De Moivre's Theorem And Its Applications

Q: What is De Moivre's Theorem?

De Moivre's Theorem states that (cos θ + i sin θ)^n = cos nθ + i sin nθ for any integer n. In Euler form, (e^(iθ))^n = e^(inθ). It's the essential tool for computing powers and roots of complex numbers.

Q: Does De Moivre's Theorem work for rational or fractional powers?

Yes, but with a caveat. If n = p/q (lowest terms), then (cos θ + i sin θ)^n has q distinct values, and cos nθ + i sin nθ is only one of them. To get all values, use (2kπ + θ)/n for k = 0, 1, ..., q−1.

Q: How do you find the nth roots of a complex number?

Convert to polar form z = r(cos θ + i sin θ). Then z^(1/n) = r^(1/n)[cos((2kπ+θ)/n) + i sin((2kπ+θ)/n)] for k = 0, 1, ..., n−1. This gives all n distinct roots.

Q: What are the nth roots of unity?

The nth roots of unity are the n solutions of x^n = 1: 1, α, α², ..., α^(n−1) where α = cos(2π/n) + i sin(2π/n). Geometrically, they form vertices of a regular n-gon inscribed in the unit circle.

Q: What are the cube roots of unity?

The cube roots of unity are the three solutions of z³ = 1: 1, ω = (−1 + i√3)/2, and ω² = (−1 − i√3)/2. They form an equilateral triangle on the unit circle.

Q: What is the value of 1 + ω + ω²?

1 + ω + ω² = 0. This follows from ω³ − 1 = (ω − 1)(ω² + ω + 1) = 0; since ω ≠ 1, the second factor must be zero. This is one of the most-used identities in JEE.

Q: Are ω and ω² conjugates of each other?

Yes. Since ω = (−1+i√3)/2 and ω² = (−1−i√3)/2 differ only in sign of the imaginary part, ω̄ = ω² and (ω²)̄ = ω.

Q: How to factorize x³ − 1 using cube roots of unity?

x³ − 1 = (x − 1)(x − ω)(x − ω²) = (x − 1)(x² + x + 1). Similarly, x³ + 1 = (x + 1)(x + ω)(x + ω²) = (x + 1)(x² − x + 1).

Q: What is 1 + ωⁿ + ω²ⁿ equal to?

If n is a multiple of 3: value is 3 (all three terms = 1). Otherwise: value is 0 (the three terms are 1, ω, ω² in some order).

MathsComplex NumbersFor JEE aspirants

De Moivre's Theorem states that for any integer $n$ , $(cos θ + i sin θ)^{n} = cos n θ + i sin n θ$ . It is the fundamental bridge between trigonometry and complex numbers, and powers three major applications: computing $n^{th}$ roots of complex numbers, deriving the $n^{th}$ roots of unity (including the JEE-favourite cube roots of unity $1, ω, ω^{2}$ ), and expanding multiple angles like $cos n θ$ and $sin n θ$ . This concept covers the theorem, its extension to rational exponents, and all key applications.

Key Formulas - Quick Reference

$(cos θ + i sin θ)^{n} = cos n θ + i sin n θ (n \in Z)$
$cos n θ = cos^{n} θ -^{n} C_{2} cos^{n - 2} θ sin^{2} θ + \dots$
$sin n θ =^{n} C_{1} cos^{n - 1} θ sin θ -^{n} C_{3} cos^{n - 3} θ sin^{3} θ + \dots$
$n^{th}$ roots of $z = r cis θ$ : $z^{1/ n} = r^{1/ n} cis \frac{2 k π + θ}{n}, k = 0, 1, \dots, n - 1$
$n^{th}$ roots of unity: $1, α, α^{2}, \dots, α^{n - 1}$ where $α = cis (2 π / n)$
Sum of $n^{th}$ roots of unity $= 0$ ; product $= - 1$ (n even) or $+ 1$ (n odd)
Cube roots of unity: $1, ω = \frac{- 1 + i 3}{2}, ω^{2} = \frac{- 1 - i 3}{2}$
$ω^{3} = 1$ and $1 + ω + ω^{2} = 0$
$1 + ω^{n} + ω^{2 n} = 3$ if $3 ∣ n$ ; else $= 0$

1. Statement of De Moivre's Theorem

Theorem If

n

is any integer, then:

(cos θ + i sin θ)^{n} = cos n θ + i sin n θ

In Euler form: $(e^{i θ})^{n} = e^{in θ}$ - the theorem is essentially the exponent law for $e^{i θ}$ .

