π CHAPTER 4: COMPLEX NUMBERS
Subject: Mathematics | Class: XI | Kendriya Vidyalaya Sangathan
π INTRODUCTION
A complex number is a number of the form a + ib, where a and b are real numbers and i = √(-1).
πΉ Key Definitions:
- i (Imaginary Unit): i = √(-1), i² = -1
- Real Part: Re(z) = a
- Imaginary Part: Im(z) = b
- Pure Imaginary: a = 0, b ≠ 0
- Purely Real: b = 0
πΉ Powers of i
- i¹ = i
- i² = -1
- i³ = -i
- i⁴ = 1
- i⁵ = i (pattern repeats every 4 powers)
General: i4n = 1, i4n+1 = i, i4n+2 = -1, i4n+3 = -i
➕ ALGEBRA OF COMPLEX NUMBERS
πΉ Addition
(a + ib) + (c + id) = (a + c) + i(b + d)
πΉ Subtraction
(a + ib) - (c + id) = (a - c) + i(b - d)
πΉ Multiplication
(a + ib)(c + id) = (ac - bd) + i(ad + bc)
πΉ Division
(a + ib)/(c + id) = [(a + ib)(c - id)]/[(c + id)(c - id)]
= (ac + bd)/(c² + d²) + i(bc - ad)/(c² + d²)
π’ CONJUGATE AND MODULUS
πΉ Conjugate of a Complex Number
If z = a + ib, then conjugate of z is z̅ = a - ib
Properties of Conjugate:
- z + z̅ = 2a (Purely Real)
- z - z̅ = 2ib (Purely Imaginary)
- z · z̅ = a² + b²
- (z₁ + z₂)̅ = z̅₁ + z̅₂
- (z₁ · z₂)̅ = z̅₁ · z̅₂
- (z₁/z₂)̅ = z̅₁/z̅₂
πΉ Modulus of a Complex Number
If z = a + ib, then modulus of z is |z| = √(a² + b²)
Properties of Modulus:
- |z| = |z̅|
- |z₁ · z₂| = |z₁| · |z₂|
- |z₁/z₂| = |z₁|/|z₂| (z₂ ≠ 0)
- |z₁ + z₂| ≤ |z₁| + |z₂| (Triangle Inequality)
- |z|² = z · z̅
π ARGUMENT AND POLAR FORM
πΉ Argument of a Complex Number
If z = a + ib, then arg(z) = ΞΈ where tan ΞΈ = b/a
πΉ Polar Form
z = r(cos ΞΈ + i sin ΞΈ) where r = |z| and ΞΈ = arg(z)
πΉ Euler's Form:
z = r·eiΞΈ = r(cos ΞΈ + i sin ΞΈ)
πΉ Argument of z in Different Quadrants
| Quadrant | Condition | arg(z) |
|---|---|---|
| I | a > 0, b > 0 | ΞΈ = tan⁻¹(b/a) |
| II | a < 0, b > 0 | ΞΈ = Ο - tan⁻¹(|b/a|) |
| III | a < 0, b < 0 | ΞΈ = Ο + tan⁻¹(|b/a|) |
| IV | a > 0, b < 0 | ΞΈ = -tan⁻¹(|b/a|) |
π MCQ QUESTIONS
Q1. The value of i528 is:
(a) 1 (b) -1 (c) i (d) -i
Answer: (a) 1
Solution: 528 ÷ 4 = 132 (remainder 0) ⇒ i528 = 1
Q2. [i¹⁹ + (1/i)²⁵]² is equal to:
(a) 4 (b) -4 (c) i (d) -i
Answer: (b) -4
Solution: i¹⁹ = i³ = -i, (1/i)²⁵ = i²⁵ = i
(-i + i)² = 0 ⇒ But wait, check carefully:
i¹⁹ = i³ = -i, (1/i)²⁵ = (-i)²⁵ = -i
(-i - i)² = (-2i)² = 4i² = -4
Q3. i⁵⁹² + i⁵⁹⁰ + i⁵⁸⁸ + i⁵⁸⁶ + i⁵⁸⁴ is equal to:
(a) -1 (b) 1 (c) -2i (d) i
Answer: (a) -1
Solution: 592÷4=148(r=0)→1, 590÷4=147(r=2)→-1, 588÷4=147(r=0)→1, 586÷4=146(r=2)→-1, 584÷4=146(r=0)→1
Sum = 1 - 1 + 1 - 1 + 1 = 1
Q4. in + in+1 + in+2 + in+3 is equal to:
(a) 0 (b) 1 (c) -1 (d) 2
Answer: (a) 0
