Complex Numbers | Ronny Best Mathematics — CBC Senior Secondary Kenya

Section 1 · Motivation

Why Do Complex Numbers Exist?

Senior Secondary

Mathematics evolved to solve increasingly difficult problems. Real numbers cannot solve every equation — when we try to find the square root of a negative number, real arithmetic fails. Complex numbers were invented specifically to close this gap, and they turned out to describe the physical world more accurately than real numbers alone ever could.

The Problem That Real Numbers Cannot Solve

Consider the equation x² = −1. No real number satisfies this: squaring any positive number gives a positive result; squaring any negative number also gives a positive result; and 0² = 0. There is no real x for which x² = −1. For centuries, mathematicians dismissed such equations as having "no solution" and moved on.

In the 16th century, Italian mathematicians Niccolò Tartaglia, Gerolamo Cardano, and later Rafael Bombelli discovered that even when solving real cubic equations, intermediate steps required the square root of negative numbers — and the final answer was always a real number. The imaginary quantities cancelled out at the end. This forced mathematicians to take them seriously. Cardano called them "sophistic" (fictitious), but Bombelli showed they obeyed consistent rules and could be manipulated algebraically.

By the 18th century, Euler introduced the notation i = √(−1), and Gauss gave complex numbers their modern geometric interpretation. Today, complex numbers are indispensable in electrical engineering (alternating current analysis), quantum mechanics (wave functions), fluid dynamics, signal processing, and control systems.

CBC Connection: The KICD curriculum frames complex numbers as an extension of the real number system — a natural progression from real numbers studied in earlier grades. Understanding why we extend the number system is a core competency under CBC's emphasis on mathematical reasoning, not just computation.

Section 2 · Foundations

The Imaginary Unit & Standard Form

Senior Secondary

We define a new number i, called the imaginary unit, with the property that i² = −1. From this single definition, an entire number system is built.

Definitions

Core Definitions

Imaginary unit i = √(−1) ⟹ i² = −1

Standard form z = a + bi where a, b ∈ ℝ

Real part Re(z) = a

Imaginary part Im(z) = b (note: b is real, not bi)

Conjugate z̄ = a − bi

Modulus (magnitude) |z| = √(a² + b²) (always ≥ 0)

The Cyclic Powers of i

A crucial pattern: successive powers of i repeat with a cycle of 4. You must memorise this cycle, as it is tested directly in KCSE examinations.

Powers of i — Cycle of 4

i¹= i

i²= −1

i³= i² × i = −i

i⁴= i² × i² = (−1)(−1) = 1

i⁵= i⁴ × i = i (cycle repeats)

iⁿ= i^(n mod 4) — divide n by 4 and use the remainder

Critical distinction: The imaginary part of z = 3 + 4i is 4, not 4i. Im(z) is always a real number — the coefficient of i. This is a common source of errors in examinations.

Worked Examples — Imaginary Unit

Worked Examples · Section 2

Example 1 Simplify i²³

Step 1

Divide the exponent by 4 to find the remainder: 23 ÷ 4 = 5 remainder 3

Step 2

So i²³ = i³ = −i (using the cycle: i¹=i, i²=−1, i³=−i, i⁴=1)

i²³ = −i

Example 2 If z = 5 − 3i, state Re(z), Im(z), z̄, and |z|.

Identify

z = 5 − 3i is in standard form a + bi with a = 5 and b = −3

Parts

Re(z) = 5 | Im(z) = −3 (not −3i)

Conjugate

z̄ = 5 − (−3)i = 5 + 3i (change only the sign of the imaginary part)

Modulus

|z| = √(5² + (−3)²) = √(25 + 9) = √34 ≈ 5.83

Re(z) = 5 · Im(z) = −3 · z̄ = 5 + 3i · |z| = √34

Example 3 Solve x² + 9 = 0 over the complex numbers.

Rearrange

x² = −9

Square root

x = ±√(−9) = ±√(9 × −1) = ±3√(−1) = ±3i

Check

(3i)² = 9i² = 9(−1) = −9 ✓ (−3i)² = 9i² = −9 ✓

x = 3i or x = −3i

Section 3 · Core Operations

Arithmetic of Complex Numbers

Senior Secondary

Complex numbers obey the same laws of algebra as real numbers — commutativity, associativity, distributivity — with one extra rule: whenever i² appears, replace it with −1. This single substitution is the key to all complex arithmetic.

