Complex Numbers and Quadratic Equations

Complex Numbers: Definition and Basic Concepts

A **complex number** is a number of the form **z = a + ib**, where a and b are real numbers. Here:

**a** is the **real part**, denoted **Re(z)**

**b** is the **imaginary part**, denoted **Im(z)**

**i** is the imaginary unit, defined by **i² = –1**

**Key Point:** The imaginary unit i is introduced to solve equations like x² + 1 = 0, which have no solution in the real number system.

**Example:** For z = 2 + 5i, we have Re(z) = 2 and Im(z) = 5.

**Equality of Complex Numbers:** Two complex numbers z₁ = a + ib and z₂ = c + id are equal if and only if:

a = c (real parts equal)

b = d (imaginary parts equal)

**Worked Example:** If 4x + i(3x – y) = 3 + i(–6), find x and y.

Equating real parts: 4x = 3 → x = 3/4

Equating imaginary parts: 3x – y = –6

Substituting x = 3/4: 3(3/4) – y = –6 → 9/4 – y = –6 → y = 33/4

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Addition and Subtraction of Complex Numbers

**Addition:** For z₁ = a + ib and z₂ = c + id:

**z₁ + z₂ = (a + c) + i(b + d)**

**Example:** (2 + 3i) + (–6 + 5i) = (2 – 6) + i(3 + 5) = –4 + 8i

**Properties of Addition:**

**Closure Law:** The sum of two complex numbers is always a complex number

**Commutative Law:** z₁ + z₂ = z₂ + z₁

**Associative Law:** (z₁ + z₂) + z₃ = z₁ + (z₂ + z₃)

**Additive Identity:** 0 + 0i (denoted 0) exists such that z + 0 = z for any z

**Additive Inverse:** For z = a + ib, the inverse is –z = –a – ib such that z + (–z) = 0

**Subtraction:** **z₁ – z₂ = z₁ + (–z₂) = (a – c) + i(b – d)**

**Example:** (6 + 3i) – (2 – i) = (6 + 3i) + (–2 + i) = 4 + 4i

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Multiplication of Complex Numbers

**Formula:** For z₁ = a + ib and z₂ = c + id:

**z₁z₂ = (ac – bd) + i(ad + bc)**

**Derivation:**

(a + ib)(c + id) = ac + iad + ibc + i²bd = ac + iad + ibc – bd = (ac – bd) + i(ad + bc)

**Example:** (3 + 5i)(2 + 6i) = (3·2 – 5·6) + i(3·6 + 5·2) = (6 – 30) + i(18 + 10) = –24 + 28i

**Properties of Multiplication:**

**Closure Law:** z₁z₂ is always a complex number

**Commutative Law:** z₁z₂ = z₂z₁

**Associative Law:** (z₁z₂)z₃ = z₁(z₂z₃)

**Multiplicative Identity:** 1 + 0i (denoted 1) exists such that z·1 = z

**Multiplicative Inverse:** For non-zero z = a + ib, the inverse is z⁻¹ = (a – ib)/(a² + b²)

**Distributive Law:** z₁(z₂ + z₃) = z₁z₂ + z₁z₃ and (z₁ + z₂)z₃ = z₁z₃ + z₂z₃

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Division of Complex Numbers

**Formula:** For z₁ and z₂ (z₂ ≠ 0):

**z₁/z₂ = z₁/z₂ × (z₂̄/z₂̄) = (z₁·z₂̄)/|z₂|²**

where z₂̄ is the conjugate of z₂.

