Precalculus — Chapter 8

Complex Numbers

The imaginary unit i, powers of i, arithmetic with complex numbers, conjugates, polar form, and De Moivre's Theorem — with worked examples.

Definition of Complex Numbers

Imaginary Unit

i = √(−1)

i² = −1

Complex Number

a + bi

a, b ∈ ℝ

Real & Imaginary Parts

Re = a, Im = b

b = 0 → real number

Every real number is a complex number (with b = 0). The complex number system extends the reals to allow square roots of negative numbers. On the complex plane, the real part is the x-axis and the imaginary part is the y-axis.

Powers of i — The Cycle of 4

i

−1

−i

i⁴

1

Finding i^n for any n

1. Divide n by 4 and find the remainder.

2. Remainder 0 → 1, remainder 1 → i, remainder 2 → −1, remainder 3 → −i

Example: Simplify i²⁷

27 ÷ 4 = 6 remainder 3

i²⁷ = i³ = −i

Arithmetic with Complex Numbers

Addition

(a + bi) + (c + di) = (a + c) + (b + d)i

Add real parts together, add imaginary parts together.

(3 + 2i) + (1 − 5i) = 4 − 3i

Subtraction

(a + bi) − (c + di) = (a − c) + (b − d)i

Subtract real parts, subtract imaginary parts.

(5 + 3i) − (2 + 7i) = 3 − 4i

Multiplication

(a + bi)(c + di) = (ac − bd) + (ad + bc)i

FOIL, then replace i² = −1 to simplify.

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

Division

(a + bi)/(c + di): multiply top and bottom by (c − di)

Multiply by the conjugate of the denominator. The denominator becomes c² + d².

(1 + 2i)/(3 − i) × (3 + i)/(3 + i) = (3 + i + 6i + 2i²)/10 = (1 + 7i)/10

Complex Conjugates

Definition

conjugate of (a + bi) = (a − bi)

Flip the sign of the imaginary part only.

Key Property

(a + bi)(a − bi) = a² + b²

Product of conjugates is always a real number.

Why conjugates matter:

  • Dividing complex numbers (rationalize the denominator)
  • Complex roots of polynomials always come in conjugate pairs
  • Computing |z|² = z · z̄ = a² + b²

Absolute Value (Modulus)

|a + bi| = √(a² + b²)

The distance from the origin to the point (a, b) on the complex plane

Example

|3 + 4i| = √(3² + 4²)

= √(9 + 16) = √25

= 5

Properties

|z₁ · z₂| = |z₁| · |z₂|

|z₁ / z₂| = |z₁| / |z₂|

|z̄| = |z|

Polar Form

z = a + bi = r(cos θ + i·sin θ) = r·cis θ

r = |z| = √(a² + b²) ← modulus

θ = arctan(b/a) ← argument (adjust for quadrant)

a = r·cos θ ← rectangular from polar

b = r·sin θ ← rectangular from polar

Quadrant Adjustment for θ

arctan(b/a) gives a value in [−90°, 90°]. Adjust: if a < 0, add 180°. If a > 0 and b < 0, add 360°. Always check that (r·cos θ, r·sin θ) gives back the original (a, b).

Example: Convert 2 + 2i to polar form

r = √(2² + 2²) = √8 = 2√2

θ = arctan(2/2) = arctan(1) = 45°

z = 2√2 · cis 45°

Check: 2√2·cos 45° = 2√2·(√2/2) = 2 ✓, 2√2·sin 45° = 2 ✓

Multiplication in Polar Form

r₁·cis θ₁ × r₂·cis θ₂ = r₁r₂·cis(θ₁ + θ₂)

Multiply moduli, add arguments

Division: divide moduli, subtract arguments

Example

2·cis 30° × 3·cis 60°

= (2 × 3)·cis(30° + 60°)

= 6·cis 90° = 6i

De Moivre's Theorem

(r·cis θ)ⁿ = rⁿ·cis(nθ)

Raise modulus to the nth power; multiply argument by n

De Moivre's Theorem makes computing large powers of complex numbers simple. Convert to polar form first, apply the theorem, then convert back to rectangular form if needed.

