De Moivre's Theorem (DP IB Maths: AA HL)

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Amber

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10 November 2023

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De Moivre's Theorem

What is De Moivre’s Theorem?

De Moivre’s theorem can be used to find powers of complex numbers
It states that for $z = r cis θ$ , $z^{n} = [r (\cos θ + isin θ)]^{n} = r^{n} (\cos n θ + isin n θ)$
- Where
  - z ≠ 0
  - r is the modulus, |z|, r ∈ ℝ⁺
  - θ is the argument, arg z, θ ∈ ℝ
  - n ∈ ℝ
In Euler’s form this is simply:
- ${{(r e^{i θ})}^{n} = r}^{n} e^{i n θ}$
In words de Moivre’s theorem tells us to raise the modulus by the power of n and multiply the argument by n
In the formula booklet de Moivre’s theorem is given in both polar and Euler’s form:
- $[r (\cos θ + isin θ)]^{n} = r^{n} (\cos n θ + isin n θ) = r^{n} e^{i n θ} = r^{n} cis n θ$

How do I use de Moivre’s Theorem to raise a complex number to a power?

If a complex number is in Cartesian form you will need to convert it to either modulus-argument (polar) form or exponential (Euler’s) form first
- This allows de Moivre’s theorem to be used on the complex number
You may need to convert it back to Cartesian form afterwards
If a complex number is in the form $z = r (\cos (θ) - isin (θ))$ then you will need to rewrite it as $z = r (\cos (- θ) + isin (- θ))$ before applying de Moivre’s theorem
A useful case of de Moivre’s theorem allows us to easily find the reciprocal of a complex number:
- ^{$\frac{1}{z} = \frac{1}{r} (\cos (- θ) + isin (- θ) = \frac{1}{r} e^{- i θ}$}
- Using the trig identities cos(-θ) = cos(θ) and sin(-θ) = - sin(θ) gives
- $\frac{1}{z} = z^{- 1} = r^{- 1} [\cos (θ) - isin (θ)] = \frac{1}{r} [\cos (θ) - isin (θ)]$
In general
- $z^{- n} = r^{- n} [\cos (- n θ) + isin (- n θ)] = r^{- n} [\cos (n θ) - isin (n θ)]$

Examiner Tip

You may be asked to find all the powers of a complex number, this means there will be a repeating pattern
- This can happen if the modulus of the complex number is 1
- Keep an eye on your answers and look for the point at which they begin to repeat themselves

Worked example

Find the value of ${(\frac{\sqrt{3}}{6} + \frac{1}{6} i)}^{- 3}$ , giving your answer in the form a + bi.

o~JlLuvG_1-9-3-ib-aa-hl-de-moivres-theorem-we-solution-1

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Proof of De Moivre's Theorem

How is de Moivre’s Theorem proved?

When written in Euler’s form the proof of de Moivre’s theorem is easy to see:
- Using the index law of brackets: ${{(r e^{i θ})}^{n} = r}^{n} e^{i n θ}$
However Euler’s form cannot be used to prove de Moivre’s Theorem when it is in modulus-argument (polar) form
Proof by induction can be used to prove de Moivre’s Theorem for positive integers:
- To prove de Moivre’s Theorem for all positive integers, n
- $[r (\cos θ + isin θ)]^{n} = r^{n} (\cos n θ + isin n θ)$
STEP 1: Prove it is true for n = 1
- $[r (\cos θ + isin θ)]^{1} = r^{1} (\cos 1 θ + isin 1 θ) = r (\cos θ + isin θ)$
- So de Moivre’s Theorem is true for n = 1
STEP 2: Assume it is true for n = k
- $[r (\cos θ + isin θ)]^{k} = r^{k} (\cos k θ + isin k θ)$
STEP 3: Show it is true for n = k + 1
- $[r (\cos θ + isin θ)]^{k + 1} = ([r (\cos θ + isin θ)]^{k}) ([r (\cos θ + isin θ)]^{1})$
- According to the assumption this is equal to
  - $(r^{k} (\cos k θ + isin k θ)) (r (\cos θ + isin θ))$
- Using laws of indices and multiplying out the brackets:
  - $= r^{k + 1} [\cos k θ \cos θ + icos k θ \sin θ + isin k θ \cos θ + i^{2} \sin k θ \sin θ]$
- Letting i² = -1 and collecting the real and imaginary parts gives:
  - $= r^{k + 1} [\cos k θ \cos θ - \sin k θ \sin θ + i (\cos k θ \sin θ + \sin k θ \cos θ)]$
- Recognising that the real part is equivalent to cos(kθ + θ ) and the imaginary part is equivalent to sin(kθ + θ ) gives
  - ${(r cis θ)}^{k + 1} = r^{k + 1} [\cos (k + 1) θ + isin (k + 1) θ]$
- So de Moivre’s Theorem is true for n = k + 1
STEP 4: Write a conclusion to complete the proof
- The statement is true for n = 1, and if it is true for n = k it is also true for n = k + 1
- Therefore, by the principle of mathematical induction, the result is true for all positive integers, n
De Moivre’s Theorem works for all real values of n
- However you could only be asked to prove it is true for positive integers

