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Complex Numbers A-Level Pure 3

Grade 12 · Pure Mathematics 3 · Cambridge A-Level 9709 · Age 17–18

Welcome to Complex Numbers — Pure 3

Complex numbers extend the real number line into a two-dimensional plane, enabling solutions to all polynomial equations regardless of discriminant. They underpin quantum mechanics, signal processing, control theory, and electrical engineering — making this one of the most powerful extensions of mathematics you will encounter at A-Level.

z = a + bi  |  |z| = √(a² + b²)  |  arg(z) = arctan(b/a)  |  z* = a − bi

From Cartesian form to the Argand diagram, from modulus-argument form to conjugate pairs — every idea builds on the last to create a unified and elegant theory.

Learning Objectives

  • Understand i = √(−1) and i² = −1, and compute powers of i
  • Perform addition, subtraction, multiplication of complex numbers in the form a + bi
  • Find the complex conjugate z* = a − bi and use it in division
  • Solve quadratic equations with negative discriminant giving complex roots
  • Represent complex numbers on an Argand diagram
  • Find the modulus |z| = √(a² + b²) and interpret it geometrically
  • Find the argument arg(z) correctly in all four quadrants using principal argument convention
  • Convert between Cartesian form and modulus-argument form r(cosθ + i·sinθ)
  • Use properties |z₁z₂| = |z₁||z₂| and arg(z₁z₂) = arg(z₁) + arg(z₂)
  • Show that complex roots of real polynomials occur in conjugate pairs and apply this to solve cubics and quartics

Topic Overview

Complex Arithmetic

Add, subtract, multiply and simplify using i² = −1

Conjugates

z* = a − bi — essential for division and real polynomial roots

Argand Diagram

Geometric representation: point (a, b) in the complex plane

Modulus & Argument

Distance from origin and angle from positive real axis

Modulus-Argument Form

z = r(cosθ + i·sinθ) — multiply moduli, add arguments

Roots of Polynomials

Complex roots come in conjugate pairs for real polynomials

Learn 1 — Introduction to Complex Numbers

The imaginary unit i is defined so that i² = −1. This single definition unlocks an entirely new number system.

The Imaginary Unit

i = √(−1)    so    i² = −1

The Complex Number z = a + bi

a = Re(z) is the real part  |  b = Im(z) is the imaginary part
When b = 0, z is a real number. When a = 0, z is purely imaginary.

Addition and Subtraction

(a + bi) + (c + di) = (a + c) + (b + d)i
(a + bi) − (c + di) = (a − c) + (b − d)i
Rule: add real parts together and imaginary parts together separately.

Multiplication

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

Expand like double brackets (FOIL), then replace i² with −1:

Example: Expand (3 + 2i)(1 + 4i)

= 3·1 + 3·4i + 2i·1 + 2i·4i
= 3 + 12i + 2i + 8i²
= 3 + 14i + 8(−1)
= 3 − 8 + 14i = −5 + 14i

Powers of i (Cyclic Pattern)

i¹ = i  |  i² = −1  |  i³ = −i  |  i⁴ = 1  |  i⁵ = i   (repeats every 4)
To find iⁿ: divide n by 4, use the remainder. Remainder 0 → 1, remainder 1 → i, remainder 2 → −1, remainder 3 → −i.

Example: Simplify (2 + 3i) − (1 − i) + i²

= 2 + 3i − 1 + i + (−1)
= (2 − 1 − 1) + (3 + 1)i
= 0 + 4i = 4i

Learn 2 — Complex Conjugate and Division

The Complex Conjugate

If z = a + bi, then z* = a − bi
The conjugate simply flips the sign of the imaginary part. The real part is unchanged.
Key result: z · z* = (a + bi)(a − bi) = a² + b² = |z|²

Example: Verify z · z* is real for z = 3 + 4i

z* = 3 − 4i
(3 + 4i)(3 − 4i) = 9 − 12i + 12i − 16i²
= 9 − 16(−1) = 9 + 16 = 25 (real and positive ✓)

Division of Complex Numbers

To divide, multiply both numerator and denominator by the conjugate of the denominator. This makes the denominator real.

