Unit 3 · Chapter 3.1

3.1Introduction to Complex Numbers

Define i = √(−1), write complex numbers in a + bi form, perform addition, subtraction, multiplication, and division, and find complex conjugates.

Complex numbers extend the real number system to solve equations like x² = −1. They are essential in electrical engineering, quantum mechanics, signal processing, and the study of polynomial roots.

Essential Question

Why do mathematicians extend the real number system to include imaginary numbers, and how do complex numbers behave under arithmetic operations?

Lesson Overview

The real numbers cannot solve every polynomial equation. For example, x² = −1 has no real solution because no real number squared gives a negative result. To fill this gap, mathematicians defined the imaginary unit i = √(−1), giving us the complex number system. Every complex number has the standard form a + bi, where a is the real part and b is the imaginary part. In this lesson you will learn to add, subtract, multiply, and divide complex numbers, find conjugates, compute the modulus, and solve equations with complex solutions.

Key Definitions

  • Imaginary unit: i = √(−1), so i² = −1
  • Complex number: a + bi (a, b ∈ ℝ)
  • Real part: a; Imaginary part: b
  • Complex conjugate of a+bi is a−bi
  • Modulus: |a+bi| = √(a²+b²)

Learning Targets

  • Write complex numbers in a + bi form
  • Add, subtract, multiply, and divide complex numbers
  • Simplify powers of i using the cycle i, −1, −i, 1
  • Find the conjugate and modulus of a complex number
  • Solve quadratic equations with complex solutions

The Complex Plane

ReIm-3-3-2-2-1-11122333+2i−1+3i2−2i−3−i0

Points plotted as (Re, Im)

Powers of i

i¹ = ii² = −1i³ = −ii⁴ = 1Powersof icycle of 4

Cycle repeats every 4 steps

Conjugate Product

−abi+abia−biabi(a+bi)(a−bi) = a² + b²

Middle terms cancel

Worked Examples

Example 1

Write √(−25) in terms of i and simplify.

√(−25) = √(25 · −1)

= √25 · √(−1)

= 5 · i

Answer:5i
Example 2

Add (3 + 4i) + (−1 + 2i).

Group real parts: 3 + (−1) = 2

Group imaginary parts: 4i + 2i = 6i

Answer:2 + 6i
Example 3

Multiply (2 + 3i)(1 − 4i).

FOIL: 2·1 + 2·(−4i) + 3i·1 + 3i·(−4i)

= 2 − 8i + 3i − 12i²

Replace i² = −1: = 2 − 8i + 3i − 12(−1)

= 2 + 12 + (−8 + 3)i

= 14 − 5i

Answer:14 − 5i
Example 4

Divide (3 + 2i) ÷ (1 − i). Write in a + bi form.

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

(3 + 2i)(1 + i) / [(1 − i)(1 + i)]

Denominator: 1² + 1² = 2

Numerator: 3 + 3i + 2i + 2i² = 3 + 5i + 2(−1) = 1 + 5i

= (1 + 5i) / 2

Answer:1/2 + (5/2)i
Example 5

Solve x² + 9 = 0 over the complex numbers.

x² = −9

x = ±√(−9)

= ±√9 · √(−1)

= ±3i

Answer:x = 3i or x = −3i

Guided Practice

Guided Problem 1

Simplify √(−49).

Hint: Factor out −1 under the radical, then use √(−1) = i.

Guided Problem 2

Subtract (5 − 3i) − (2 + 7i).

Hint: Distribute the minus sign, then combine real parts and imaginary parts separately.

Guided Problem 3

Multiply (4 − i)(4 + i).

Hint: This is a conjugate pair. Use (a − bi)(a + bi) = a² + b².

Guided Problem 4

Divide (2 + i) ÷ (3 − 2i).

Hint: Multiply top and bottom by the conjugate of the denominator: (3 + 2i).

Guided Problem 5

Simplify i¹⁵.

Hint: Divide the exponent by 4 and use the remainder: i¹ = i, i² = −1, i³ = −i, i⁴ = 1.

Key Vocabulary

Imaginary unit (i)

The number defined by i = √(−1), so i² = −1. It is the foundation of the complex number system.

Example: i² = −1; i³ = −i; i⁴ = 1

Complex number

A number of the form a + bi where a and b are real numbers. a is the real part and b is the imaginary part.

Example: 3 + 4i, −2 − i, 0 + 5i = 5i

Complex conjugate

The conjugate of a + bi is a − bi. Conjugates have the same real part but opposite imaginary parts.

Example: Conjugate of 3 + 2i is 3 − 2i

Modulus (absolute value)

The distance from the origin to a complex number in the complex plane. |a + bi| = √(a² + b²).

Example: |3 + 4i| = √(9 + 16) = 5

Standard form

A complex number written as a + bi with the real part first and the imaginary part second.

Example: Write 2i + 5 in standard form: 5 + 2i

Argand diagram

A coordinate plane where the horizontal axis represents the real part and the vertical axis represents the imaginary part of a complex number.

Example: 3 + 2i is plotted at the point (3, 2)

Check Your Understanding

Interactive Practice — 5 Questions

1

What is the value of i²?

2

What is (2 + 3i) + (4 − i)?

3

What is the product (3 + i)(3 − i)?

4

Which is the complex conjugate of −5 + 2i?

5

What is i²³?

Independent Practice

Independent Practice

1

Simplify √(−64).

2

Write 7 − √(−16) in standard form a + bi.

3

Add (−3 + 5i) + (6 − 2i).

4

Subtract (4 + i) − (−1 + 3i).

5

Multiply (2 + 5i)(3 − i).

⚠️

Common Mistakes

Writing √(−9) = −3 (treating the negative as a sign, not under the radical).

√(−9) = √9 · √(−1) = 3i. Always factor out i from a negative radicand.

Forgetting to replace i² with −1 after multiplying: (1 + i)² = 1 + 2i + i² = 1 + 2i + i².

Always substitute i² = −1: 1 + 2i + (−1) = 2i.

Dividing by a complex number without multiplying by the conjugate: (2 + i)/(1 + i) ≠ 2.

Multiply numerator and denominator by the conjugate of the denominator to get a real denominator.

Confusing the modulus with the imaginary part: |3 + 4i| = 4.

|3 + 4i| = √(3² + 4²) = √(9 + 16) = √25 = 5. Use the distance formula.

💡

Math Tips

📌

The powers of i cycle with period 4: i¹ = i, i² = −1, i³ = −i, i⁴ = 1. To simplify iⁿ, find n mod 4.

🔄

Multiplying a complex number by its conjugate always gives a non-negative real number: (a+bi)(a−bi) = a² + b².

📌

When dividing complex numbers, the conjugate trick turns the denominator into a real number, making simplification straightforward.

📐

The modulus |a + bi| is the Pythagorean distance from the origin to the point (a, b) in the Argand diagram.

🧮

If the discriminant b² − 4ac is negative, the quadratic has two complex conjugate roots: x = (−b ± i√|Δ|) / (2a).