2. Expansions of Multiple Angles

Writing the binomial expansion of $(cos θ + i sin θ)^{n}$ and equating real and imaginary parts:

Cos nθ formula

cos n θ = cos^{n} θ -^{n} C_{2} cos^{n - 2} θ sin^{2} θ +^{n} C_{4} cos^{n - 4} θ sin^{4} θ - \dots

Sin nθ formula

sin n θ =^{n} C_{1} cos^{n - 1} θ sin θ -^{n} C_{3} cos^{n - 3} θ sin^{3} θ +^{n} C_{5} cos^{n - 5} θ sin^{5} θ - \dots

Tan nθ formula (dividing)

tan n θ = \frac{^{n} C _{1} tan θ - ^{n} C _{3} tan ^{3} θ + ^{n} C _{5} tan ^{5} θ - \dots}{1 - ^{n} C _{2} tan ^{2} θ + ^{n} C _{4} tan ^{4} θ - \dots}

3. De Moivre's Theorem for Rational Exponents

n

is a rational number, one of the values of

(cos θ + i sin θ)^{n}

cos n θ + i sin n θ

. Specifically, let

n = p / q

where

p, q

are integers with

q > 0

and

g cd (p, q) = 1

. Then

(cos θ + i sin θ)^{n}

has

q

distinct values.

nth Roots of a Complex Number

If $z = r (cos θ + i sin θ)$ and $n$ is a positive integer, then:

z^{1/ n} = r^{1/ n} [cos \frac{2 k π + θ}{n} + i sin \frac{2 k π + θ}{n}], k = 0, 1, 2, \dots, n - 1

Note Any

n

consecutive values of

k

will serve - the roots are periodic mod

n

4. Application: nth Roots of Unity

One of the most important applications of De Moivre's Theorem is solving $x^{n} = 1$ .

Setting $x^{n} = 1 = cos 2 k π + i sin 2 k π$ (for integer $k$ ):

x = cos \frac{2 k π}{n} + i sin \frac{2 k π}{n}, k = 0, 1, 2, \dots, n - 1

Let $α = cos \frac{2 π}{n} + i sin \frac{2 π}{n}$ . Then by De Moivre's Theorem, the $k^{th}$ root is $α^{k}$ :

The n roots of unity

1, α, α^{2}, α^{3}, \dots, α^{n - 1}

Sum of the Roots

1 + α + α^{2} + \dots + α^{n - 1} = \frac{1 - α ^{n}}{1 - α} = 0 (∵ α^{n} = 1)

Consequently:

k = 0 \sum n - 1 cos \frac{2 k π}{n} = 0 and k = 0 \sum n - 1 sin \frac{2 k π}{n} = 0

The sum of the $n^{th}$ roots of unity is zero.

Product of the Roots

1 \cdot α \cdot α^{2} \dots α^{n - 1} = α^{n (n - 1) /2} = cos {π (n - 1)} + i sin {π (n - 1)}

Result

If $n$ is even: product $= - 1$
If $n$ is odd: product $= + 1$

Geometric Interpretation

Note The points representing

n

roots of unity are located at the vertices of a regular polygon of $n$ sides inscribed in a unit circle centred at the origin, with one vertex on the positive real axis.

Figure 6: The n-th roots of unity form a regular n-gon on the unit circle (shown here for n = 6)

5. Application: Cube Roots of Unity

Definition and Values (n = 3)

Definition The cube roots of unity are the three solutions of the equation

z^{3} = 1

. Using De Moivre's Theorem with

n = 3

z_{k} = cos \frac{2 k π}{3} + i sin \frac{2 k π}{3}, k = 0, 1, 2

Computing:

$k = 0$ : $z_{0} = cos 0 + i sin 0 = 1$
$k = 1$ : $z_{1} = cos \frac{2 π}{3} + i sin \frac{2 π}{3} = - \frac{1}{2} + i \frac{3}{2} = \frac{- 1 + i 3}{2}$
$k = 2$ : $z_{2} = cos \frac{4 π}{3} + i sin \frac{4 π}{3} = - \frac{1}{2} - i \frac{3}{2} = \frac{- 1 - i 3}{2}$

These are conventionally denoted $1$ , $ω$ , and $ω^{2}$ .