Rules for the Four Operations

Let z₁ = a + bi and z₂ = c + di

Addition z₁ + z₂ = (a + c) + (b + d)i — add real to real, imaginary to imaginary

Subtraction z₁ − z₂ = (a − c) + (b − d)i

Multiplication z₁ × z₂ = (ac − bd) + (ad + bc)i — expand then replace i² = −1

Division z₁ ÷ z₂ = (z₁ × z̄₂) / (c² + d²) — multiply top and bottom by the conjugate of z₂

Useful identity z × z̄ = a² + b² = |z|² (always a positive real number)

Why multiply by the conjugate when dividing? The denominator c + di is complex. Multiplying both numerator and denominator by c − di turns the denominator into (c + di)(c − di) = c² + d² — a real number — making the result expressible in standard form a + bi.

✓ Multiplication: FOIL, then replace i² = −1

(a + bi)(c + di) = ac + adi + bci + bdi² = (ac − bd) + (ad + bc)i

✓ Division: Multiply by the conjugate

Conjugate of (c + di) is (c − di). The denominator becomes c² + d².

Worked Examples — Arithmetic

Worked Examples · Section 3

Example 4 Given z₁ = 4 + 3i and z₂ = 1 − 2i, find z₁ + z₂ and z₁ − z₂.

Add

z₁ + z₂ = (4 + 1) + (3 + (−2))i = 5 + i

Subtract

z₁ − z₂ = (4 − 1) + (3 − (−2))i = 3 + 5i = 3 + 5i

z₁ + z₂ = 5 + i | z₁ − z₂ = 3 + 5i

Example 5 Multiply (2 + 3i)(1 − 4i).

Expand

= 2(1) + 2(−4i) + 3i(1) + 3i(−4i)

Collect

= 2 − 8i + 3i − 12i²

Replace

Replace i² = −1: = 2 − 5i − 12(−1) = 2 + 12 − 5i = 14 − 5i

(2 + 3i)(1 − 4i) = 14 − 5i

Example 6 Divide (3 + 4i) ÷ (1 + 2i). Give the answer in standard form.

Conjugate

The conjugate of the denominator (1 + 2i) is (1 − 2i)

Multiply

Numerator: (3 + 4i)(1 − 2i) = 3 − 6i + 4i − 8i² = 3 − 2i + 8 = 11 − 2i

Denominator

(1 + 2i)(1 − 2i) = 1² + 2² = 5

Result

(3 + 4i) ÷ (1 + 2i) = (11 − 2i) / 5 = 11/5 − (2/5)i = 2.2 − 0.4i

(3 + 4i) ÷ (1 + 2i) = 2.2 − 0.4i

Example 7 Show that z × z̄ = |z|² for z = 3 − 5i.

Conjugate

z = 3 − 5i ⟹ z̄ = 3 + 5i

Product

z × z̄ = (3 − 5i)(3 + 5i) = 9 + 15i − 15i − 25i² = 9 + 25 = 34

Modulus²

|z|² = (√(9 + 25))² = 34 ✓ The identity holds.

z × z̄ = 34 = |z|² ✓

Section 4 · Geometry

The Argand Diagram

Senior Secondary

Jean-Robert Argand (1806) showed that complex numbers can be plotted on a plane — the complex plane or Argand diagram — with real parts on the horizontal axis and imaginary parts on the vertical axis. This geometric view transforms algebraic operations into visual, intuitive ones.

The Complex Plane

Every complex number z = a + bi is a point (a, b) in the plane, or equivalently a position vector from the origin to that point. The horizontal axis is called the real axis (Re), and the vertical axis is the imaginary axis (Im).

Geometric Interpretations

Point representation z = a + bi ↔ point (a, b) in the plane

Modulus = distance |z| = distance from origin to z = √(a² + b²)

Argument = angle arg(z) = θ = angle from positive real axis (measured anticlockwise)

Conjugate reflection z̄ = reflection of z in the real axis

Addition = vector sum z₁ + z₂ = parallelogram law of vector addition

Multiplication by i iz = rotation of z by 90° anticlockwise

Argument conventions: The argument θ is typically given in radians and expressed in the range −π < θ ≤ π (principal argument). When computing arg(z) = arctan(b/a), you must use the correct quadrant — do not simply evaluate arctan and accept the calculator's answer without checking which quadrant the point (a, b) lies in.

Computing the Argument — Quadrant Rules

arg(z) = θ based on quadrant of (a, b)

Q1: a > 0, b > 0 θ = arctan(b/a) (positive, in (0, π/2))

Q2: a < 0, b > 0 θ = π − arctan(|b/a|) (in (π/2, π))

Q3: a < 0, b < 0 θ = −π + arctan(|b/a|) (in (−π, −π/2))

Q4: a > 0, b < 0 θ = −arctan(|b/a|) (in (−π/2, 0))

On axes: a=0, b>0 θ = π/2 | a=0, b<0: θ = −π/2

Worked Examples — Argand Diagram

Worked Examples · Section 4

Example 8 Find the modulus and argument of z = −1 + √3 i.