**Worked Example:** Find (6 + 3i)/(2 – i).

Multiply numerator and denominator by conjugate of denominator: (2 + i)

Numerator: (6 + 3i)(2 + i) = 12 + 6i + 6i + 3i² = 12 + 12i – 3 = 9 + 12i

Denominator: (2 – i)(2 + i) = 4 + 1 = 5

Result: (9 + 12i)/5 = 9/5 + 12i/5

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Powers of the Imaginary Unit i

**Base Powers:**

i¹ = i

i² = –1

i³ = i² · i = –1 · i = –i

i⁴ = (i²)² = (–1)² = 1

i⁵ = i⁴ · i = i

i⁶ = i⁴ · i² = –1

**Cyclical Pattern:** Powers of i repeat with period 4.

**General Formula:** For any integer k:

**i^(4k) = 1**

**i^(4k+1) = i**

**i^(4k+2) = –1**

**i^(4k+3) = –i**

**Worked Example:** Find i³⁵.

35 = 4(8) + 3, so 35 ≡ 3 (mod 4)

Therefore, i³⁵ = i³ = –i

**Key Point:** Always reduce the exponent modulo 4 to find the power of i.

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Square Roots of Negative Real Numbers

**Fundamental Result:** For any positive real number a:

**√(–a) = i√a**

**Explanation:** If √(–a) = x, then x² = –a. Setting x = ib where b > 0, we get (ib)² = i²b² = –b² = –a, so b = √a.

**Examples:**

√(–1) = i

√(–4) = 2i

√(–9) = 3i

**Critical Property:** The rule √a · √b = √(ab) does NOT hold when both a and b are negative.

**Counterexample:**

If we assume √(–1) · √(–1) = √(–1 × –1) = √1 = 1

But i · i = i² = –1, which contradicts 1 = –1

**Correct Approach:** When both a < 0 and b < 0, compute separately: √a = i√|a| and √b = i√|b|, then multiply.

√(–1) · √(–1) = i · i = –1 ✓

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Modulus and Conjugate of a Complex Number

**Modulus (Absolute Value):** For z = a + ib:

**|z| = √(a² + b²)**

This represents the distance from the point (a, b) to the origin in the Argand plane.

**Example:** |3 + 4i| = √(9 + 16) = √25 = 5

**Conjugate:** For z = a + ib:

**z̄ = a – ib**

The conjugate is obtained by replacing i with –i.

**Example:** Conjugate of 2 – 5i is 2 + 5i.

**Geometric Interpretation:** z and z̄ are mirror images of each other about the real axis in the Argand plane.

**Key Properties:**

z · z̄ = (a + ib)(a – ib) = a² + b² = |z|²

z̄̄ = z (conjugate of conjugate is the original)

|z̄| = |z|

Re(z) = (z + z̄)/2

Im(z) = (z – z̄)/(2i)

**Properties with Multiple Complex Numbers:**

(z₁z₂)̄ = z̄₁ · z̄₂

(z₁ + z₂)̄ = z̄₁ + z̄₂

(z₁/z₂)̄ = z̄₁/z̄₂ (provided z₂ ≠ 0)

|z₁z₂| = |z₁| · |z₂|

|z₁/z₂| = |z₁|/|z₂|

**Multiplicative Inverse using Conjugate:**

**z⁻¹ = z̄/|z|²**

**Worked Example:** Find the multiplicative inverse of 2 – 3i.

z̄ = 2 + 3i

|z|² = 4 + 9 = 13

z⁻¹ = (2 + 3i)/13 = 2/13 + 3i/13

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Important Algebraic Identities for Complex Numbers

**Sum of Squares:**

**(z₁ + z₂)² = z₁² + 2z₁z₂ + z₂²**

**Difference of Squares:**

**(z₁ – z₂)² = z₁² – 2z₁z₂ + z₂²**

**Sum of Cubes:**

**(z₁ + z₂)³ = z₁³ + 3z₁²z₂ + 3z₁z₂² + z₂³**

**Difference of Cubes:**

**(z₁ – z₂)³ = z₁³ – 3z₁²z₂ + 3z₁z₂² – z₂³**

**Difference of Squares (Factorization):**

**(z₁² – z₂²) = (z₁ + z₂)(z₁ – z₂)**

**Worked Example:** Express (5 – 3i)³ in the form a + ib.

Using (a – b)³ = a³ – 3a²b + 3ab² – b³

(5 – 3i)³ = 5³ – 3(5²)(3i) + 3(5)(3i)² – (3i)³

= 125 – 225i + 3(5)(9)(–1) – 27i³

= 125 – 225i – 135 – 27(–i)

= 125 – 135 – 225i + 27i

= –10 – 198i

**Key Point:** These identities work for complex numbers just as they do for real numbers due to the field properties of complex numbers.

