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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'.
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De Moivre and Polar Inequations
What is it?
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'.
In simple words
- 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.
Important terms
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'.
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.
What you should learn
- 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.
Examples
- (cis pi/6)6 = cis pi = -1
- Three cube roots
- r^n first
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).
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.
Recommended for Class 11
Full school-level explanation
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'.
Powers and roots, then regions:
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)
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).
Remember
- n theta
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.
Remember
- centre and radius
Examples
- |z-1|<=2 closed disc
- |z|=1 unit circle
- Arg z = pi/4 a ray
How to answer a de moivre and polar inequations question
How it works
- 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.
Remember
- power: r^n and n theta
- |z-z0| = r is a circle
- circle centre (1,1)
- = circle, < disc
- n theta
- centre and radius
- power: r^n and n theta
- |z-z0| = r is a circle
How the ideas connect
Worked: |z - (1+i)| = 2. centre 1+i, which is the point (1,1) radius 2 a circle (not a disc: equality)
Deeper understanding
Go beyond memorising De Moivre and Polar Inequations. Connect the definition to reasons, relationships, exceptions, and exam wording.
How the ideas connect
Worked: |z - (1+i)| = 2. centre 1+i, which is the point (1,1) radius 2 a circle (not a disc: equality)
Common mistakes
Exam focus
Core Idea
1. DE MOIVRE
2. CIRCLES AND DISCS
Worked Example
Exam focus
Quick revision
- power: r^n and n theta
- |z-z0| = r is a circle
- circle centre (1,1)
- = circle, < disc
- 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 th
- 1. DE MOIVRE: Integer n can be negative: use 1r|n| and -|n| theta. To find z^3 = 8i, write 8i in polar, then t
Equality is the boundary. < is inside. Do not mix disc and circle.
For long answers on De Moivre and Polar Inequations: define clearly → explain the main parts → give one example → state one exam caution.
Exam Focus
Exam Focus
Important points
- power: r^n and n theta
- |z-z0| = r is a circle
- circle centre (1,1)
- = circle, < disc
Important formulas
- power: r^n and n theta
- |z-z0| = r is a circle
- circle centre (1,1)
Common exam traps
- Equality is the boundary. < is inside. Do not mix disc and circle.
- Do not use a remembered formula before checking that its conditions fit the question.
Long question structure
Definition → key parts → example → exam caution
Common Mistakes
Common Mistakes
- Mistake
Equality is the boundary. < is inside. Do not mix disc and circle.
- Mistake
Do not use a remembered formula before checking that its conditions fit the question.
- Mistake
Do not hide sign changes or restrictions; write them on the line where they are used.
- Mistake
Equality is the boundary. < is inside. Do not mix disc and circle.
Revision Zone
60-Second Revision
Quick cues from this topic — use them for last-minute revision.
power: r^n and n theta
|z-z0| = r is a circle
circle centre (1,1)
= circle, < disc
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
|z-z0| = r is a circle
- 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
- Formula
- |z-z0| = r is a circle
- Example
- (cis pi/6)6 = cis pi = -1
- Critical warning
- Equality is the boundary. < is inside. Do not mix disc and circle.
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Exam Practice
Define
Q1. Define De Moivre and Polar Inequations.
Explain
Q2. Explain: DE MOIVRE.
Example
Q3. Give one clear example of De Moivre and Polar Inequations and say why it fits.
Mistake
Q4. State one common mistake students make about De Moivre and Polar Inequations.
Exam note
Q5. Write a short exam note on De Moivre and Polar Inequations: definition, two points, one example.
Sources & References
School-standard facts for Matric and Intermediate. Original explanations — not copied from textbooks.