Example: Find (1 + i)⁶

Step 1: Convert 1 + i to polar form

r = √(1² + 1²) = √2, θ = 45°

1 + i = √2 · cis 45°

Step 2: Apply De Moivre's with n = 6

(√2 · cis 45°)⁶ = (√2)⁶ · cis(6 × 45°)

= 8 · cis 270°

Step 3: Convert back

cis 270° = cos 270° + i·sin 270° = 0 + i·(−1) = −i

(1 + i)⁶ = 8 · (−i) = −8i

Worked Examples

Example 1: Divide (3 + 2i) ÷ (1 − i)

Multiply numerator and denominator by the conjugate of (1 − i), which is (1 + i):

(3 + 2i)(1 + i) / [(1 − i)(1 + i)]

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

Denominator: 1² + 1² = 2

(3 + 2i)/(1 − i) = (1 + 5i)/2 = 1/2 + 5i/2

Example 2: Convert 2 + 2i to polar form

r = √(2² + 2²) = √8 = 2√2

θ = arctan(2/2) = arctan(1) = 45° (first quadrant, no adjustment needed)

2 + 2i = 2√2 · cis 45°

Example 3: Find (1 + i)⁶ using De Moivre's Theorem

1 + i = 2√2 · cis 45° (from Example 2 but note r = √2)

r = √2, θ = 45°

(√2)⁶ · cis(6 × 45°) = 8 · cis 270°

cos 270° = 0, sin 270° = −1

(1 + i)⁶ = 8(0 − i) = −8i

Solving Equations with Complex Solutions

Real equations can have complex solutions. Whenever a quadratic has a negative discriminant (b² − 4ac < 0), the solutions are complex conjugates.

Example 1: Solve x² = −4

x² = −4

x = ±√(−4) = ±√(4 · −1)

x = ±2i

Example 2: Solve x² + 2x + 5 = 0

Discriminant: 4 − 20 = −16

x = (−2 ± √(−16))/2

x = (−2 ± 4i)/2

x = −1 ± 2i

Key Pattern

Complex roots of real polynomials always come in conjugate pairs. If −1 + 2i is a root, then −1 − 2i is also a root. This is the Conjugate Root Theorem.

Quick Reference

Fundamentals

i = √(−1), i² = −1

i¹=i, i²=−1, i³=−i, i⁴=1

i^n: remainder of n÷4

|a+bi| = √(a²+b²)

conj(a+bi) = a−bi

(a+bi)(a−bi) = a²+b²

Polar Form

r = √(a²+b²)

θ = arctan(b/a) + quadrant fix

z = r·cis θ = r(cosθ + i·sinθ)

z₁z₂ = r₁r₂·cis(θ₁+θ₂)

z^n = rⁿ·cis(nθ)

De Moivre's Theorem

Frequently Asked Questions

What is the imaginary unit i?

The imaginary unit i is defined as i = √(−1), so i² = −1. A complex number has the form a + bi where a is the real part and b is the imaginary part. The powers of i cycle with period 4: i¹ = i, i² = −1, i³ = −i, i⁴ = 1. To find i^n, divide n by 4 and use the remainder: remainder 0 → 1, remainder 1 → i, remainder 2 → −1, remainder 3 → −i.

How do you divide complex numbers?

To divide complex numbers, multiply both numerator and denominator by the conjugate of the denominator. The conjugate of (a + bi) is (a − bi). For example, to compute (3 + 2i)/(1 − i): multiply top and bottom by (1 + i) to get (3 + 2i)(1 + i)/[(1 − i)(1 + i)] = (3 + 3i + 2i + 2i²)/(1 + 1) = (3 + 5i − 2)/2 = (1 + 5i)/2 = 1/2 + 5i/2. The denominator always becomes a² + b², a real number.

What is De Moivre's Theorem?

De Moivre's Theorem states that for a complex number in polar form z = r·cis θ (where cis θ means cos θ + i·sin θ), raising it to the nth power gives: (r·cis θ)^n = r^n·cis(nθ). This makes computing powers of complex numbers straightforward — you raise the modulus r to the power n and multiply the argument θ by n. For example, (1 + i)^6: convert to polar form 2^(1/2)·cis 45°, then apply De Moivre's: (√2)^6·cis(6 × 45°) = 8·cis 270° = −8i.

Related Topics

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