Complex Numbers — NCERT Class 11 Maths

How to Use This Page

Attempt each NCERT exercise problem on your own first, then compare your method with the step-by-step solutions below. Every question from Chapter 4 (Complex Numbers) is worked out with full steps — apply the definitions, perform the algebra, and verify your answer.

Key Concepts (Quick Recap)

• Complex number: z = a + ib, where a is real part Re(z) and b is imaginary part Im(z)
• Equality: a + ib = c + id ⟺ a = c and b = d
• Addition: (a + ib) + (c + id) = (a + c) + i(b + d)
• Multiplication: (a + ib)(c + id) = (ac − bd) + i(ad + bc)
• Conjugate: z̄ = a − ib
• Modulus: |z| = √(a² + b²)
• Polar form: z = r(cos θ + i sin θ), where r = |z|, θ = arg(z)
• Properties: z z̄ = |z|², |z₁z₂| = |z₁||z₂|, |z₁ + z₂| ≤ |z₁| + |z₂|
• Quadratic with complex roots: x = (−b ± i√|D|) / 2a when D < 0

Exercise 4.1

Q1. Express the following in a + ib form: (i) (5 + 3i) + (2 − 5i) (ii) (5 + 3i) − (2 − 5i) (iii) (3 + 4i)(2 − i) (iv) (1 − i)³

(i) (5 + 3i) + (2 − 5i) = (5 + 2) + i(3 − 5) = 7 − 2i

(ii) (5 + 3i) − (2 − 5i) = (5 − 2) + i(3 + 5) = 3 + 8i

(iii) (3 + 4i)(2 − i) = 6 − 3i + 8i − 4i² = 6 + 5i + 4 = 10 + 5i

(iv) (1 − i)³: First find (1 − i)² = 1 − 2i + i² = 1 − 2i − 1 = −2i Then (1 − i)³ = (−2i)(1 − i) = −2i + 2i² = −2i − 2 = −2 − 2i

Q2. Express (1 − i / 1 + i) in the form a + ib.

Multiply numerator and denominator by the conjugate of the denominator (1 − i):

(1 − i)/(1 + i) × (1 − i)/(1 − i) = (1 − i)²/(1² − i²) = (1 − 2i + i²)/(1 + 1) = (1 − 2i − 1)/2 = −2i/2 = −i (i.e., 0 − i)

Q3. Express (5 + √2i)/(1 − √2i) in the form a + ib.

Multiply by conjugate of denominator: (5 + √2i)/(1 − √2i) × (1 + √2i)/(1 + √2i) = (5 + 5√2i + √2i + 2i²) / (1 + 2) = (5 + 6√2i − 2) / 3 = (3 + 6√2i) / 3 = 1 + 2√2i

Q4. Express (1 − i)⁴ in the form a + ib.

(1 − i)² = 1 − 2i + i² = −2i (1 − i)⁴ = (−2i)² = 4i² = −4 (1 − i)⁴ = −4 + 0i

Q5. Express (1/3 + 3i) − (7 + i/3) in the form a + ib.

(1/3 + 3i) − (7 + i/3) = (1/3 − 7) + i(3 − 1/3) = (1/3 − 21/3) + i(9/3 − 1/3) = −20/3 + i(8/3) = −20/3 + (8/3)i

Q6. Express (−5 + 3i)/(1 + 6i) in the form a + ib.

Multiply by conjugate of denominator: (−5 + 3i)(1 − 6i) / (1 + 36) = (−5 + 30i + 3i − 18i²) / 37 = (−5 + 33i + 18) / 37 = (13 + 33i) / 37 = 13/37 + (33/37)i

Q7. Find the multiplicative inverse of 4 + 3i.

Multiplicative inverse = 1/(4 + 3i) = (4 − 3i)/((4 + 3i)(4 − 3i)) = (4 − 3i)/(16 + 9) = (4 − 3i)/25 = 4/25 − (3/25)i

Q8. Find the multiplicative inverse of √5 + 3i.

