de Moivre’s Theorem and its Applications

de Moivre’s Theorem and its Applications

Abraham de Moivre (1667–1754) was one of the mathematicians to use complex numbers in trigonometry.

The formula (cosθ + i sinθ )ⁿ = (cos nθ + i sin nθ ) known by his name, was instrumental in bringing trigonometry out of the realm of geometry and into that of analysis.

1. de Moivre's Theorem

de Moivre’s Theorem 

Given any complex number cosθ + i sinθ and any integer n,

(cosθ + i sinθ )n = cos nθ + i sin nθ .

Corollary

(1) (cosθ - i sinθ )n  = cos nθ - i sin nθ 

(2) (cosθ + i sinθ )-n  = cos nθ - i sin nθ 

(3) (cosθ - i sinθ )-n = cos nθ + i sin nθ 

(4) sinθ + i cosθ = i (cosθ - i sinθ ) .

Now let us apply de Moivre’s theorem to simplify complex numbers and to find solution of equations.

Example 2.28

If z = (cosθ + i sinθ ) , show that zⁿ + 1/ zⁿ = 2 cos nθ and zⁿ – [1/ zⁿ] = 2i sin nθ .

Solution

Let z = (cosθ + i sinθ ) . 

By de Moivre’s theorem ,

zⁿ = (cosθ + i sinθ )ⁿ = cos nθ + i sin nθ 

Example 2.29

Similarly, 

Solution

Example 2.30

Solution

Example 2.31

Simplify 

(i) (1+ i)¹⁸

(ii) (-√3 + 3i)³¹ .

Solution

(i) (1+ i)¹⁸

Let 1+ i = r (cosθ + i sinθ ) . Then, we get

(ii) (-√3 + 3i)³¹ .

Let -√3 + 3i = r (cosθ + i sinθ ) . Then, we get

Raising power 31 on both sides,

2. Finding n^th roots of a complex number

de Moivre’s formula can be used to obtain roots of complex numbers. Suppose n is a positive integer and a complex number ω is n th root of z denoted by z^{1/ n} , then we have

ωⁿ = z                     …………(1)

Let ω = ρ (cosϕ + i sinϕ ) and

z = r (cosθ + i sinθ ) = r (cos(θ + 2kπ ) + i sin (θ + 2kπ )),   k ∈ Z

Since w is the n^th root of z , then

ωⁿ = z

⇒ ρⁿ (cosϕ + i sinϕ)ⁿ = r (cos(θ + 2kπ ) + i sin (θ + 2kπ )) ,   k ∈ Z

By de Moivre’s theorem,

⇒ ρⁿ (cosnϕ + i sinnϕ) = r (cos (θ + 2kπ ) + i sin (θ + 2kπ )), k ∈ Z

Comparing the moduli and arguments, we get

Although there are infinitely many values of  k , the distinct values of  ω  are obtained when k = 0,1, 2, 3,K, n −1. When k = n, n +1, n + 2,K we get the same roots at regular intervals (cyclically). Therefore the nth roots of complex number z = r (cosθ + i sinθ ) are

If we set ω =  the formula for the  n ^th   roots of  a complex number has a nice geometric interpretation, as shown in Figure. Note that because | ω | = ⁿ√r the n roots all have the same modulus ⁿ√r they all lie on a circle of radius ⁿ√r with centre at the origin. Furthermore, the n roots are equally spaced along the circle, because successive n roots have arguments that differ by 2π/n .

Remark

(i) General form of de Moivre's Theorem

If x is rational, then cos xθ + i sin xθ is one of the values of (cosθ + i sinθ )^x .

(ii) Polar form of unit circle

Let z  = e^iθ  = cosθ + i sinθ . Then, we get

|z|² = |cosθ + i sinθ|2

⇒  | x + iy|² = cos²θ + sin²θ = 1

⇒ x² + y² = 1.

Therefore, |z| = 1 represents a unit circle (radius one) centre at the origin. 

3. The nth  roots of unity

The solutions of the equation zⁿ = 1 , for positive values of integer n , are the n roots of the unity.

