Strand 3 — Number

Complex Numbers

5th Year · 6th Year (Leaving Cert)

  • By the end of this lesson students will be able to define and perform operations with complex numbers in the form a + bi.
  • By the end of this lesson students will be able to represent complex numbers geometrically on an Argand diagram.
  • By the end of this lesson students will be able to calculate the modulus and argument of a complex number and express it in polar form (HL).
  • By the end of this lesson students will be able to apply De Moivre's Theorem to find powers and roots of complex numbers (HL).
  • By the end of this lesson students will be able to solve problems involving roots of unity (HL).

Key concepts

Introduction to Complex Numbers (a + bi form)

A complex number is a number of the form z = a + bi, where 'a' and 'b' are real numbers, and 'i' is the imaginary unit, defined by i² = -1. 'a' is called the real part of z, denoted Re(z), and 'b' is called the imaginary part of z, denoted Im(z). Complex numbers extend the real number system, allowing us to find roots of negative numbers. Operations such as addition, subtraction, multiplication, and division are performed by treating 'i' as a variable, remembering that i² = -1. For division, we multiply the numerator and denominator by the complex conjugate of the denominator.

i² = -1 z = a + bi Re(z) = a Im(z) = b Complex conjugate: z̄ = a - bi
Argand Diagram

An Argand diagram is a graphical representation of complex numbers. The complex number z = a + bi is represented as a point (a, b) in a Cartesian coordinate system, where the horizontal axis represents the real part (Real axis) and the vertical axis represents the imaginary part (Imaginary axis). This plane is often referred to as the complex plane.

Modulus and Argument

The modulus of a complex number z = a + bi, denoted |z| or r, is the distance from the origin to the point (a, b) on the Argand diagram. It is always a non-negative real number. The argument of a complex number z, denoted arg(z) or θ, is the angle (in radians) that the line segment from the origin to the point (a, b) makes with the positive real axis, measured anti-clockwise. The principal argument is typically restricted to the interval (-π, π]. To find θ, we use tan θ = b/a and adjust for the quadrant in which the complex number lies.

Modulus: |z| = r = √(a² + b²) Argument: tan θ = b/a (adjust for quadrant) Principal argument: -π < θ ≤ π
Polar Form (HL)

A complex number z = a + bi can also be expressed in polar form (or modulus-argument form) using its modulus r and argument θ. From the Argand diagram, we can see that a = r cos θ and b = r sin θ. Substituting these into z = a + bi gives the polar form. This form is particularly useful for multiplication, division, powers, and roots of complex numbers.

z = r(cos θ + i sin θ)
De Moivre's Theorem (HL)

De Moivre's Theorem provides a formula for finding powers of complex numbers in polar form. For any integer n, the nth power of a complex number z = r(cos θ + i sin θ) is given by raising the modulus to the power of n and multiplying the argument by n. This theorem is also fundamental for finding the nth roots of complex numbers.

[r(cos θ + i sin θ)]ⁿ = rⁿ(cos nθ + i sin nθ) For roots: z^(1/n) = r^(1/n) [cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)] for k = 0, 1, ..., n-1
Roots of Unity (HL)

The nth roots of unity are the solutions to the equation zⁿ = 1. These roots can be found using De Moivre's Theorem by expressing 1 in polar form (1 = 1(cos(0) + i sin(0))). There are always n distinct nth roots of unity. Geometrically, these roots are equally spaced points on the unit circle in the Argand diagram, forming the vertices of a regular n-sided polygon. The sum of the nth roots of unity is always 0 (for n > 1), and their product is (-1)^(n-1).

zⁿ = 1 z_k = cos((2kπ)/n) + i sin((2kπ)/n) for k = 0, 1, ..., n-1

Key facts to remember

  • 1The imaginary unit i is defined by i² = -1.
  • 2A complex number z = a + bi has real part Re(z) = a and imaginary part Im(z) = b.
  • 3The complex conjugate of z = a + bi is z̄ = a - bi.
  • 4The modulus of z = a + bi is |z| = √(a² + b²).
  • 5The argument of z, arg(z) = θ, is found using tan θ = b/a, adjusted for the quadrant of z. The principal argument is in (-π, π].
  • 6The polar form of z is r(cos θ + i sin θ), where r = |z| and θ = arg(z).
  • 7De Moivre's Theorem states [r(cos θ + i sin θ)]ⁿ = rⁿ(cos nθ + i sin nθ) for any integer n.
  • 8To find the nth roots of a complex number w, use z_k = |w|^(1/n) [cos((arg(w) + 2kπ)/n) + i sin((arg(w) + 2kπ)/n)] for k = 0, 1, ..., n-1.

Worked examples

Example 1

Given z₁ = 2 + 3i and z₂ = 1 - i, calculate: (a) z₁ + z₂ (b) z₁z₂ (c) z₁/z₂ (d) The modulus and principal argument of z₁.

