MathematicsChapter 1: Complex Numbers
De Moivre and Polar Inequations
De Moivre: [r (cos theta + i sin theta)]^n = r^n (cos n theta + i sin n theta) for integer n (and the school extension to roots).
Learning objectives
- State the main ideas and notation used in de moivre and polar inequations.
- Apply the rules of de moivre and polar inequations in a clear, step-by-step solution.
- Check signs, conditions, and the final answer for mathematical accuracy.
Definition
De Moivre: [r (cos theta + i sin theta)]n = rn (cos n theta + i sin n theta) for integer n (and the school extension to roots). nth roots: r1n} and theta + 2 pi kn for k = 0,1,...,n-1. Inequations: |z - z0| < r is the open disc centre z0 radius r. |z - z0| = r is the circle. Meaning: polar language turns powers into 'stretch and spin'.
Mathematical form
n = rn (cos n theta + i sin n theta) for in
k = 0
Where / symbols
- DE MOIVRE AND POLAR INEQUATIONS
- De Moivre: [r (cos theta + i sin theta)]^n = r^n (cos n theta + i sin n theta) for integer n (and the school extension to roots). nth roots: r^{1/n} and (theta + 2 pi k)/n for k = 0,1,...,n-1. Inequations: |z - z0| < r is the open disc centre z0 radius r. |z - z0| = r is the circle. Meaning: polar language turns powers into 'stretch and spin'.
- 1. DE MOIVRE
- Integer n can be negative: use 1/r^{|n|} and -|n| theta. To find z^3 = 8i, write 8i in polar, then take cube roots (three answers).
- 2. CIRCLES AND DISCS
- |z| < 2: open disc radius 2 at 0. |z-i| = 1: circle centre i. Re(z) > 0: right half-plane. Sketch first, then inequalities.
Relationship
- |z-z0| = r is a circle
Overview
Powers and roots, then regions:
De Moivre
Integer n can be negative: use 1r|n| and -|n| theta. To find z^3 = 8i, write 8i in polar, then take cube roots (three answers).
Examples (cis pi/6)^6 = cis pi = -1 · Three cube roots · r^n first
Circles and Discs
|z| < 2: open disc radius 2 at 0. |z-i| = 1: circle centre i. Re(z) > 0: right half-plane. Sketch first, then inequalities.
Examples |z-1|<=2 closed disc · |z|=1 unit circle · Arg z = pi/4 a ray
WHAT — DE MOIVRE AND POLAR INEQUATIONS
De Moivre: [r (cos theta + i sin theta)]n = rn (cos n theta + i sin n theta) for integer n (and the school extension to roots). nth roots: r1n} and theta + 2 pi kn for k = 0,1,...,n-1. Inequations: |z - z0| < r is the open disc centre z0 radius r. |z - z0| = r is the circle. Meaning: polar language turns powers into 'stretch and spin'.
RULE — 1. DE MOIVRE
Integer n can be negative: use 1r|n| and -|n| theta. To find z^3 = 8i, write 8i in polar, then take cube roots (three answers).
METHOD — 2. CIRCLES AND DISCS
|z| < 2: open disc radius 2 at 0. |z-i| = 1: circle centre i. Re(z) > 0: right half-plane. Sketch first, then inequalities.
WORKED EXAMPLE
Worked: |z - (1+i)| = 2. centre 1+i, which is the point (1,1) radius 2 a circle (not a disc: equality)
How to answer a de moivre and polar inequations question
- Identify the given quantities, notation, and required result.
- Choose the matching rule and write it before substitution.
- Work one algebraic step at a time, keeping brackets and signs clear.
- Check restrictions, simplify the result, and verify it where possible.
Worked example
- |z - (1+i)| = 2.
- centre 1+i, which is the point (1,1)
- radius 2
- a circle (not a disc: equality)
In simple words
Equality is the boundary. < is inside. Do not mix disc and circle.
Key points
- power: r^n and n theta
- |z-z0| = r is a circle
- n theta
- centre and radius
Exam preparation
Exam Focus
Quick revision
Revision Zone
Review the key facts, formulas and checks before you move on.De Moivre and Polar Inequations
Sticky Notes · Quick Revision
Core Idea
- De Moivre: [r (cos theta + i sin theta)]^n = r^n (cos n theta + i sin n theta) for integer n (and the school extension to roots).
- nth roots: r1n and theta + 2 pi kn for k = 0,1,...,n-1.
- Inequations: |z - z0| < r is the open disc centre z0 radius r.
1. DE MOIVRE
- Integer n can be negative: use 1r|n| and -|n| theta.
- To find z^3 = 8i, write 8i in polar, then take cube roots (three answers).
2. CIRCLES AND DISCS
- |z| < 2: open disc radius 2 at 0.
- |z-i| = 1: circle centre i.
- Re(z) > 0: right half-plane.
- Sketch first, then inequalities.
Worked Example
- Worked: |z - (1+i)| = 2.
- centre 1+i, which is the point (1,1)
- radius 2
- a circle (not a disc: equality)
Common Misconception
- Equality is the boundary. < is inside. Do not mix disc and circle.
FBISE Exam Check
- State the rule for de moivre and polar inequations before substituting values.
- Show the main algebra steps, keep exact values where possible, and check every condition.
60-second revision
- De Moivre: [r (cos theta + i sin theta)]^n = r^n (cos n theta + i sin n theta) for integer n (and the school extension to roots).
- nth roots: r1n and theta + 2 pi kn for k = 0,1,...,n-1.
- Inequations: |z - z0| < r is the open disc centre z0 radius r.
- Equality is the boundary. < is inside. Do not mix disc and circle.
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