Expansion of Number Systems: The Imaginary Unit i and Complex Numbers
The expansion of number systems throughout mathematical history has been propelled by the necessity to solve equations that possess no solutions in existing domains. Within the real number system, equations of the form x² + 1 = 0 or x² + 16 = 0 cannot be solved because the square of any real number is non-negative. To resolve this fundamental limitation, mathematicians introduced the imaginary unit i, formally defined such that i² = -1. This breakthrough establishes the set of complex numbers, denoted by C, containing all numbers of the algebraic form a + bi, where a and b are real numbers. In this lesson, Secondary 1 scholars explore the cyclic powers of i, algebraic operations on complex numbers, and the fundamental properties of complex conjugates.
1. Definition of the Imaginary Unit i and its Cyclic Powers
The imaginary unit i is defined as the mathematical entity whose square equals -1: i² = -1, which implies i = √(-1). Consequently, for any positive real number k, we have √(-k) = i√k. For example, √(-25) = i√25 = 5i.
The Cyclic Pattern of Powers of i (Modulus 4)
Because integer powers of i repeat in cycles of four, any power iⁿ can be determined by evaluating the remainder when n is divided by 4:
Remainder 1
Remainder 2
Remainder 3
Remainder 0
General Law: For any integer n and remainder r ∈ {0, 1, 2, 3}, i⁴ⁿ⁺ʳ = iʳ.
Examples: i²⁴ = i⁴⁽⁶⁾ = 1 | i⁶³ = i⁴⁽¹⁵⁾⁺³ = i³ = -i | i⁻¹ = 1/i = i⁴/i = i³ = -i | i⁻² = -1.
2. The Set of Complex Numbers C and Equality
A complex number is defined as an expression z = a + bi, where a, b ∈ ℝ. Here, a = Re(z) is the real part, and b = Im(z) is the imaginary part.
- If b = 0, then z = a, which is a purely real number (thus ℝ ⊂ ℂ).
- If a = 0 and b ≠ 0, then z = bi, which is called a pure imaginary number.
Criterion of Equality of Two Complex Numbers
Two complex numbers are equal if and only if their real parts are equal AND their imaginary parts are equal:
a + bi = c + di ⟺ a = c and b = d.
Example: If (2x - 3) + (3y + 1)i = 7 + 10i, then 2x - 3 = 7 ⟹ 2x = 10 ⟹ x = 5, and 3y + 1 = 10 ⟹ 3y = 9 ⟹ y = 3.
3. Arithmetic Operations & Complex Conjugates
| Operation | Algebraic Definition | Illustrative Exemplar |
|---|---|---|
| Addition | (a + bi) + (c + di) = (a + c) + (b + d)i | (4 + 3i) + (2 - 5i) = 6 - 2i |
| Subtraction | (a + bi) - (c + di) = (a - c) + (b - d)i | (7 + 6i) - (3 + 8i) = 4 - 2i |
| Multiplication | (a + bi)(c + di) = (ac - bd) + (ad + bc)i | (2 + 3i)(1 - 4i) = 2 - 8i + 3i - 12i² = 2 - 5i + 12 = 14 - 5i |
| Complex Conjugate | If z = a + bi, then conjugate z̄ = a - bi | z = 5 - 4i ⟹ z̄ = 5 + 4i |
| Conjugate Product | z · z̄ = (a + bi)(a - bi) = a² + b² ∈ ℝ⁺₀ | (3 + 2i)(3 - 2i) = 3² + 2² = 9 + 4 = 13 |
(3 + 2i) / (2 - i) = [(3 + 2i)(2 + i)] / [(2 - i)(2 + i)] = (6 + 3i + 4i + 2i²) / (4 + 1) = (4 + 7i) / 5 = 4/5 + (7/5)i.
Algebraic Mastery Workshop
Problem 1: Simplifying Cyclic Powers of i
Evaluate in simplest form: E = i²⁰²⁶ + i⁻¹⁵.
• 2026 ÷ 4 = 506 remainder 2 ⟹ i²⁰²⁶ = i² = -1.
• For i⁻¹⁵, add the nearest multiple of 4 greater than 15 (which is 16): i⁻¹⁵ = i⁻¹⁵ · i¹⁶ = i¹ = i.
• Thus, E = -1 + i.
Problem 2: Complex Conjugate Multiplication
Simplify into the standard form a + bi: (1 + i)⁸.
• Notice that (1 + i)² = 1 + 2i + i² = 1 + 2i - 1 = 2i.
• Therefore, (1 + i)⁸ = [(1 + i)²]⁴ = (2i)⁴ = 2⁴ · i⁴ = 16(1) = 16.
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