1/(√5 + 3i) = (√5 − 3i)/((√5)² + 9) = (√5 − 3i)/(5 + 9) = (√5 − 3i)/14 = √5/14 − (3/14)i

Q9. Find the multiplicative inverse of −i.

1/(−i) = −1/i = −i/(i · i) = −i/(−1) = i

Q10. Express (3 − √−16)(2 + √−25) in the form a + ib.

√−16 = 4i and √−25 = 5i

(3 − 4i)(2 + 5i) = 6 + 15i − 8i − 20i² = 6 + 7i + 20 = 26 + 7i

Exercise 4.2

Q1. Find the modulus and argument of each of the following: (i) z = 1 − i√3 (ii) z = −√3 + i (iii) z = −1 − i

(i) z = 1 − i√3. a = 1, b = −√3.

arg(z) = tan⁻¹(−√3/1) = −π/3. Since a > 0, b < 0 (Q4), argument = −π/3

(ii) z = −√3 + i. a = −√3, b = 1.

θ = tan⁻¹(1/(−√3)) = tan⁻¹(−1/√3). Since a < 0, b > 0 (Q2), argument = π − π/6 = 5π/6

(iii) z = −1 − i. a = −1, b = −1.

θ = tan⁻¹(−1/−1) = tan⁻¹(1) = π/4. Since a < 0, b < 0 (Q3), argument = π + π/4 ... wait, that's 5π/4 but let me recalculate. In Q3: arg = −(π − π/4) = −3π/4. Actually, arg = π + tan⁻¹(b/a) where tan⁻¹ gives the reference angle. tan⁻¹(1) = π/4. In Q3: arg = −π + π/4 ... No. The standard: arg = π + tan⁻¹(b/a) for Q3. = π + tan⁻¹(−1/−1) but we take the acute angle π/4 and adjust for Q3. Argument = −3π/4 (or equivalently 5π/4 − 2π = −3π/4 is the principal value in (−π, π])

Q2. Convert the following into polar form: (i) 1 + i (ii) −1 + i (iii) −1 − i (iv) 1 − i

(i) z = 1 + i. r = √2, θ = π/4. Polar form: √2(cos π/4 + i sin π/4)

(ii) z = −1 + i. r = √2, θ = 3π/4. Polar form: √2(cos 3π/4 + i sin 3π/4)

(iii) z = −1 − i. r = √2, θ = −3π/4. Polar form: √2(cos(−3π/4) + i sin(−3π/4))

(iv) z = 1 − i. r = √2, θ = −π/4. Polar form: √2(cos(−π/4) + i sin(−π/4))

Q3. Find the square root of −9.

−9 = 9(−1) = 9i² √(−9) = ±3i

Q4. Find the square root of −7 − 24i.

Let √(−7 − 24i) = x + iy. Then: (x + iy)² = −7 − 24i x² − y² + 2ixy = −7 − 24i

Real: x² − y² = −7 ... (1) Imaginary: 2xy = −24 ⟹ xy = −12 ... (2)

From (1) and (3): 2x² = 25 − 7 = 18 ⟹ x² = 9 ⟹ x = ±3 From (2): if x = 3, y = −4; if x = −3, y = 4

Since xy = −12 < 0, x and y have opposite signs.

√(−7 − 24i) = ±(3 − 4i)

Q5. Find the square root of 1 + 4√3i.

Let √(1 + 4√3i) = x + iy. x² − y² = 1 and 2xy = 4√3 ⟹ xy = 2√3 x² + y² = √(1 + 48) = 7

From x² − y² = 1 and x² + y² = 7: 2x² = 8 ⟹ x² = 4 ⟹ x = ±2 y = 2√3/x: if x = 2, y = √3; if x = −2, y = −√3

Since xy = 2√3 > 0, x and y have the same sign.