Solution: in(1 + i + i² + i³) = in(1 + i - 1 - i) = 0
Q5. The conjugate of i-35 is:
(a) 1 (b) -1 (c) i (d) -i
Answer: (c) i
Solution: i-35 = 1/i³⁵ = 1/i³ = 1/(-i) = i
Conjugate of i = -i
Q6. If z₁ = 3 + 2i and z₂ = 2 - 4i, then |z₁ + z₂|² + |z₁ - z₂|² is equal to:
(a) 11 (b) 22 (c) 55 (d) 66
Answer: (d) 66
Solution: z₁ + z₂ = 5 - 2i, |z₁+z₂|² = 25+4 = 29
z₁ - z₂ = 1 + 6i, |z₁-z₂|² = 1+36 = 37
Sum = 29 + 37 = 66
Q7. The value of i4 + i8 + i12 + ... + i4n is:
(a) n (b) 0 (c) 1 (d) -1
Answer: (a) n
Solution: i⁴ = 1, i⁸ = 1, ... i4n = 1
Sum = n × 1 = n
Q8. If z = 1 + i, then z² is equal to:
(a) 2i (b) -2i (c) 1 - i (d) -1 - i
Answer: (a) 2i
Solution: z² = (1 + i)² = 1 + 2i + i² = 1 + 2i - 1 = 2i
Q9. If z = 1 - i, then z̅ is:
(a) 1 + i (b) -1 + i (c) 1 - i (d) -1 - i
Answer: (a) 1 + i
Q10. The modulus of 1 + i is:
(a) 1 (b) √2 (c) 2 (d) √3
Answer: (b) √2
Solution: |1 + i| = √(1² + 1²) = √2
Q11. The argument of 1 + i is:
(a) Ο/4 (b) Ο/3 (c) Ο/2 (d) Ο/6
Answer: (a) Ο/4
Solution: tan ΞΈ = 1/1 = 1 ⇒ ΞΈ = Ο/4
Q12. If z = 1 + i, then z³ is equal to:
(a) -2 + 2i (b) 2 - 2i (c) -2 - 2i (d) 2 + 2i
Answer: (a) -2 + 2i
Solution: z³ = (1+i)²(1+i) = (2i)(1+i) = 2i + 2i² = 2i - 2 = -2 + 2i
Q13. The real part of (1 + i)² is:
(a) 0 (b) 1 (c) -1 (d) 2
Answer: (a) 0
Solution: (1+i)² = 2i, real part = 0
Q14. If z = 2 + 3i, then z · z̅ is equal to:
(a) 13 (b) 5 (c) 6 (d) 12
Answer: (a) 13
Solution: z · z̅ = (2+3i)(2-3i) = 4 + 9 = 13
Q15. The multiplicative inverse of 2 + i is:
(a) (2 - i)/5 (b) (2 + i)/5 (c) (2 - i)/3 (d) (2 + i)/3
Answer: (a) (2 - i)/5
Solution: 1/(2+i) = (2-i)/((2+i)(2-i)) = (2-i)/5
π§ ASSERTION & REASON QUESTIONS
Q26. Assertion: The complex number i + i² is a real number.
Reason: i² = -1
Answer: (a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.
Explanation: i + i² = i - 1 = -1 + i, which is not purely real. Wait, let's re-evaluate:
i + i² = i - 1 = -1 + i (has imaginary part). Actually i + i² = -1 + i, which is not real.
Correction: The correct answer is (d) Assertion is incorrect, reason is correct.
Q27. Assertion: The conjugate of 2 + 3i is 2 - 3i.
Reason: The conjugate of z = a + ib is a - ib.
Answer: (a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.
Q28. Assertion: |z|² = z · z̅
Reason: z · z̅ = a² + b²
Answer: (a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.
Q29. Assertion: The modulus of 3 + 4i is 5.
Reason: |z| = √(a² + b²)
Answer: (a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.
Explanation: |3+4i| = √(9+16) = 5 ✓
Q30. Assertion: (1 + i)⁴ is a real number.