(a + bi) / (c + di) = [(a + bi)(c − di)] / [(c + di)(c − di)] = [(a + bi)(c − di)] / (c² + d²)

Example: Compute (3 + 2i) / (1 − i) in the form a + bi

Multiply top and bottom by the conjugate of (1 − i), which is (1 + i):
= (3 + 2i)(1 + i) / [(1 − i)(1 + i)]
Denominator: (1 − i)(1 + i) = 1 + 1 = 2
Numerator: (3 + 2i)(1 + i) = 3 + 3i + 2i + 2i² = 3 + 5i − 2 = 1 + 5i
Answer: (1 + 5i) / 2 = 0.5 + 2.5i

Example: Compute (2 + i) / (3 + 4i)

Multiply by (3 − 4i)/(3 − 4i):
Denominator: 9 + 16 = 25
Numerator: (2 + i)(3 − 4i) = 6 − 8i + 3i − 4i² = 6 − 5i + 4 = 10 − 5i
Answer: (10 − 5i)/25 = 0.4 − 0.2i
Common error: Students multiply by the denominator itself rather than its conjugate. Always use the conjugate to make the denominator real.

Learn 3 — Argand Diagram

The Argand diagram represents complex numbers geometrically. The horizontal axis is the real axis, the vertical axis is the imaginary axis.

Plotting Complex Numbers

z = a + bi is plotted as the point (a, b) or as a vector from the origin to (a, b).
Real axis = horizontal  |  Imaginary axis = vertical

Modulus

|z| = √(a² + b²)

The modulus is the distance from the origin to the point (a, b) — found by Pythagoras's theorem.

Argument

arg(z) = angle from the positive real axis, measured anticlockwise
Principal argument: −π < arg(z) ≤ π (i.e. −180° < θ ≤ 180°)

Finding arg(z) by quadrant:
• Quadrant 1 (a > 0, b > 0): arg(z) = arctan(b/a)
• Quadrant 2 (a < 0, b > 0): arg(z) = π − arctan(|b/a|)
• Quadrant 3 (a < 0, b < 0): arg(z) = −π + arctan(|b/a|)
• Quadrant 4 (a > 0, b < 0): arg(z) = −arctan(|b/a|)
Key trap: arctan(b/a) only gives the reference angle. You MUST adjust it based on which quadrant z lies in.

Example: Find |z| and arg(z) for z = −1 + √3·i

|z| = √(1 + 3) = √4 = 2
Point is in Quadrant 2 (a = −1 < 0, b = √3 > 0)
Reference angle = arctan(√3/1) = arctan(√3) = 60° = π/3
arg(z) = 180° − 60° = 120° = 2π/3

Example: Find arg(z) for z = −2 − 2i

Point is in Quadrant 3 (both negative)
Reference angle = arctan(2/2) = arctan(1) = 45°
arg(z) = −180° + 45° = −135° (principal argument)

Learn 4 — Modulus-Argument Form

Any complex number can be written in terms of its modulus r and argument θ. This form makes multiplication, division, and powers much easier.

Modulus-Argument Form

z = r(cosθ + i·sinθ)    where r = |z| and θ = arg(z)
Converting from Cartesian: a = r·cosθ, b = r·sinθ
Converting to Cartesian: r = √(a² + b²), θ = arg(z) using quadrant method

Multiplication in Mod-Arg Form

|z₁z₂| = |z₁| · |z₂|     arg(z₁z₂) = arg(z₁) + arg(z₂)

To multiply: multiply the moduli and add the arguments.

Division in Mod-Arg Form

|z₁/z₂| = |z₁| / |z₂|     arg(z₁/z₂) = arg(z₁) − arg(z₂)

Example: Find (1 + i)⁴ using mod-arg form

|1 + i| = √2, arg(1 + i) = 45° = π/4
So 1 + i = √2 · (cos45° + i·sin45°)
(1 + i)⁴ = (√2)⁴ · (cos(4 × 45°) + i·sin(4 × 45°))
= 4 · (cos180° + i·sin180°)
= 4 · (−1 + 0·i) = −4

Example: Express z = 2(cos(π/3) + i·sin(π/3)) in the form a + bi

a = 2·cos(π/3) = 2 × 1/2 = 1
b = 2·sin(π/3) = 2 × √3/2 = √3
z = 1 + √3·i
Remember: cos is the coefficient of the REAL part and sin is the coefficient of the IMAGINARY part in mod-arg form.

