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CBSE · Class 11 · Mathematics

Complex Numbers and Their Algebra (Ch 4: 4.2 and 4.3)

4 concepts · 32 practice questions with answers · Free to read

Imaginary unit and complex numbers

From daily life

There is no real number whose square is −1, yet engineers need one for alternating current. So mathematicians invented one.

Simple definition

The imaginary unit i has i² = −1. A complex number is z = a + ib with real a and b; a is the real part Re(z) and b the imaginary part Im(z). Two complex numbers are equal when both real parts and both imaginary parts match.

Formula: z = a + ib, i² = −1

RealImaginaryz = 3 + 2i

In the figure: z = 3 + 2i is the point (3, 2): real part along the x-axis, imaginary part along the y-axis.

Rule 1: Identify parts — Write in the form a + ib.

  • Easy: z = 4 − 5i: Re z = 4, Im z = −5.
  • Medium: z = 7: Re z = 7, Im z = 0.
  • Exam-type: z = −3i: Re z = 0, Im z = −3.

Rule 2: Equality and square roots of negatives — √(−a) = i√a for a > 0.

  • Easy: x + iy = 2 − 3i gives x = 2, y = −3.
  • Medium: √(−16) = 4i.
  • Exam-type: √(−12) = 2i√3 ≈ 3.464i.
Remember

Every real number is a complex number with imaginary part 0.

Common mistake

Writing √(−a) × √(−b) = √(ab). Use i first: √(−4) × √(−9) = 2i × 3i = −6.

Memory clue

i is the square root of −1.

Practice questions

Easy
  1. 1Write the real and imaginary parts of 6 + 5i.
  2. 2Find √(−25).
  3. 3If x + iy = 8 − 2i, find x and y.
Medium
  1. 4If (a + 1) + i(b − 2) = 5 + 3i, find a and b.
  2. 5Simplify √(−4) × √(−9).
  3. 6Simplify √(−8).
Challenging
  1. 7If 2x − 3 + i(y + 1) = 7 − 4i, find x and y.
  2. 8For what real value of k is (k − 2) + i(3k − 6) purely real?
Show answers (try them yourself first)
  1. 1Re = 6, Im = 5
  2. 25i
  3. 3x = 8, y = −2
  4. 4a = 4, b = 5
  5. 5−6
  6. 62i√2
  7. 7x = 5, y = −5
  8. 83k − 6 = 0, so k = 2

Powers of i

From daily life

Spinning an arrow by a quarter-turn each time: after 4 quarter-turns it points back where it started. Powers of i repeat in cycles of 4.

Simple definition

i¹ = i, i² = −1, i³ = −i, i⁴ = 1 and then it repeats. For any integer n, iⁿ depends only on the remainder of n when divided by 4.

Formula: i⁴ = 1

Rule 1: Find iⁿ — Divide n by 4 and look at the remainder.

  • Easy: i⁹ = i (9 = 4 × 2 + 1).
  • Medium: i⁵⁰ = -1 (50 = 4 × 12 + 2).
  • Exam-type: i⁻¹ = −i (since 1/i = −i).

Rule 2: Sum of powers — Four consecutive powers add to 0.

  • Easy: i + i² + i³ + i⁴ = 0.
  • Medium: 1 + i + i² + i³ = 0.
  • Exam-type: i¹⁰⁰ + i¹⁰¹ + i¹⁰² + i¹⁰³ = 0.
Remember

i⁴ᵏ = 1, i⁴ᵏ⁺¹ = i, i⁴ᵏ⁺² = −1, i⁴ᵏ⁺³ = −i.

Common mistake

Treating i² as +1.

Memory clue

The remainder tells the power: 0 → 1, 1 → i, 2 → −1, 3 → −i.

Practice questions

Easy
  1. 1Find i⁶.
  2. 2Find i²⁷.
  3. 3Find i⁻³.
Medium
  1. 4Find i³⁰.
  2. 5Find the value of i¹⁰ + i²⁰ + i³⁰ + i⁴⁰.
  3. 6Simplify 1/i.
Challenging
  1. 7Simplify i⁷ + i⁹.
  2. 8Find the smallest positive n for which iⁿ = 1.
Show answers (try them yourself first)
  1. 1−1
  2. 2−i
  3. 3i
  4. 4-1
  5. 50
  6. 6−i
  7. 70
  8. 84

Addition, subtraction and multiplication

From daily life

Add rupees and paise separately: 3 + 2i is like "3 rupees and 2 paise" except the second part is a different kind of number.

