De Moivre's Theorem

In the field of complex numbers, De Moivre’s Theorem is one of the most important and useful theorems which connects complex numbers and trigonometry. It is also helpful for obtaining relationships between trigonometric functions of multiple angles. De Moivre’s Theorem is also known as “De Moivre’s Identity” and “De Moivre’s Formula”. The name of the theorem is after the name of the great Mathematician De Moivre, who made many contributions to the field of Mathematics, mainly in the areas of the theory of probability and algebra.

Table of Contents

• Formula
• Proof
• Uses
• Problems

De Moivre’s Formula

Mathematical Statement: For any real number x, we have

(cos x + i sin x)ⁿ = cos(nx) + i sin(nx)

OR

Where n is a positive integer and “ i “ is the imaginary part, and i = √(-1). Also, assume i² = -1.

Remark: The result can be shown true when n is a negative integer and even n is a rational number.

Also, Read =>

• Complex Numbers
• Inverse Trigonometric Functions

Aspirants are advised, before starting this section, to revise and get familiar with the argand diagram and polar form of complex numbers. This section introduces De Moivre’s theorem statement, proof of the theorem and some of its consequences.

De Moivre’s Theorem Proof

Apply mathematical induction to prove De Moivre’s theorem.

We know, (cos x + i sin x)ⁿ = cos(nx) + i sin(nx) …(i)

Step 1: For n = 1, we have

(cos x + i sin x)¹ = cos(1x) + i sin(1x) = cos(x) + i sin(x)

Which is true.

Step 2: Assume that formula is true for n = k.

(cos x + i sin x)^k = cos(kx) + i sin(kx) ….(ii)

Step 3: Prove that the result is true for n = k + 1.

(cos x + i sin x)^k+1 = (cos x + i sin x)^k (cos x + i sin x)

= (cos (kx) + i sin (kx)) (cos x + i sin x) [Using (i)]

= cos (kx) cos x − sin(kx) sinx + i (sin(kx) cosx + cos(kx) sinx)

= cos {(k+1)x} + i sin {(k+1)x}

=> (cos x + i sin x)^k+1 = cos {(k+1)x} + i sin {(k+1)x}

Hence, the result is proven.

Since the theorem is true for n = 1 and n = k + 1, it is true ∀ n ≥ 1.

Uses of De Moivre’s Theorem

To find the roots of complex numbers

If z is a complex number, and z = r(cos x + i sin x) [In polar form]

Then, the nth roots of z are:

Where k = 0, 1, 2,….., (n − 1)

If k = 0, the above formula reduces to

To obtain relationships between powers of trigonometric functions and trigonometric angles.

We use the polar form of complex numbers to represent a complex number using trigonometry.

The parameters of the rectangular and polar forms are as follows:

a = r cos x and b = r sin x

Where

This implies, z = a + ib = r(cos x + i sin x).

Raising to a Power

Example: Evaluate ( 1 + i )¹⁰⁰⁰.

Solution:

Let z = 1 + i

We have to represent z in the form of r(cos θ + i sin θ).

Here,

Argument = θ = arc(tan (1/1) = arc tan(1) = π/4

Absolute value = r

Applying DeMoivre’s theorem, we get

z¹⁰⁰⁰ = [√2{cos(π/4) + i sin(π/4)}]¹⁰⁰⁰

= 2¹⁰⁰⁰ {cos(1000π/4) + i sin(1000π/4)}

= 2¹⁰⁰⁰ {1 + i (0)}

= 2¹⁰⁰⁰

Problems on De Moivre’s Identity

Problem 1: Evaluate (2 + 2i)⁶

Solution: Let z = 2 + 2i

Here, r = 2√2 and θ = 45 degrees

Since z lies in the first quadrant, sinθ and cosθ functions are positive.

Applying De Moivre’s Theorem,

z⁶ = (2 + 2i)⁶ = (2√2)⁶ [cos 45⁰ + i sin 45⁰]⁶

= (2√2)⁶ [cos 270⁰ + i sin 270⁰]⁶

= – 512i

Problem 2: Express five fifth‐roots of (√3 + i) in trigonometric form.

Solution:

We know, z = a + ib = r(cos x + i sin x)

Where

So,

Here, r = 2 and θ = 30 degrees

Therefore, z = 2[cos(30⁰ + 360⁰ k) + i sin cos(30⁰ + 360⁰ k)]

Applying the nth root theorem,

z^1/5 = {2[cos(30⁰ + 360⁰ k) + i sin cos(30⁰ + 360⁰ k)]}^1/5

= 2^1/5 [cos((30⁰ + 360⁰ k)/5) + i sin cos((30⁰ + 360⁰ k)/5)] …(1)

Where k = 0,1,2,3,4

At k = 0; (1)=> z₁ = 2^1/5 [cos 6⁰ + i sin 6⁰]

At k = 1; (1)=> z₁ = 2^1/5 [cos 78⁰ + i sin 78⁰]

At k = 2; (1)=> z₁ = 2^1/5 [cos 150⁰ + i sin 150⁰]

At k = 3; (1)=> z₁ = 2^1/5 [cos 222⁰ + i sin 222⁰]

At k = 4; (1)=> z₁ = 2^1/5 [cos 294⁰ + i sin 294⁰]

Problem 3: Express

Solution:

= (cos 4θ + i sin 4θ ) / (cos 4θ – i sin 4θ )

By rationalising the fraction, we have

= (cos 4θ + i sin 4θ )² / (cos² 4θ – i sin² 4θ )

= cos 8θ + i sin 8θ

Problem 4: If

Solution:

Problem 5: The value of

Solution: 

Problem 6: If

Solution: 

De Moivre’s Theorem – Video Lesson