Learn 5 — Complex Roots of Polynomial Equations

The Conjugate Root Theorem

If z = p + qi is a root of a real polynomial, then z* = p − qi is also a root.
Why? If all coefficients of the polynomial are real, substituting z* whenever z appeared produces the conjugate of zero — which is still zero. So complex roots of real polynomials always occur in conjugate pairs.

Quadratics with Complex Roots

When the discriminant b² − 4ac < 0, the quadratic has two complex conjugate roots.

Example: Solve z² + 2z + 5 = 0

Using the quadratic formula: z = (−2 ± √(4 − 20)) / 2
= (−2 ± √(−16)) / 2
= (−2 ± 4i) / 2
z = −1 + 2i or z = −1 − 2i (a conjugate pair ✓)

Using Conjugate Pairs to Solve Cubics and Quartics

If a complex root is known, its conjugate is also a root. Together they give a real quadratic factor.

If z = a + bi is a root of a real polynomial, then:
(z − (a + bi))(z − (a − bi)) = z² − 2az + (a² + b²) is a real quadratic factor.

Example: Given 2 − i is a root of z³ + az² + bz + 5 = 0, find a and b

Since coefficients are real, 2 + i is also a root.
Real quadratic factor: (z − (2−i))(z − (2+i)) = z² − 4z + 5
Divide z³ + az² + bz + 5 by (z² − 4z + 5):
z³ + az² + bz + 5 = (z² − 4z + 5)(z + c) for some constant c
Expand: z³ + cz² − 4z² − 4cz + 5z + 5c = z³ + (c−4)z² + (5−4c)z + 5c
Constant term: 5c = 5 → c = 1
So a = c − 4 = −3 and b = 5 − 4c = 1: a = −3, b = 1

Worked Examples

Example 1 — Simplify (2 + 3i)(1 − 2i) + i²

Expand (2 + 3i)(1 − 2i): 2 − 4i + 3i − 6i² = 2 − i − 6(−1) = 2 − i + 6 = 8 − i
Add i²: 8 − i + (−1) = 7 − i

Example 2 — Express (3 + 4i)/(2 − i) in the form a + bi

Multiply by conjugate of denominator (2 + i)/(2 + i):
Denominator: (2 − i)(2 + i) = 4 + 1 = 5
Numerator: (3 + 4i)(2 + i) = 6 + 3i + 8i + 4i² = 6 + 11i − 4 = 2 + 11i
Answer: (2 + 11i)/5 = 0.4 + 2.2i

Example 3 — Find |z| and arg(z) for z = −1 + √3·i

|z| = √(1 + 3) = √4 = 2
Quadrant 2: reference angle = arctan(√3/1) = 60°
arg(z) = 180° − 60° = 120° (or 2π/3 radians)

Example 4 — Express z = 2(cos(π/3) + i·sin(π/3)) in the form a + bi

a = 2·cos(60°) = 2 × 0.5 = 1
b = 2·sin(60°) = 2 × (√3/2) = √3
z = 1 + √3·i

Example 5 — Given 2 − i is a root of z³ + az² + bz + 5 = 0, find a and b

Since real polynomial, 2 + i is also a root.
Real quadratic factor: (z−(2−i))(z−(2+i)) = z² − 4z + 5
Polynomial = (z² − 4z + 5)(z + 1) [constant term check: 5×1=5 ✓]
Expand: z³ − 4z² + 5z + z² − 4z + 5 = z³ − 3z² + z + 5
a = −3, b = 1

Example 6 — Solve z² + 2z + 5 = 0 and display roots

Discriminant = 4 − 20 = −16 < 0, so complex roots.
z = (−2 ± √(−16)) / 2 = (−2 ± 4i) / 2
Roots: z = −1 + 2i and z = −1 − 2i
On Argand diagram: both points at x = −1, at y = ±2 (mirror images in real axis)

Example 7 — Find (1 + i)⁶ using modulus-argument form

|1 + i| = √2, arg(1 + i) = π/4
(1 + i)⁶ = (√2)⁶ · (cos(6π/4) + i·sin(6π/4))
= 8 · (cos(3π/2) + i·sin(3π/2))
= 8 · (0 + i(−1)) = −8i