Geometric Interpretation

The three cube roots of unity lie at the vertices of an equilateral triangle inscribed in the unit circle, with circumcentre at the origin and circumradius

1

. One vertex is at

(1, 0)

; the other two are at angles

\pm 2 π /3

Figure 7: Cube roots of unity 1, ω, ω² form an equilateral triangle on the unit circle

Fundamental Properties of ω

Property 1 - Cube Identity

ω^{3} = 1

Therefore

ω^{3 k} = 1

for any integer

k

Property 2 - Sum Identity

1 + ω + ω^{2} = 0

This follows because

ω

satisfies

ω^{2} + ω + 1 = 0

(the factor of

ω^{3} - 1 = (ω - 1) (ω^{2} + ω + 1)

Property 3 - Conjugates

\overset{ω}{ˉ} = ω^{2}

and

\overset{ˉ}{ω^{2}} = ω

. So

ω

and

ω^{2}

are complex conjugates of each other.

Powers of ω

Using $ω^{3} = 1$ , the powers of $ω$ cycle with period 3:

$n (mod 3)$	$ω^{n}$
$0$	$1$
$1$	$ω$
$2$	$ω^{2}$

Sum Formula for Powers

1 + ω^{n} + ω^{2 n} = {3 if n is a multiple of 3 0 if n is not a multiple of 3

Factorizations Using ω

$x^{3} - 1 = (x - 1) (x - ω) (x - ω^{2}) = (x - 1) (x^{2} + x + 1)$
$x^{3} + 1 = (x + 1) (x + ω) (x + ω^{2}) = (x + 1) (x^{2} - x + 1)$
$x^{2} + x + 1 = (x - ω) (x - ω^{2})$
$x^{2} + x y + y^{2} = (x - ω y) (x - ω^{2} y)$
$x^{3} + y^{3} = (x + y) (x + ω y) (x + ω^{2} y)$
$x^{3} - y^{3} = (x - y) (x - ω y) (x - ω^{2} y)$

Solved Example 1

Given

z_{1} + z_{2} + z_{3} = A

z_{1} + z_{2} ω + z_{3} ω^{2} = B

z_{1} + z_{2} ω^{2} + z_{3} ω = C

(where

ω

is a cube root of unity), express

z_{1}, z_{2}, z_{3}

in terms of

A, B, C

Solution:

Finding $z_{1}$ : Add all three equations. The $z_{2}$ terms sum to $z_{2} (1 + ω + ω^{2}) = 0$ , similarly for $z_{3}$ :

A + B + C = 3 z_{1} \Rightarrow z_{1} = \frac{A + B + C}{3}

Finding $z_{2}$ : Multiply the equations by $1, ω^{2}, ω$ respectively and add:

A + B ω^{2} + C ω = 3 z_{2} \Rightarrow z_{2} = \frac{A + B ω ^{2} + C ω}{3}

Finding $z_{3}$ : Multiply by $1, ω, ω^{2}$ and add:

z_{3} = \frac{A + B ω + C ω ^{2}}{3}

Solved Example 2

ω

is a cube root of unity, find the value of

\frac{a + bω + c ω ^{2}}{b + c ω + a ω ^{2}} + \frac{a + bω + c ω ^{2}}{c + aω + b ω ^{2}}

Solution:

Multiply first term by $ω^{2} / ω^{2}$ , second by $ω / ω$ :

= \frac{ω ^{2} ( a + bω + c ω ^{2} )}{ω ^{2} ( b + c ω + a ω ^{2} )} + \frac{ω ( a + bω + c ω ^{2} )}{ω ( c + aω + b ω ^{2} )}

Numerator of first: $a ω^{2} + b ω^{3} + c ω^{4} = a ω^{2} + b + c ω$ (using $ω^{3} = 1, ω^{4} = ω$ ).

Denominator of first: $b ω^{2} + c ω^{3} + a ω^{4} = b ω^{2} + c + aω$ .

So first term $= \frac{a ω ^{2} + b + c ω}{ω ^{2} ( b + c ω + a ω ^{2} )} = \frac{1}{ω ^{2}}$ .

