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Definition, Formula, Solved Example Problems - de Moivre’s Theorem and its Applications | 12th Mathematics : UNIT 2 : Complex Numbers

Chapter: 12th Mathematics : UNIT 2 : Complex Numbers

de Moivre’s Theorem and its Applications

de Moivre’s formula can be used to obtain roots of complex numbers.

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 - i sin  

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

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

(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 and zn – [1/ zn] = 2i sin .

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 nth 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 z1/ n , then we have

ωn = z                     …………(1)

Let ω = ρ (cosϕ + i sinϕ ) and

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

Since w is the nth root of z , then

ωn = z

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

By de Moivre’s theorem,

ρn (cos+ i sin) = r (cos (θ + 2 ) + i sin (θ + 2 )), 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 | = nr the n roots all have the same modulus nr they all lie on a circle of radius nr 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  = eiθ  = 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) = ei2kπ , k = 0, 1, 2,….. 

Using deMoivre’s theorem, we find the nth 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 nth root of unity. From equation (1), the complex numbers 1,ω,ω2 ,... ...,ωn-1 are nth roots of unity. The complex numbers 1,ω,ω2 ,... ...,ωn-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 nth roots of unity 1,ω,ω2 ,... ...,ωn-1 are in geometric progression with common ratio ω 

Therefore 1+ ω + ω2 +… + ωn-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 +… + ωn-1 = 0

The product of n, nth roots of unit is

1ωω2 ... ...ωn-1 = ω0+1+2+3+... ...+(n-1) = ω[(n-1)n]/2

The product of all the nth roots of unity is

1ωω2 ... ...ωn-1 = (-1)n-1


Note

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

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

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

(4) All the n roots of nth 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 11/3 . Let z = 11/3 then z3 = 1.

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

z3 = cos(0 + 2kπ) + i sin(0 + 2kπ) = ei2kπ , k = 0, 1, 2,... 


 

Example 2.33

Find the fourth roots of unity.

Solution

We have to find 11/4. Let z =11/4 . Then z4 = 1 .

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

z4 = cos (0 + 2kπ ) + i sin (0 + 2kπ ) = ei2kπ , k = 0, 1, 2,...


Note

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


(ii) The complex number ze 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 +

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 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.

 

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12th Mathematics : UNIT 2 : Complex Numbers


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