In polar form the equation zⁿ = 1 can be written as

zⁿ = cos (0 + 2kπ) + i sin (0 + 2kπ) = e^i2kπ , k = 0, 1, 2,….. 

Using deMoivre’s theorem, we find the n^th roots of unity from the equation given below:

Given a positive integer n , a complex number z is called an n ^th root of unity if and only if zⁿ = 1.

If we denote the complex number by ω , then

Therefore ω is an n^th root of unity. From equation (1), the complex numbers 1,ω,ω² ,... ...,ω^n-1 are n^th roots of unity. The complex numbers 1,ω,ω² ,... ...,ω^n-1 are the points in the complex plane and are the vertices of a regular polygon of n sides inscribed in a unit circle as shown in Fig 2.45. Note that because the n th roots all have the same modulus 1, they will lie on a circle of radius 1 with centre at the origin. Furthermore, the n roots are equally spaced along the circle, because successive n th roots have arguments that differ by 2π/n .

The n^th roots of unity 1,ω,ω² ,... ...,ω^n-1 are in geometric progression with common ratio ω 

Therefore 1+ ω + ω² +… + ω^n-1 = 1- ωⁿ / 1- ω = 0  since ωⁿ = 1 and ω ≠ 1 .

The sum of all the nth roots of unity is

 1+ ω + ω² +… + ω^n-1 = 0

The product of n, n^th roots of unit is

1ωω² ... ...ω^n-1 = ω^{0+1+2+3+... ...+(n-1)} = ω^[(n-1)n]/2

The product of all the n^th roots of unity is

1ωω² ... ...ω^n-1 = (-1)^n-1

Note

(1) All the n roots of n^th roots unity are in Geometrical Progression

(2) Sum of the n roots of n^th roots unity is always equal to zero.

(3) Product of the n roots of n^th roots unity is equal to (-1)^n-1 .

(4) All the n roots of n^th roots unity lie on the circumference of a circle whose centre is at the origin and radius equal to 1 and these roots divide the circle into n equal parts and form a polygon of n sides.

Example 2.32

Find the cube roots of unity. 

Solution

We have to find 1^1/3 . Let z = 1^1/3 then z³ = 1.

In polar form, the equation z³ = 1 can be written as

z³ = cos(0 + 2kπ) + i sin(0 + 2kπ) = e^i2kπ , k = 0, 1, 2,... 

Example 2.33

Find the fourth roots of unity.

Solution

We have to find 1^1/4. Let z =1^1/4 . Then z⁴ = 1 .

In polar form, the equation z⁴ = 1 can be written as

z⁴ = cos (0 + 2kπ ) + i sin (0 + 2kπ ) = e^i2kπ , k = 0, 1, 2,...

Note

(i) In this chapter the letter ω is used for n^th roots of unity. Therefore the value of ω is depending on n as shown in following table.

(ii) The complex number ze^iθ is a rotation of z by θ radians in the counter clockwise direction about the origin.

Example 2.34

Solve the equation z³ + 8i = 0 , where z ∈ C. 

Solution

Let z³ + 8i = 0 . Then, we get

z³ = -8i

Example 2.35

Find all cube roots of √3 + i 

Solution

We have to find (√3 + i)^1/3. Let z³ = √3 + i = r (cosθ + i sinθ )

Example 2.36

Suppose z₁ , z₂, and z₃ are the vertices of an equilateral triangle inscribed in the circle |z| = 2. If z₁ = 1+ i√3 , then find z₂ and z₃.

Solution

|z| = 2 represents the circle with centre (0, 0) and radius 2.

Let A, B, and C be the vertices of the given triangle. Since the vertices z₁ , z₂ , and z₃ form an equilateral triangle inscribed in the circle |z| = 2 , the sides of this triangle AB, BC, and CA subtend 2π/3 radians (120 degree) at the origin (circumcenter of the triangle).

(The complex number z e^iθ is a rotation of z by θ radians in the counter clockwise direction about the origin.)

Therefore, we can obtain z₂ and z₃ by the rotation of z₁ by 2π/3 and 4 π/3 respectively.

Given that 

Therefore, z₂ = -2, and z₃ = 1- i√3.