Example 8 — Given z₁ = 3 + 4i and z₂ = 1 − 2i, find |z₁/z₂| and arg(z₁/z₂)

|z₁| = √(9 + 16) = 5  |  |z₂| = √(1 + 4) = √5
|z₁/z₂| = 5/√5 = √5 ≈ 2.236
arg(z₁) = arctan(4/3) ≈ 53.13°  |  arg(z₂) = arctan(−2/1) ≈ −63.43°
arg(z₁/z₂) = 53.13° − (−63.43°) = 116.57°

Common Mistakes

Mistake 1 — i² = +1 instead of −1

Wrong: (2 + 3i)(1 + i) = 2 + 2i + 3i + 3i² = 2 + 5i + 3 = 5 + 5i
Correct: i² = −1, so 3i² = −3. Answer: 2 + 5i − 3 = −1 + 5i

Mistake 2 — Using arctan(b/a) directly for arg(z) without adjusting for quadrant

Wrong: For z = −1 + i, arg(z) = arctan(1/(−1)) = arctan(−1) = −45°
Correct: z is in Quadrant 2. Reference angle = 45°, so arg(z) = 180° − 45° = 135°

Mistake 3 — Conjugate of (a + bi) is (−a − bi)

Wrong: z* of (3 + 2i) is (−3 − 2i)
Correct: z* = 3 − 2i. Only the sign of the imaginary part changes, not the real part.

Mistake 4 — Dividing by the denominator instead of the conjugate

Wrong: (2 + i)/(1 + i) × (1 + i)/(1 + i) → denominator becomes (1 + i)² not real
Correct: multiply by (1 − i)/(1 − i) — the conjugate — making denominator 1² + 1² = 2

Mistake 5 — Assuming a real polynomial can have just one complex root

Wrong: "z = 3 + 2i is the only complex root of this cubic with real coefficients"
Correct: Complex roots of real polynomials always come in conjugate pairs. So z = 3 − 2i must also be a root.

Mistake 6 — Swapping real and imaginary parts in mod-arg form

Wrong: z = r(cosθ + i·sinθ) → real part is r·sinθ
Correct: Real part = r·cosθ. Imaginary part = r·sinθ. cos comes first (real), sin comes second (imaginary).

Mistake 7 — Adding moduli instead of multiplying them

Wrong: |z₁z₂| = |z₁| + |z₂|
Correct: |z₁z₂| = |z₁| × |z₂|. Moduli multiply. Similarly |z₁/z₂| = |z₁|/|z₂|.

Key Formulas

FormulaExpressionNotes
Complex numberz = a + bia = Re(z), b = Im(z)
Complex conjugatez* = a − biFlip sign of imaginary part only
Modulus|z| = √(a² + b²)Distance from origin on Argand diagram
Argumentarg(z) = angle from +ve real axisPrincipal: −π < θ ≤ π
z · z*a² + b² = |z|²Always real and non-negative
Modulus-argument formz = r(cosθ + i·sinθ)r = |z|, θ = arg(z)
Multiplication (mod-arg)|z₁z₂| = |z₁||z₂|Multiply moduli
Multiplication (mod-arg)arg(z₁z₂) = arg(z₁) + arg(z₂)Add arguments
Division (mod-arg)|z₁/z₂| = |z₁|/|z₂|Divide moduli
Division (mod-arg)arg(z₁/z₂) = arg(z₁) − arg(z₂)Subtract arguments
Conjugate root theoremIf p + qi is a root, so is p − qiReal polynomial coefficients only
Real quadratic factor(z − (a+bi))(z − (a−bi)) = z² − 2az + (a²+b²)Always has real coefficients
Quadratic formulaz = (−b ± √(b²−4ac)) / 2aIf b²−4ac < 0: complex roots
Powers of ii, −1, −i, 1 (repeating cycle of 4)Use remainder when dividing index by 4

Proof Bank

Proof 1 — z · z* = |z|²

Let z = a + bi, so z* = a − bi.

z · z* = (a + bi)(a − bi)
= a² − abi + abi − b²i²
= a² − b²(−1)
= a² + b²

But |z| = √(a² + b²), so |z|² = a² + b².

