Complex Numbers - Leaving Cert Maths Revision

Introduction

• Lecturer: Anya from Exam Revision.ie
• Topic: Complex numbers for Leaving Cert Maths (Higher and Ordinary levels)
• Exam Note: Complex numbers appear on Paper 1 for both higher and ordinary levels
• 2022 Exam Update: Important to be aware of topic choices due to changes in exam format

Topics Overview

• High-Level and Ordinary Level breakdown
• High-Level Paper 1: Complex numbers consistently appear in Section A
• Ordinary Level Paper 1: Frequently appears in Section A, more predictable placement
• Advice: Preparing for complex numbers is crucial as it is likely to be on the exam

Basics of Complex Numbers

• Definition: A number in the form a + bi
  • a: Real part
  • b: Imaginary part
• Junior Cycle Recap: Natural, rational, irrational, and real numbers
• Imaginary Unit (i): i = √(-1) and i² = -1
• Reason for Complex Numbers: Solving equations without real solutions (e.g., x² = -1)
• Key Uses:
  • Addition, subtraction, multiplication, division in rectangular form
  • Representations on Argand diagrams
  • Modulus (distance from the origin)
  • Complex conjugates (useful for division)

Ordinary Level Curriculum

• Knowledge Required:
  • Number systems inclusion of complex numbers
  • Operations in rectangular form: addition, multiplication, subtraction, division
  • Argand diagram illustration
  • Modulus calculation (distance from origin)
  • Complex conjugate calculation

Higher Level Curriculum

• Additional Requirements:
  • Applying conjugate root theorem
  • Converting between rectangular and polar forms
  • Using De Moivre’s Theorem
  • Proof by induction for De Moivre’s theorem
  • Finding nth roots of unity
  • Applications in trigonometry and polynomial roots

Ordinary Level Example Problem

Plotting Points on Argand Diagram

1. Plot Points: Given numbers and their positions on the Argand diagram (

Modulus Calculation

1. Example: |z₃| = √(1² + 5²) = √26
2. Investigate: |z₃| vs |z₁| + |z₂|
3. Conclusion: √26 ≠ 2√13

Division Using Conjugates

1. Example: (z₁ / z₂) using complex conjugate of the denominator
2. Calculation:

• Multiply numerator and denominator by the conjugate
  • Simplify to get -i

Polar Form and De Moivre’s Theorem

• Conversion to Polar Form:
  • Identify modulus |z| and argument θ
  • Use formula: r * (cosθ + i sinθ)
• Using De Moivre’s Theorem:
  • For expressions like (1 + i)^8, rewrite in polar form then apply the theorem to simplify
• Example Calculation: (1 + i)^8 = 16*

Conjugate Root Theorem

• Definition: If a + bi is a root, then a - bi is also a root for polynomials with real coefficients
• Example: Given one root, automatically know the conjugate root

General Polar Form for Roots

• Formula: z = r^1/n [cos(θ/n + 2kπ/n) + i sin(θ/n + 2kπ/n)] for multiple roots
• Example Process: Given w² = v, find values by converting to polar, then general polar form, then using De Moivre’s theorem

Unity Roots

• Equation: To find nth roots, e.g., cube roots of unity solve z^3 = 1
• Universal Setup: z = e^(2kπi/3) for k = 0, 1, 2

Complex Division and Conjugates

• Method: Multiply by conjugate to simplify expressions
• Example: a + bi equals sqrt of a complex number by equating reals and imaginaries

Conclusion

• Additional Resources: Students encouraged to try Ordinary Level questions for practice
• Discount Code: 20% off with code