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Argand Plane and Geometric Representation

**Argand Plane (Complex Plane):** A plane where each point (x, y) represents the complex number z = x + iy.

x-coordinate → real part

y-coordinate → imaginary part

**Key Axes:**

**Real Axis:** The horizontal axis, containing points of the form a + 0i

**Imaginary Axis:** The vertical axis, containing points of the form 0 + bi

**Geometric Properties:**

**Modulus as Distance:** |z| = √(x² + y²) is the distance from point P(x, y) to the origin O(0, 0)

**Conjugate as Reflection:** z = x + iy and z̄ = x – iy correspond to points P(x, y) and Q(x, –y), which are mirror images about the real axis

**Example Representation:**

2 + 4i → point A(2, 4)

–2 + 3i → point B(–2, 3)

–5 – 2i → point E(–5, –2)

**Distance Between Two Complex Numbers:**

The distance between z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂ is:

**d = |z₁ – z₂| = √[(x₁ – x₂)² + (y₁ – y₂)²]**

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Worked Problems and Common Techniques

**Expressing Complex Expressions in a + ib Form:**

**Technique 1: Multiply by Conjugate of Denominator**

For (5 + 2i)/(1 – 2i):

Multiply by (1 + 2i)/(1 + 2i)

Numerator: (5 + 2i)(1 + 2i) = 5 + 10i + 2i + 4i² = 5 + 12i – 4 = 1 + 12i

Denominator: (1 – 2i)(1 + 2i) = 1 + 4 = 5

Result: (1 + 12i)/5 = 1/5 + 12i/5

**Technique 2: Using Powers of i**

For i⁻³⁵:

i⁻³⁵ = 1/i³⁵ = 1/i³ (since 35 ≡ 3 mod 4)

i³ = –i

i⁻³⁵ = 1/(–i) = –1/i = –1/i × i/i = –i/i² = –i/(–1) = i

**Technique 3: Algebraic Expansion**

For (1 – i)⁴:

(1 – i)² = 1 – 2i + i² = 1 – 2i – 1 = –2i

(1 – i)⁴ = [(1 – i)²]² = (–2i)² = 4i² = –4

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Important Exam-Relevant Points

**Do NOT Assume:**

√(ab) = √a · √b when both a and b are negative

Complex numbers can be ordered (cannot say one is "greater than" another)

**Always Remember:**

i² = –1 is fundamental; use it to simplify any expression with i²

The modulus is always non-negative and real

A complex number equals another iff both real and imaginary parts are equal

Conjugate pairs are essential for division and finding inverses

**Verification Technique:**

If z · z⁻¹ = 1 + 0i, then z⁻¹ is correct.

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Summary of Key Formulas

| **Concept** | **Formula** |

|---|---|

| Complex number | z = a + ib |

| Equality | a + ib = c + id ⟺ a = c, b = d |

| Addition | (a + ib) + (c + id) = (a+c) + i(b+d) |

| Subtraction | (a + ib) – (c + id) = (a–c) + i(b–d) |

| Multiplication | (a + ib)(c + id) = (ac–bd) + i(ad+bc) |

| Division | z₁/z₂ = z₁z̄₂/|z₂|² |

| Modulus | |z| = √(a² + b²) |

| Conjugate | z̄ = a – ib |

| Property | z · z̄ = |z|² |

| Multiplicative inverse | z⁻¹ = z̄/|z|² |

| Powers of i (period 4) | i⁴ᵏ=1, i⁴ᵏ⁺¹=i, i⁴ᵏ⁺²=–1, i⁴ᵏ⁺³=–i |