Therefore z · z* = |z|² ✓

Consequence: z · z* is always real and non-negative. It equals zero only when z = 0.

Proof 2 — Complex roots of real polynomials occur in conjugate pairs

Let P(z) = aₙzⁿ + aₙ₋₁zⁿ⁻¹ + ... + a₁z + a₀ where all aₖ are real.

Suppose P(α) = 0 for some complex number α = p + qi.

Claim: P(α*) = 0 also.

Key lemma: For any complex numbers z, w and real constant c:
(z + w)* = z* + w*     (zw)* = z* · w*     c* = c (since c is real)

Proof of claim:
P(α*) = aₙ(α*)ⁿ + ... + a₁α* + a₀
= aₙ(αⁿ)* + ... + a₁α* + a₀*   [using (zw)* = z*w*, and aₖ* = aₖ]
= (aₙαⁿ + ... + a₁α + a₀)*
= (P(α))*
= 0* = 0 ✓

Therefore α* is also a root. Complex roots occur in conjugate pairs.

Argand Diagram Visualiser

Enter a complex number z = a + bi. Click Plot to see the point on the Argand diagram, its modulus, argument, and conjugate.

Enter values and click Plot.

Exercise 1 — Complex Arithmetic

Exercise 2 — Conjugate and Division

Exercise 3 — Modulus and Argument

Exercise 4 — Modulus-Argument Form

Exercise 5 — Complex Roots of Polynomials

Practice — Mixed Questions (30 Questions)

Challenge — Hard Questions (15 Questions)

Exam Style Questions

Question 1 [4 marks]

The complex number z satisfies the equation z² − 6z + 13 = 0. Find the two roots, expressing each in the form a + bi where a and b are real numbers.

Discriminant = 36 − 52 = −16 M1
z = (6 ± √(−16))/2 = (6 ± 4i)/2 M1
z = 3 + 2i and z = 3 − 2i A1 A1

Question 2 [5 marks]

Given that 1 + 2i is a root of the equation z³ − 3z² + 7z − 5 = 0, find the other two roots.

Since real coefficients, 1 − 2i is also a root. B1
Real quadratic factor: (z−1)² + 4 = z² − 2z + 5 M1
Divide z³ − 3z² + 7z − 5 by (z² − 2z + 5): M1
Quotient = z − 1, so third root is z = 1 A1
All three roots: z = 1 + 2i, z = 1 − 2i, z = 1 A1

Question 3 [4 marks]

Express (4 + 3i)/(2 − i) in the form a + bi, showing your working clearly.

Multiply numerator and denominator by (2 + i) M1
Denominator: (2−i)(2+i) = 5 A1
Numerator: (4+3i)(2+i) = 8 + 4i + 6i + 3i² = 8 + 10i − 3 = 5 + 10i M1
Answer: (5 + 10i)/5 = 1 + 2i A1

Question 4 [3 marks]

Find the modulus and argument of z = −√3 + i. Give the argument in radians in the range −π < θ ≤ π.

|z| = √(3 + 1) = √4 = 2 B1
z is in Quadrant 2. Reference angle = arctan(1/√3) = π/6 M1
arg(z) = π − π/6 = 5π/6 A1

Question 5 [5 marks]

The complex numbers z₁ = 2(cos(π/6) + i·sin(π/6)) and z₂ = 3(cos(π/3) + i·sin(π/3)).
(a) Find |z₁z₂| and arg(z₁z₂).
(b) Express z₁z₂ in the form a + bi.

(a) |z₁z₂| = 2 × 3 = 6 B1
arg(z₁z₂) = π/6 + π/3 = π/6 + 2π/6 = π/2 B1
(b) z₁z₂ = 6(cos(π/2) + i·sin(π/2)) = 6(0 + i) = 6i M1 A1
i.e., a = 0, b = 6 A1

Question 6 [4 marks]

Show that z = 2 + i satisfies z² − 4z + 5 = 0. Hence write down the other root of this equation.