I(a) z₁ + z₂ = (2 + 3i) + (1 - i) = (2 + 1) + (3 - 1)i = 3 + 2i
II(b) z₁z₂ = (2 + 3i)(1 - i) = 2(1) + 2(-i) + 3i(1) + 3i(-i) = 2 - 2i + 3i - 3i² = 2 + i - 3(-1) = 2 + i + 3 = 5 + i
III(c) z₁/z₂ = (2 + 3i)/(1 - i) Multiply numerator and denominator by the conjugate of the denominator (1 + i): = [(2 + 3i)(1 + i)] / [(1 - i)(1 + i)] Numerator: (2)(1) + (2)(i) + (3i)(1) + (3i)(i) = 2 + 2i + 3i + 3i² = 2 + 5i - 3 = -1 + 5i Denominator: 1² - i² = 1 - (-1) = 1 + 1 = 2 So, z₁/z₂ = (-1 + 5i)/2 = -1/2 + 5/2 i
IV(d) For z₁ = 2 + 3i: Modulus |z₁| = √(2² + 3²) = √(4 + 9) = √13 Argument: tan θ = 3/2. Since Re(z₁) > 0 and Im(z₁) > 0, z₁ is in the first quadrant. θ = tan⁻¹(3/2) ≈ 0.9828 radians (to 4 decimal places) Principal argument of z₁ is 0.9828 radians.

Answer

(a) 3 + 2i (b) 5 + i (c) -1/2 + 5/2 i (d) |z₁| = √13, arg(z₁) ≈ 0.9828 radians

Always simplify i² to -1 in calculations.

Example 2

Express z = -1 + √3i in polar form. Hence, use De Moivre's Theorem to find z⁶ in the form a + bi.

IFirst, find the modulus r and argument θ for z = -1 + √3i.
IIr = |z| = √((-1)² + (√3)²) = √(1 + 3) = √4 = 2
IIITo find θ, we have tan θ = (√3)/(-1) = -√3. Since Re(z) = -1 < 0 and Im(z) = √3 > 0, z is in the second quadrant.
IVThe reference angle is tan⁻¹(√3) = π/3. In the second quadrant, θ = π - π/3 = 2π/3.
VSo, the polar form of z is 2(cos(2π/3) + i sin(2π/3)).
VINow, use De Moivre's Theorem to find z⁶:
VIIz⁶ = [2(cos(2π/3) + i sin(2π/3))]⁶
VIIIz⁶ = 2⁶(cos(6 * 2π/3) + i sin(6 * 2π/3))
9z⁶ = 64(cos(4π) + i sin(4π))
10Since cos(4π) = 1 and sin(4π) = 0 (as 4π is equivalent to 0 or 2π on the unit circle),
11z⁶ = 64(1 + 0i) = 64 + 0i
12Therefore, z⁶ = 64.

Answer

Polar form: 2(cos(2π/3) + i sin(2π/3)) z⁶ = 64

Remember to correctly identify the quadrant for the argument to ensure the principal argument is found.

Example 3

Find the three cube roots of 8i in the form a + bi.

ILet z³ = 8i. First, express 8i in polar form.
IIFor w = 8i, r = |8i| = √(0² + 8²) = 8.
IIIThe point (0, 8) is on the positive imaginary axis, so the argument θ = π/2.
IVThus, 8i = 8(cos(π/2) + i sin(π/2)).
VTo find the cube roots, we use the formula for roots: z_k = r^(1/n) [cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)] for k = 0, 1, 2.
VIHere, r = 8, n = 3, θ = π/2.
VIIz_k = 8^(1/3) [cos((π/2 + 2kπ)/3) + i sin((π/2 + 2kπ)/3)]
VIIIz_k = 2 [cos((π + 4kπ)/6) + i sin((π + 4kπ)/6)]
9For k = 0:
10z₀ = 2 [cos(π/6) + i sin(π/6)] = 2(√3/2 + i(1/2)) = √3 + i
11For k = 1:
12z₁ = 2 [cos((π + 4π)/6) + i sin((π + 4π)/6)] = 2 [cos(5π/6) + i sin(5π/6)] = 2(-√3/2 + i(1/2)) = -√3 + i
13For k = 2:
14z₂ = 2 [cos((π + 8π)/6) + i sin((π + 8π)/6)] = 2 [cos(9π/6) + i sin(9π/6)] = 2 [cos(3π/2) + i sin(3π/2)] = 2(0 + i(-1)) = -2i
15The three cube roots of 8i are √3 + i, -√3 + i, and -2i.

Answer

The three cube roots of 8i are √3 + i, -√3 + i, and -2i.

Always remember to add 2kπ to the argument before dividing by n when finding roots.

Common mistakes

  • Incorrectly determining the argument θ by not considering the quadrant of the complex number.
  • Forgetting to multiply by the complex conjugate when dividing complex numbers, or making errors in the multiplication.
  • Not including the '2kπ' term when applying De Moivre's Theorem to find roots, leading to only one root instead of n distinct roots.
  • Confusing the imaginary part 'b' with 'bi' (e.g., stating Im(3+4i) = 4i instead of 4).
  • Errors in trigonometric values for common angles (e.g., sin(π/2), cos(π), etc.) when converting to/from polar form.

Exam tips

  • Always sketch an Argand diagram for complex numbers, especially when finding the argument, to visually confirm the quadrant and avoid errors.
  • Ensure your calculator is in radian mode for HL questions involving arguments, unless the question specifically asks for degrees.
  • When finding roots, clearly list all values of k (from 0 to n-1) and calculate each root separately to ensure all distinct roots are found.
  • Show all steps clearly, particularly when converting between Cartesian and polar forms or applying De Moivre's Theorem, as partial marks are awarded for correct method.
  • Practise using the complex number functions on your calculator (if allowed) to check your answers, but always show full manual working in the exam.

Ready to practise?

Try a problem on this topic

Snap a photo or type a question — get step-by-step working instantly.