Identify

a = −1, b = √3. Point is in Quadrant 2 (negative real, positive imaginary)

Modulus

|z| = √(a² + b²) = √(1 + 3) = √4 = 2

Basic angle

arctan(|b/a|) = arctan(√3/1) = arctan(√3) = π/3 (i.e. 60°)

Quadrant 2

θ = π − π/3 = 2π/3 (i.e. 120°)

|z| = 2 | arg(z) = 2π/3 (or 120°)

Example 9 Find the modulus and argument of z = 3 − 3i.

Identify

a = 3, b = −3. Point is in Quadrant 4 (positive real, negative imaginary)

Modulus

|z| = √(9 + 9) = √18 = 3√2

Argument

arctan(|−3/3|) = arctan(1) = π/4. Q4 ⟹ θ = −π/4 (i.e. −45°)

|z| = 3√2 | arg(z) = −π/4 (or −45°)

Section 5 · Advanced Form

Polar Form & Euler's Formula

Advanced

Standard form z = a + bi expresses a complex number in terms of its rectangular coordinates. Polar form expresses the same number in terms of its distance from the origin (modulus r) and its angle from the positive real axis (argument θ). For multiplication, division, and powers, polar form is far more efficient.

Polar & Exponential Form

Polar form z = r(cos θ + i sin θ) abbreviated as z = r cis θ

Parameters r = |z| = √(a²+b²) θ = arg(z)

Converting: polar → rectangular a = r cos θ b = r sin θ

Euler's formula e^(iθ) = cos θ + i sin θ (Leonhard Euler, 1748)

Exponential form z = r e^(iθ)

Multiplication in polar z₁z₂ = r₁r₂ cis(θ₁ + θ₂) — multiply moduli, add arguments

Division in polar z₁/z₂ = (r₁/r₂) cis(θ₁ − θ₂) — divide moduli, subtract arguments

Euler's Identity — considered the most beautiful equation in mathematics:

e^(iπ) + 1 = 0

This single equation connects five of the most fundamental constants in mathematics: e (Euler's number), i (imaginary unit), π (pi), 1 (the multiplicative identity), and 0 (the additive identity). It follows directly from Euler's formula: e^(iπ) = cos π + i sin π = −1 + 0 = −1, so e^(iπ) + 1 = 0.

Worked Examples — Polar Form

Worked Examples · Section 5

Example 10 Convert z = 1 + √3 i to polar form and exponential form.

Modulus

r = |z| = √(1² + (√3)²) = √(1 + 3) = √4 = 2

Argument

a = 1 > 0, b = √3 > 0 → Quadrant 1. θ = arctan(√3/1) = arctan(√3) = π/3

Polar form

z = 2(cos π/3 + i sin π/3) = 2 cis(π/3)

Exponential

z = 2e^(iπ/3)

z = 2 cis(π/3) = 2e^(iπ/3)

Example 11 If z₁ = 3 cis(π/4) and z₂ = 2 cis(π/6), find z₁z₂ and z₁/z₂ in polar form.

Multiply

z₁z₂ = (3 × 2) cis(π/4 + π/6) = 6 cis(3π/12 + 2π/12) = 6 cis(5π/12)

Divide

z₁/z₂ = (3/2) cis(π/4 − π/6) = (3/2) cis(3π/12 − 2π/12) = (3/2) cis(π/12)

z₁z₂ = 6 cis(5π/12) | z₁/z₂ = (3/2) cis(π/12)

Section 6 · Advanced

De Moivre's Theorem

Advanced

Abraham de Moivre (1730) discovered a powerful theorem that makes computing powers and roots of complex numbers straightforward. In rectangular form, raising z = a + bi to the 8th power would require 7 multiplications. In polar form, it takes one step.

De Moivre's Theorem — for all integer and rational n

[r(cos θ + i sin θ)]ⁿ = rⁿ(cos nθ + i sin nθ)

In words: raise the modulus to the power n, and multiply the argument by n. This applies for all integer values of n (positive, negative, and zero) and can be extended to rational values of n for finding roots.

Applications of De Moivre's Theorem

Application 1: Powers of complex numbers

Convert to polar form, apply the theorem, convert back if needed.

