions 5 questions
Q22 1 Mark

Statement 1: i³ = −i.

Statement 2: i⁴ = 1.

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Correct answer: Option 1 — Both statements are true.
Q23 1 Mark

Statement 1: The modulus of a complex number is always non-negative.

Statement 2: The modulus equals zero only when the complex number is zero.

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Correct answer: Option 1 — Both statements are true.
Q24 1 Mark

Statement 1: Every quadratic equation has roots in C.

Statement 2: Complex roots always occur in conjugate pairs for quadratics with real coefficients.

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Correct answer: Option 1 — Both statements are true.
Q25 1 Mark

Statement 1: The conjugate of a sum equals the sum of the conjugates.

Statement 2: The conjugate of a product equals the product of the conjugates.

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Correct answer: Option 1 — Both statements are true.
Q26 1 Mark

Statement 1: Every real number is a complex number.

Statement 2: Not every complex number is a real number.

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Correct answer: Option 1 — Both statements are true.
Case Study / Passage Questions 3 questions
Q27 3 Marks
In an AC circuit the impedance of two components is given by Z₁ = 3 + 4i ohms and Z₂ = 2 − i ohms (where i represents the imaginary unit). The engineer needs the total impedance for a series and a parallel connection.
  1. The total impedance for the series connection Z₁ + Z₂ equals:
    A1 + 3i
    B5 + 3i
    C5 − 5i
    D5 + 5i
  2. The magnitude |Z₁| equals:
    A√34
    B5
    C√25
    D√50
  3. Compute the magnitude of Z₂.
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1. Option 2 — 5 + 3i
2. Option 2 — 5
3. For series: Z = Z₁ + Z₂ = (3 + 2) + (4 − 1)i = 5 + 3i. For parallel: Z = (Z₁ · Z₂)/(Z₁ + Z₂); compute numerator (3 + 4i)(2 − i) = 6 − 3i + 8i + 4 = 10 + 5i; divide by 5 + 3i. Magnitude |Z₁| = √(9 + 16) = √25 = 5 ohms.
Q28 3 Marks
A physics student solves the quadratic equation x² − 4x + 13 = 0 to find a complex amplitude. The student wants the discriminant the roots and the sum and product to verify.
  1. The discriminant equals:
    A16
    B−16
    C−36
    D36
  2. The roots of the equation are:
    A2 ± 3i
    B−2 ± 3i
    C2 ± 6i
    D4 ± 13i
  3. Verify the sum and product of the roots formulas.
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1. Option 3 — −36
2. Option 1 — 2 ± 3i
3. D = 16 − 52 = −36. Roots = (4 ± √−36)/2 = (4 ± 6i)/2 = 2 ± 3i. Sum = 4 = −b/a ✓. Product = (2 + 3i)(2 − 3i) = 4 + 9 = 13 = c/a ✓.
Q29 3 Marks
A complex number z = 1 + i is plotted on the Argand plane. The student wants to know its modulus argument and the polar form.
  1. The modulus |z| equals:
    A1
    B√2
    C2
    D√3
  2. The argument arg(z) equals:
    Aπ/6
    Bπ/4
    Cπ/3
    Dπ/2
  3. Write z in polar form.
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1. Option 2 — √2
2. Option 2 — π/4
3. |z| = √(1² + 1²) = √2. arg(z) = tan⁻¹(1/1) = π/4 (first quadrant). Polar form: z = √2 (cos π/4 + i sin π/4) = √2 e^(iπ/4).
Table-Based Questions 3 questions
Q30 3 Marks

Study the powers of i:

PowerValue
i¹i
i²−1
i³−i
i⁴1
i⁵i
  1. The value of i⁴ equals:
    A1
    Bi
    C−1
    D−i
  2. The value of i^25 equals:
    Ai
    B−i
    C−1
    D1
  3. Compute i^100 using the cyclic pattern.
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1. Option 1 — 1
2. Option 1 — i
3. Powers of i cycle with period 4: i¹ = i i² = −1 i³ = −i i⁴ = 1. For i^25 = i^(4·6 + 1) = i. The cycle property makes computing high powers of i straightforward.
Q31 3 Marks

Study the operations on complex numbers:

OperationResultType
(2 + 3i) + (1 − i)3 + 2iComplex
(2 + 3i)(1 − i)5 + iComplex
(2 + 3i)/(1 − i)(−1/2) + (5/2)iComplex
|3 + 4i|5Real
arg(1 + i)π/4Real
  1. The sum (2 + 3i) + (1 − i) equals:
    A3 + 2i
    B3 − 2i
    C1 + 2i
    D−1 + 2i
  2. The modulus |3 + 4i| equals:
    A3
    B4
    C5
    D7
  3. Compute (2 + 3i)(1 − i) step by step.
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1. Option 1 — 3 + 2i
2. Option 3 — 5
3. |3 + 4i| = √(3² + 4²) = √25 = 5. The modulus represents the distance from the origin to the point 3 + 4i on the Argand plane and is a real non-negative number.
Q32 6 Marks

For z₁ = 3 + 4i and z₂ = 1 − 2i, compute (i) z₁ + z₂, (ii) z₁ · z₂, (iii) z₁/z₂, (iv) |z₁| and arg(z₁).

Complex numberRealImaginary
z₁34
z₂1−2
Picture-Based Questions 1 question
Q33 3 Marks

Study the Argand diagram of z = 3 + 4i and its conjugate z̄ and answer:

Complex Numbers and Quadratic Equations figure
  1. The modulus |z| = |3 + 4i| equals:
    A3
    B4
    C5
    D7
  2. z and z̄ are reflections of each other about the:
    Ax-axis
    By-axis
    Cline y = x
    Dorigin
  3. Compute |z|² using z · z̄ and verify the result.
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1. Option 3 — 5
2. Option 1 — x-axis
3. |z|² = z · z̄ = (3 + 4i)(3 − 4i) = 9 − (4i)² = 9 + 16 = 25, so |z| = 5. Geometrically, |z| is the distance from the origin to the point z on the Argand plane — here a 3-4-5 right-triangle gives |z| = 5.
Previous chapter Binomial Theorem Next chapter Conic Sections

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