Reason: (1 + i)² = 2i
Answer: (a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.
Explanation: (1+i)⁴ = (2i)² = 4i² = -4 (real) ✓
π VERY SHORT ANSWER TYPE QUESTIONS
Q31. Find the value of i35.
Answer: i35 = i32 · i³ = 1 · (-i) = -i
Q32. Find the value of i102.
Answer: 102 ÷ 4 = 25 (remainder 2) ⇒ i102 = i² = -1
Q33. Find the conjugate of 3 - 4i.
Answer: Conjugate of 3 - 4i is 3 + 4i
Q34. Find the modulus of 5 + 12i.
Answer: |5 + 12i| = √(25 + 144) = √169 = 13
Q35. Find the argument of 1 + i.
Answer: arg(1 + i) = tan⁻¹(1/1) = Ο/4
Q36. Write 1 + i in polar form.
Answer: r = √2, ΞΈ = Ο/4
z = √2(cos Ο/4 + i sin Ο/4)
Q37. Find the value of (1 + i)².
Answer: (1 + i)² = 1 + 2i + i² = 1 + 2i - 1 = 2i
Q38. Find the value of (1 - i)².
Answer: (1 - i)² = 1 - 2i + i² = 1 - 2i - 1 = -2i
Q39. If a = cos ΞΈ + i sin ΞΈ, find the value of (1+a)/(1-a).
Answer: (1 + cosΞΈ + i sinΞΈ)/(1 - cosΞΈ - i sinΞΈ)
= [2cos²(ΞΈ/2) + 2i sin(ΞΈ/2)cos(ΞΈ/2)]/[2sin²(ΞΈ/2) - 2i sin(ΞΈ/2)cos(ΞΈ/2)]
= [cos(ΞΈ/2) + i sin(ΞΈ/2)]/[sin(ΞΈ/2) - i cos(ΞΈ/2)]
= [cos(ΞΈ/2) + i sin(ΞΈ/2)]/[cos(Ο/2 - ΞΈ/2) + i sin(Ο/2 - ΞΈ/2)]
= eiΞΈ/2 / ei(Ο/2 - ΞΈ/2) = ei(ΞΈ - Ο/2) = i cot(ΞΈ/2)
Q40. What is the smallest positive integer n for which (1 + i)2n = (1 - i)2n?
Answer: (1 + i)² = 2i, (1 - i)² = -2i
(2i)n = (-2i)n ⇒ in = (-i)n ⇒ in = (-1)nin
⇒ (-1)n = 1 ⇒ n is even
Smallest positive even integer is 2
Q41. If z₁, z₂, z₃ are complex numbers such that |z₁| = |z₂| = |z₃| = |1/z₁ + 1/z₂ + 1/z₃| = 1, then find the value of |z₁ + z₂ + z₃|.
Answer: |1/z₁ + 1/z₂ + 1/z₃| = |(z₂z₃ + z₁z₃ + z₁z₂)/(z₁z₂z₃)|
= |z₁z₂ + z₂z₃ + z₃z₁| = 1 (since |z₁z₂z₃| = 1)
|z₁ + z₂ + z₃|² = (z₁+z₂+z₃)(z̅₁+z̅₂+z̅₃)
= 3 + (z₁z̅₂ + ... ) + (z̅₁z₂ + ...)
= 3 + 2Re(z₁z̅₂ + z₂z̅₃ + z₃z̅₁)
Using the given condition, we get |z₁ + z₂ + z₃| = 1
Q42. If z₁ = 2 - i and z₂ = -2 + i, find:
(i) Re(z₁z₂) (ii) Im(1/z₁)
Answer: (i) z₁z₂ = (2-i)(-2+i) = -4 + 2i + 2i - i² = -4 + 4i + 1 = -3 + 4i
Re(z₁z₂) = -3
(ii) 1/z₁ = 1/(2-i) = (2+i)/5 = 2/5 + i/5
Im(1/z₁) = 1/5
π SHORT ANSWER TYPE QUESTIONS
Q43. If (x + iy)³ = u + iv, then show that u/x + v/y = 4(x² - y²).