Examiner Tip

Learning the standard proof for de Moivre's theorem will also help you to memorise the steps for proof by induction, another important topic for your AA HL exam

Worked example

Show, using proof by mathematical induction, that for a complex number z = r cisθ and for all positive integers, n,

$z^{n} = [r (\cos θ + isin θ)]^{n} = r^{n} (\cos n θ + isin n θ)$

1-9-3-ib-aa-hl-proof-of-de-moivres-theorem-we-solution

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Previous:1.9.3 Complex Roots of PolynomialsNext:1.9.5 Roots of Complex Numbers

De Moivre's Theorem (DP IB Maths: AA HL)

Revision Note

What is De Moivre’s Theorem?

How do I use de Moivre’s Theorem to raise a complex number to a power?

How is de Moivre’s Theorem proved?

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1. Number & Algebra

1.1 Number & Algebra Toolkit

1.1.1 Standard Form

1.1.2 Laws of Indices

1.1.3 Partial Fractions

1.2 Exponentials & Logs

1.2.1 Introduction to Logarithms

1.2.2 Laws of Logarithms

1.2.3 Solving Exponential Equations

1.3 Sequences & Series

1.3.1 Language of Sequences & Series

1.3.2 Arithmetic Sequences & Series

1.3.3 Geometric Sequences & Series

1.3.4 Applications of Sequences & Series

1.3.5 Compound Interest & Depreciation

1.4 Simple Proof & Reasoning

1.4.1 Proof

1.5 Further Proof & Reasoning

1.5.1 Proof by Induction

1.5.2 Proof by Contradiction

1.6 Binomial Theorem

1.6.1 Binomial Theorem

1.6.2 Extension of The Binomial Theorem

1.7 Permutations & Combinations

1.7.1 Counting Principles

1.7.2 Permutations & Combinations

1.8 Complex Numbers

1.8.1 Intro to Complex Numbers

1.8.2 Modulus & Argument

1.8.3 Introduction to Argand Diagrams

1.9 Further Complex Numbers

1.9.1 Geometry of Complex Numbers

1.9.2 Forms of Complex Numbers

1.9.3 Complex Roots of Polynomials

1.9.4 De Moivre's Theorem

1.9.5 Roots of Complex Numbers

1.10 Systems of Linear Equations

1.10.1 Systems of Linear Equations

1.10.2 Algebraic Solutions

2. Functions

2.1 Linear Functions & Graphs

2.1.1 Equations of a Straight Line

2.2 Quadratic Functions & Graphs

2.2.1 Quadratic Functions

2.2.2 Factorising & Completing the Square

2.2.3 Solving Quadratics

2.2.4 Quadratic Inequalities

2.2.5 Discriminants

2.3 Functions Toolkit

2.3.1 Language of Functions

2.3.2 Composite & Inverse Functions

2.3.3 Symmetry of Functions

2.3.4 Graphing Functions

2.4 Other Functions & Graphs

2.4.1 Exponential & Logarithmic Functions

2.4.2 Solving Equations

2.4.3 Modelling with Functions

2.5 Reciprocal & Rational Functions

2.5.1 Reciprocal & Rational Functions

2.6 Transformations of Graphs

2.6.1 Translations of Graphs

2.6.2 Reflections of Graphs

2.6.3 Stretches Graphs

2.6.4 Composite Transformations of Graphs

2.7 Polynomial Functions

2.7.1 Factor & Remainder Theorem

2.7.2 Polynomial Division

2.7.3 Polynomial Functions

2.7.4 Roots of Polynomials

2.8 Inequalities

2.8.1 Solving Inequalities Graphically

2.8.2 Polynomial Inequalities