Simple definition

For z₁ = a + ib and z₂ = c + id: z₁ + z₂ = (a + c) + i(b + d); z₁ − z₂ = (a − c) + i(b − d); z₁z₂ = (ac − bd) + i(ad + bc). Multiply as with brackets and replace i² by −1.

Formula: z₁z₂ = (ac − bd) + i(ad + bc)

Rule 1: Add and subtract — Add the real parts and the imaginary parts separately.

  • Easy: (3 + 2i) + (1 − 4i) = 4 − 2i.
  • Medium: (3 + 2i) − (1 − 4i) = 2 + 6i.
  • Exam-type: (5 − i) + (−5 + i) = 0.

Rule 2: Multiply — Expand, then use i² = −1.

  • Easy: (3 + 2i)(1 − 4i) = 11 − 10i.
  • Medium: (2 + i)(2 − i) = 5.
  • Exam-type: (1 + i)² = 2i.
Remember

Properties such as commutative, associative and distributive laws hold for complex numbers.

Common mistake

Writing (a + ib)(c + id) = ac + ibd (forgetting the cross terms).

Memory clue

Multiply like brackets, then i² = −1.

Practice questions

Easy
  1. 1Find (4 + 3i) + (2 − 5i).
  2. 2Find (4 + 3i) − (2 − 5i).
  3. 3Find (2 + 3i)(1 + i).
Medium
  1. 4Find (3 − i)².
  2. 5Find (1 + 2i)(1 − 2i).
  3. 6Find (2 − 3i)(4 + 5i).
Challenging
  1. 7Find (1 − i)⁴.
  2. 8Find real x and y if (x + iy)(1 − i) = 2 + 4i.
Show answers (try them yourself first)
  1. 16 − 2i
  2. 22 + 8i
  3. 3−1 + 5i
  4. 48 − 6i
  5. 55
  6. 623 − 2i
  7. 7-4
  8. 8x + iy = (2 + 4i)/(1 − i) = −1 + 3i, so x = −1, y = 3

Division and reciprocal

From daily life

To remove an awkward denominator like √2 we multiply by its partner. For a complex denominator the partner is its conjugate.

Simple definition

To divide, multiply the numerator and denominator by the conjugate of the denominator: 1/(a + ib) = (a − ib)/(a² + b²). Then z₁/z₂ = z₁·z₂̄ / (c² + d²).

Formula: 1/(a + ib) = (a − ib)/(a² + b²)

Rule 1: Reciprocal — Conjugate on top, a² + b² below.

  • Easy: 1/(3 + 4i) = (3 − 4i)/25 = 0.12 − 0.16i.
  • Medium: 1/i = −i.
  • Exam-type: 1/(1 − i) = (1 + i)/2 = 0.5 + 0.5i.

Rule 2: Division — Multiply top and bottom by the conjugate of the bottom.

  • Easy: (2 + i)/(1 − i) = 0.5 + 1.5i.
  • Medium: (5 + 5i)/(3 − 4i) = −0.2 + 1.4i.
  • Exam-type: (1 + i)/(1 − i) = i.
Remember

The product of a complex number and its conjugate is always real: (a + ib)(a − ib) = a² + b².

Common mistake

Multiplying the denominator by itself instead of its conjugate.

Memory clue

Conjugate rescues the denominator.

Practice questions

Easy
  1. 1Find 1/(2 + i).
  2. 2Find (3 + 2i)/(1 + i).
  3. 3Find (4 − 3i)/(2i).
Medium
  1. 4Find (1 + 2i)/(2 − i).
  2. 5Write (3 + i)/(3 − i) in the form a + ib.
  3. 6Find the multiplicative inverse of 6 − 8i.
Challenging
  1. 7Find (2 + i)/(1 + i) + (2 − i)/(1 − i).
  2. 8Simplify ((1 + i)/(1 − i))³⁰.
Show answers (try them yourself first)
  1. 10.4 − 0.2i
  2. 22.5 − 0.5i
  3. 3−1.5 − 2i
  4. 4i
  5. 50.8 + 0.6i
  6. 60.06 + 0.08i
  7. 73
  8. 8-1