√(1 + 4√3i) = ±(2 + √3i)

Q6. Find the square root of −11 − 60i.

x² − y² = −11, 2xy = −60 ⟹ xy = −30 x² + y² = √(121 + 3600) = √3721 = 61

2x² = 61 − 11 = 50 ⟹ x² = 25 ⟹ x = ±5 y = −30/x: x = 5 ⟹ y = −6; x = −5 ⟹ y = 6

√(−11 − 60i) = ±(5 − 6i)

Q7. Find the square root of −8 − 6i.

x² − y² = −8, 2xy = −6 ⟹ xy = −3 x² + y² = √(64 + 36) = √100 = 10

2x² = 10 − 8 = 2 ⟹ ... wait, that gives x² = 1 ⟹ x = ±1. Then y = −3/x: x=1, y=−3; x=−1, y=3.

Check: (1 − 3i)² = 1 − 6i + 9i² = 1 − 6i − 9 = −8 − 6i ✓

√(−8 − 6i) = ±(1 − 3i)

Q8. Find the square root of 1 − i.

x² − y² = 1, 2xy = −1 ⟹ xy = −1/2 x² + y² = √(1 + 1) = √2

2x² = 1 + √2 ⟹ x² = (1 + √2)/2 2y² = √2 − 1 ⟹ y² = (√2 − 1)/2

x = √((1 + √2)/2), y = −√((√2 − 1)/2) (since xy < 0)

√(1 − i) = ±(√((1 + √2)/2) − i√((√2 − 1)/2))

Miscellaneous Exercise

Q1. Evaluate: i⁹ + i¹⁹.

i⁹ = i⁴ˣ²⁺¹ = i¹ = i (since i⁴ = 1) i¹⁹ = i⁴ˣ⁴⁺³ = i³ = −i

i⁹ + i¹⁹ = i + (−i) = 0

Q2. Evaluate: i⁻³⁹.

i⁻³⁹ = 1/i³⁹ = 1/i⁴ˣ⁹⁺³ = 1/i³ = 1/(−i) = i/(−i · i) = i/1 = i

Q3. Express (5 + 2i)/(1 − 3i) in a + ib form.

(5 + 2i)(1 + 3i) / (1 + 9) = (5 + 15i + 2i + 6i²) / 10 = (5 + 17i − 6) / 10 = (−1 + 17i)/10 = −1/10 + (17/10)i

Q4. Express (−√3 + √2i)/(√3 + √2i) in a + ib form.

Multiply by conjugate of denominator: (−√3 + √2i)(√3 − √2i) / (3 + 2) = (−3 + √6i + √6i − 2i²) / 5 = (−3 + 2√6i + 2) / 5 = (−1 + 2√6i) / 5 = −1/5 + (2√6/5)i

Q6. If z₁ = 2 − i, z₂ = −2 + i, find: (i) Re(z₁z₂) (ii) Im(z₁z₂)

z₁z₂ = (2 − i)(−2 + i) = −4 + 2i + 2i − i² = −4 + 4i + 1 = −3 + 4i

(i) Re(z₁z₂) = −3 (ii) Im(z₁z₂) = 4

Q7. Find the modulus and argument of z = 1 + i√3.

arg(z) = tan⁻¹(√3/1) = π/3

Q8. Solve the equation: x² − 2x + 3 = 0.

Using the quadratic formula: x = (2 ± √(4 − 12))/2 = (2 ± √(−8))/2 = (2 ± 2i√2)/2

x = 1 ± i√2

Q9. Solve: x² + 3 = 0.

x² = −3 x = ±i√3

Q10. Solve: x² + x + 1 = 0.

D = 1 − 4 = −3 x = (−1 ± √(−3))/2 = (−1 ± i√3)/2

x = (−1 + i√3)/2 or x = (−1 − i√3)/2

These are the complex cube roots of unity (ω and ω²).

Internal Links

• [Chapter 4 Guide — Complex Numbers](/school/cbse/class/11/maths/cbse-complex-numbers)
• [All Class 11 Maths Solutions](/school/cbse/class/11/maths/solutions)
• [Class 11 Maths Subject Guide](/school/cbse/class/11/maths)