Application 2: Multiple-angle identities

Expand (cos θ + i sin θ)ⁿ by binomial theorem and equate real/imaginary parts to express cos(nθ) and sin(nθ) in terms of powers of cos θ and sin θ.

Worked Examples — De Moivre's Theorem

Worked Examples · Section 6

Example 12 Compute (1 + i)⁸ using De Moivre's theorem.

Polar form

1 + i: r = √(1+1) = √2 , θ = arctan(1/1) = π/4

Write

1 + i = √2 · (cos π/4 + i sin π/4) = √2 · cis(π/4)

Apply

(1+i)⁸ = (√2)⁸ · cis(8 × π/4) = 2⁴ · cis(2π)

Simplify

cis(2π) = cos 2π + i sin 2π = 1 + 0 = 1. So (1+i)⁸ = 16 × 1 = 16

(1 + i)⁸ = 16

Example 13 Express cos 3θ in terms of cos θ only using De Moivre's theorem.

Setup

By De Moivre: cos 3θ + i sin 3θ = (cos θ + i sin θ)³

Expand

Let c = cos θ, s = sin θ. (c + is)³ = c³ + 3c²(is) + 3c(is)² + (is)³

Simplify

= c³ + 3ic²s + 3c(i²s²) + i³s³ = c³ + 3ic²s − 3cs² − is³

Collect

Real part: cos 3θ = c³ − 3cs² = cos³θ − 3cosθ sin²θ

Use identity

Replace sin²θ = 1 − cos²θ: cos 3θ = cos³θ − 3cosθ(1 − cos²θ) = 4cos³θ − 3cosθ

cos 3θ = 4cos³θ − 3cos θ

Example 14 Find (√3 − i)⁶ in standard form a + bi.

Polar

a = √3, b = −1 → r = √(3+1) = 2. Q4: θ = −arctan(1/√3) = −π/6

Apply

(√3−i)⁶ = 2⁶ · cis(6 × −π/6) = 64 · cis(−π)

Convert

cis(−π) = cos(−π) + i sin(−π) = −1 + 0 = −1

(√3 − i)⁶ = 64 × (−1) = −64

Section 7 · Extension

nth Roots of Complex Numbers

Advanced

Every non-zero complex number has exactly n distinct nth roots. These roots are equally spaced around a circle of radius r^(1/n) in the complex plane, forming a regular n-sided polygon. This is one of the most beautiful results in complex number theory.

nth Roots Formula

General formula z_k = r^(1/n) · cis((θ + 2πk)/n) for k = 0, 1, 2, …, n−1

Number of roots Exactly n distinct roots

Modulus of each root |z_k| = r^(1/n) (all roots have the same modulus)

Spacing between roots Angular gap = 2π/n radians = 360°/n

Roots of Unity

The nth roots of unity are the solutions to zⁿ = 1. These are particularly important in higher mathematics. For zⁿ = 1: r = 1, θ = 0, so z_k = cis(2πk/n) for k = 0, 1, …, n−1.

Worked Examples · Section 7

Example 15 Find all cube roots of unity (solve z³ = 1).

Setup

z³ = 1. In polar form: r = 1, θ = 0. Apply the roots formula with n = 3.

k = 0

z₀ = 1^(1/3) · cis((0 + 0)/3) = cis(0) = 1

k = 1

z₁ = cis(2π/3) = cos 120° + i sin 120° = −½ + i(√3/2)

k = 2

z₂ = cis(4π/3) = cos 240° + i sin 240° = −½ − i(√3/2)

Geometry

The three roots form an equilateral triangle inscribed in the unit circle (radius 1), equally spaced at 120° intervals.

z = 1, −½ + (√3/2)i, −½ − (√3/2)i

Example 16 Find the two square roots of z = 4i.

Polar form

4i: a = 0, b = 4. r = 4, θ = π/2 (positive imaginary axis)

Formula

z_k = 4^(1/2) · cis((π/2 + 2πk)/2) = 2 · cis((π/4 + πk)) for k = 0, 1

k = 0

z₀ = 2 cis(π/4) = 2(cos 45° + i sin 45°) = 2(1/√2 + i/√2) = √2 + i√2

k = 1

z₁ = 2 cis(5π/4) = 2(cos 225° + i sin 225°) = 2(−1/√2 − i/√2) = −√2 − i√2

Check k=0

(√2 + i√2)² = 2 + 2i√2·i√2 + ... = 2 + 4i² + 4i·... let's verify: (√2)²+(i√2)²+2(√2)(i√2) = 2 − 2 + 4i·(1/2)... = 0 + 4i·1 = 4i ✓

√(4i) = √2 + i√2 and −√2 − i√2

Example 17 Find all fourth roots of z = 16.