Answer: (x + iy)³ = x³ + 3x²(iy) + 3x(iy)² + (iy)³
= x³ + 3ix²y - 3xy² - iy³
= (x³ - 3xy²) + i(3x²y - y³)
u = x³ - 3xy², v = 3x²y - y³
u/x + v/y = (x³ - 3xy²)/x + (3x²y - y³)/y
= x² - 3y² + 3x² - y² = 4x² - 4y² = 4(x² - y²) ✓
Q44. If ((1 + i)/(1 - i))m = 1, then find the least positive integral value of m.
Answer: (1+i)/(1-i) = (1+i)²/((1-i)(1+i)) = (1+2i+i²)/2 = (2i)/2 = i
im = 1 ⇒ m = 4, 8, 12, ...
Least positive integral value = 4
Q45. Simplify (1 + i)⁴(1 + 1/i)⁴.
Answer: (1 + i)⁴ = [(1+i)²]² = (2i)² = 4i² = -4
(1 + 1/i) = (1 - i) since 1/i = -i
(1 - i)⁴ = [(1-i)²]² = (-2i)² = 4i² = -4
Product = (-4)(-4) = 16
Q46. Find the value of i⁴⁵ + i⁴⁶ + i⁴⁷ + i⁴⁸.
Answer: i⁴⁵ = i, i⁴⁶ = -1, i⁴⁷ = -i, i⁴⁸ = 1
Sum = i - 1 - i + 1 = 0
Q47. If z(2 - i) = (3 + i), then z²⁰ is equal to:
Answer: z = (3+i)/(2-i) = (3+i)(2+i)/((2-i)(2+i)) = (6+3i+2i+i²)/5 = (5+5i)/5 = 1+i
z² = (1+i)² = 2i
z²⁰ = (z²)¹⁰ = (2i)¹⁰ = 2¹⁰ · i¹⁰ = 1024 · i² = -1024
Q48. Evaluate Ξ£n=1¹³ (in + in+1) where n ∈ N.
Answer: Ξ£(in + in+1) = Ξ£ in(1 + i)
= (1+i)(i + i² + i³ + ... + i¹³)
= (1+i)(i - 1 - i + 1 + i - 1 - i + ... + i)
= (1+i)(i) = i + i² = i - 1 = -1 + i
Q49. Evaluate 1 + i² + i⁴ + i⁶ + ... + i2n.
Answer: i² = -1, i⁴ = 1, i⁶ = -1, ...
If n is odd: sum = 1 - 1 + 1 - 1 + ... + 1 = 1
If n is even: sum = 1 - 1 + 1 - 1 + ... - 1 = 0
Q50. Find the value of (1 + i)⁶ + (1 - i)³.
Answer: (1+i)² = 2i, (1+i)⁶ = (2i)³ = 8i³ = -8i
(1-i)² = -2i, (1-i)³ = (-2i)(1-i) = -2i + 2i² = -2i - 2 = -2 - 2i
Sum = -8i - 2 - 2i = -2 - 10i
✅ ANSWERS - MCQs
| Q.No. | Answer | Q.No. | Answer |
|---|---|---|---|
| 1 | (a) | 11 | (a) |
| 2 | (b) | 12 | (a) |
| 3 | (a) | 13 | (a) |
| 4 | (a) | 14 | (a) |
| 5 | (c) | 15 | (a) |
| 6 | (c) | 16 | (d) |
| 7 | (a) | 17 | (c) |
| 8 | (b) | 18 | (a) |
| 9 | (c) | 19 | (b) |
| 10 | (b) | 20 | (a) |
π KEY FORMULAS SUMMARY
| Formula | Expression |
|---|---|
| Imaginary Unit | i² = -1 |
| Powers of i | i⁴βΏ = 1, i⁴βΏ⁺¹ = i, i⁴βΏ⁺² = -1, i⁴βΏ⁺³ = -i |
| Complex Number | z = a + ib |
| Conjugate | z̅ = a - ib |
| Modulus | |z| = √(a² + b²) |
| Modulus Squared | |z|² = z · z̅ |
| Argument | tan ΞΈ = b/a |
| Polar Form | z = r(cos ΞΈ + i sin ΞΈ) |
| Euler's Form | z = r·eiΞΈ |
| Multiplication | (a+ib)(c+id) = (ac-bd) + i(ad+bc) |
| Division | (a+ib)/(c+id) = [(a+ib)(c-id)]/(c²+d²) |
π Kendriya Vidyalaya Sangathan | Class XI Mathematics
Chapter 4: Complex Numbers | Student Support Material