(2+i)² − 4(2+i) + 5 M1
= (4 + 4i + i²) − 8 − 4i + 5 M1
= (4 + 4i − 1) − 8 − 4i + 5 = 3 + 4i − 8 − 4i + 5 = 0 ✓ A1
Other root = 2 − i (conjugate of 2 + i) B1

Question 7 [3 marks]

Find the value of (1 + i)⁸, giving your answer as a real number.

|1+i| = √2, arg(1+i) = π/4 B1
(1+i)⁸ = (√2)⁸ · (cos(8π/4) + i·sin(8π/4)) M1
= 16 · (cos(2π) + i·sin(2π)) = 16 × 1 = 16 A1

Question 8 [6 marks]

The equation z⁴ − 2z³ + 6z² − 2z + 5 = 0 has a root z = i.
(a) Write down another root.
(b) Show that z² + 1 is a factor, then fully factorise the equation over the reals.
(c) Find all four roots.

(a) z = −i (conjugate) B1
(b) (z − i)(z + i) = z² + 1 is a factor. Divide: M1
z⁴ − 2z³ + 6z² − 2z + 5 = (z² + 1)(z² − 2z + 5) M1 A1
(c) z² + 1 = 0 → z = ±i A1
z² − 2z + 5 = 0 → z = (2 ± √(4−20))/2 = 1 ± 2i A1
All roots: z = i, z = −i, z = 1 + 2i, z = 1 − 2i

Past Paper Questions

Past Paper 1 — Cambridge 9709 P3 Style [4 marks]

Find the complex number w such that w + 2w* = 9 + 4i, where w* denotes the complex conjugate of w.

Let w = a + bi, so w* = a − bi M1
w + 2w* = (a + bi) + 2(a − bi) = 3a − bi M1
Real part: 3a = 9 → a = 3 A1
Imaginary part: −b = 4 → b = −4 A1
w = 3 − 4i

Past Paper 2 — Cambridge 9709 P3 Style [5 marks]

The polynomial p(z) = z³ + az + b, where a and b are real constants, has a root z = 2 + i.
Find the values of a and b, and state the third root.

Conjugate root 2 − i is also a root. B1
Quadratic factor: (z−(2+i))(z−(2−i)) = z² − 4z + 5 M1
p(z) = (z² − 4z + 5)(z + c) for some c M1
No z² term in p(z): c − 4 = 0 → c = 4. Third root z = −4. A1
Expand: a = coefficient of z = 5 + (−4)(4) = 5 − 16 = −11; b = 5×4 = 20 A1
a = −11, b = 20, third root = −4

Past Paper 3 — Cambridge 9709 P3 Style [4 marks]

Given that z = cosθ + i sinθ, show that z + z* = 2cosθ and z − z* = 2i sinθ. Hence find the real and imaginary parts of z².

z = cosθ + i sinθ, z* = cosθ − i sinθ B1
z + z* = 2cosθ ✓ and z − z* = 2i sinθ ✓ B1
z² = (cosθ + i sinθ)² = cos²θ − sin²θ + 2i sinθ cosθ M1
Re(z²) = cos²θ − sin²θ = cos(2θ)  |  Im(z²) = 2sinθcosθ = sin(2θ) A1

Past Paper 4 — Cambridge 9709 P3 Style [5 marks]

On an Argand diagram, the point A represents the complex number 4 + 2i and the point B represents 1 − i.
(a) Find |AB| where AB is treated as a complex number difference.
(b) Find arg(B − A) giving the answer in degrees to 1 d.p.

B − A = (1 − i) − (4 + 2i) = −3 − 3i M1
|B − A| = √(9 + 9) = √18 = 3√2 ≈ 4.243 A1
For −3 − 3i (Quadrant 3): reference angle = arctan(3/3) = 45° M1
arg(B − A) = −180° + 45° = −135° A1 A1

Past Paper 5 — Cambridge 9709 P3 Style [6 marks]

The complex number u = 1 + i√3.
(a) Find the modulus and argument of u.
(b) Find u³ in the form a + bi.
(c) Verify that u³ is real.

(a) |u| = √(1 + 3) = 2 B1
arg(u) = arctan(√3/1) = 60° = π/3 B1
(b) u³ = (2)³ · (cos(3 × 60°) + i·sin(3 × 60°)) M1
= 8(cos180° + i·sin180°) A1
= 8(−1 + 0·i) = −8 A1
(c) u³ = −8 has no imaginary part (Im(u³) = 0), so u³ is real ✓ B1