By symmetry, second term $= \frac{1}{ω}$ .

Total $= \frac{1}{ω ^{2}} + \frac{1}{ω} = ω + ω^{2} = - 1$ (using $1 + ω + ω^{2} = 0$ ).

Common Mistakes to Avoid

Watch out

Applying De Moivre's Theorem directly for non-integer $n$ . For rational $n = p / q$ , $(cos θ + i sin θ)^{n}$ has $q$ distinct values - $cos n θ + i sin n θ$ is only one of them.
Missing roots. The equation $z^{n} = w$ has exactly $n$ distinct roots - students find one and stop. Vary $k$ from $0$ to $n - 1$ .
Forgetting $ω^{3} = 1$ . Any $ω^{n}$ reduces via $n mod 3$ . Don't compute $ω^{100}$ directly - it's just $ω$ .
Confusing $1 + ω + ω^{2}$ vs $1 + ω^{n} + ω^{2 n}$ . The first is always $0$ . The second is $0$ only if $n$ is not a multiple of $3$ ; it's $3$ if $n$ is.
Writing $ω = e^{iπ /3}$ . Wrong! $ω = e^{i \cdot 2 π /3}$ . The angle is $2 π /3$ , not $π /3$ .
Assuming product of $n^{th}$ roots is $1$ . It's $- 1$ when $n$ is even, $+ 1$ when $n$ is odd.

Frequently Asked Questions

What is De Moivre's Theorem?

De Moivre's Theorem states that $(cos θ + i sin θ)^{n} = cos n θ + i sin n θ$ for any integer $n$ . In Euler form, $(e^{i θ})^{n} = e^{in θ}$ . It's the essential tool for computing powers and roots of complex numbers.

Does De Moivre's Theorem work for rational or fractional powers?

Yes, but with a caveat. If $n = p / q$ (lowest terms), then $(cos θ + i sin θ)^{n}$ has $q$ distinct values, and $cos n θ + i sin n θ$ is only one of them. To get all values, use $\frac{2 k π + θ}{n}$ for $k = 0, 1, \dots, q - 1$ .

How do you find the nth roots of a complex number?

Convert to polar form $z = r (cos θ + i sin θ)$ . Then $z^{1/ n} = r^{1/ n} [cos \frac{2 k π + θ}{n} + i sin \frac{2 k π + θ}{n}]$ for $k = 0, 1, \dots, n - 1$ . This gives all $n$ distinct roots.

What are the nth roots of unity?

The $n^{th}$ roots of unity are the $n$ solutions of $x^{n} = 1$ : $1, α, α^{2}, \dots, α^{n - 1}$ where $α = cos (2 π / n) + i sin (2 π / n)$ . Geometrically, they form vertices of a regular $n$ -gon inscribed in the unit circle.

What are the cube roots of unity?

The cube roots of unity are the three solutions of $z^{3} = 1$ : $1, ω = \frac{- 1 + i 3}{2}$ , and $ω^{2} = \frac{- 1 - i 3}{2}$ . They form an equilateral triangle on the unit circle.

What is the value of 1 + ω + ω²?

$1 + ω + ω^{2} = 0$ . This follows from $ω^{3} - 1 = (ω - 1) (ω^{2} + ω + 1) = 0$ ; since $ω \neq = 1$ , the second factor must be zero. This is one of the most-used identities in JEE.

Are ω and ω² conjugates of each other?

Yes. Since $ω = \frac{- 1 + i 3}{2}$ and $ω^{2} = \frac{- 1 - i 3}{2}$ differ only in sign of the imaginary part, $\overset{ω}{ˉ} = ω^{2}$ and $\overset{ˉ}{ω^{2}} = ω$ .

How to factorize x³ − 1 using cube roots of unity?

$x^{3} - 1 = (x - 1) (x - ω) (x - ω^{2}) = (x - 1) (x^{2} + x + 1)$ . Similarly, $x^{3} + 1 = (x + 1) (x + ω) (x + ω^{2}) = (x + 1) (x^{2} - x + 1)$ .

What is 1 + ωⁿ + ω²ⁿ equal to?

If $n$ is a multiple of $3$ : value is $3$ (all three terms $= 1$ ). Otherwise: value is $0$ (the three terms are $1, ω, ω^{2}$ in some order).

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