Polar

16 is a real positive number: r = 16, θ = 0. Apply n = 4.

Formula

z_k = 16^(1/4) · cis(2πk/4) = 2 · cis(πk/2) for k = 0, 1, 2, 3

Results

k=0: 2·cis(0) = 2 k=1: 2·cis(π/2) = 2i k=2: 2·cis(π) = −2 k=3: 2·cis(3π/2) = −2i

⁴√16 = 2, 2i, −2, −2i (forming a square on the complex plane)

Section 8 · Exam Preparation

Common Errors to Avoid in Examinations

All Levels

These errors appear repeatedly in KCSE scripts and Senior Secondary assessments. Study each one carefully — recognising a wrong approach is as important as knowing the correct method.

✗ Wrong Im(3 + 4i) = 4i

✓ Correct Im(3 + 4i) = 4

The imaginary part is a real number (the coefficient of i), not the term 4i itself.

✗ Wrong √(−4) × √(−9) = √36 = 6

✓ Correct √(−4) × √(−9) = 2i × 3i = 6i² = −6

The rule √a × √b = √(ab) only holds when both a and b are non-negative real numbers.

✗ Wrong arg(−1 − i) = arctan(−1/−1) = arctan(1) = π/4

✓ Correct (−1, −1) is in Q3: arg = −3π/4 (or 225°)

Always check the quadrant before stating the argument. arctan alone gives the reference angle, not the principal argument.

✗ Wrong Dividing without multiplying by conjugate: (2+i)/(1+i) = (2+i)/(1+i)

✓ Correct Multiply top and bottom by (1−i) to get a real denominator.

Division must eliminate the imaginary part from the denominator using the conjugate.

✗ Wrong (1+i)³ = 1 + i³ = 1 − i

✓ Correct Use De Moivre or expand fully: (1+i)³ = 1 + 3i + 3i² + i³ = 1+3i−3−i = −2+2i

You cannot distribute a power over addition: (a+b)ⁿ ≠ aⁿ + bⁿ.

✗ Wrong Forgetting that z³=1 has 3 solutions, only writing z = 1

✓ Correct zⁿ = w has exactly n complex solutions. Always find all n roots.

An nth degree polynomial equation has exactly n complex roots (counting multiplicity).

✗ Wrong Conjugate of 3 + 4i is −3 − 4i

✓ Correct Conjugate of 3 + 4i is 3 − 4i

The conjugate changes only the sign of the imaginary part, not the real part.

✗ Wrong In De Moivre: (2 cis θ)³ = 2 cis(3θ)

✓ Correct (2 cis θ)³ = 2³ cis(3θ) = 8 cis(3θ)

Both the modulus and the argument are affected: rⁿ for the modulus, nθ for the argument.

Section 9 · Quick Reference

Complete Formula Sheet — Complex Numbers

Reference

Use this formula sheet as a quick reference during revision. All formulas are required knowledge for the KCSE Senior Secondary Mathematics examination.

Standard Form & Parts

z = a + bi (a, b ∈ ℝ)

Re(z) = a

Im(z) = b (not bi!)

i² = −1, i³ = −i, i⁴ = 1

Modulus & Conjugate

z̄ = a − bi

|z| = √(a² + b²)

|z|² = z × z̄

|z₁z₂| = |z₁||z₂|

Arithmetic Rules

Add: add real to real, im to im

Multiply: FOIL, replace i² = −1

Divide: multiply by conjugate

Denom after conj: c² + d²

Polar Form

z = r cis θ = r(cos θ + i sin θ)

r = |z|, θ = arg(z)

a = r cos θ, b = r sin θ

z = re^(iθ) (Euler form)

Polar Arithmetic

z₁z₂ = r₁r₂ cis(θ₁+θ₂)

z₁/z₂ = (r₁/r₂) cis(θ₁−θ₂)

e^(iπ) + 1 = 0 (Euler's identity)

De Moivre's Theorem

(r cis θ)ⁿ = rⁿ cis(nθ)

Valid for all integers n

cos 3θ = 4cos³θ − 3cosθ

sin 3θ = 3sinθ − 4sin³θ

nth Roots

z_k = r^(1/n) cis((θ+2πk)/n)

k = 0, 1, 2, …, n−1

n distinct roots, equally spaced

Gap between roots: 2π/n

Roots of Unity (zⁿ=1)

z_k = cis(2πk/n)

Cube roots: 1, −½±(√3/2)i

Sum of all roots = 0

Product of all roots = (−1)ⁿ⁺¹