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.

**de Moivre’s Theorem**

Given any complex number **cos***θ *+ *i *sin*θ *and any integer *n*,

**(cos***θ *+ *i *sin*θ ***)**n** **=

**Corollary**

(1) (cos*θ *- *i *sin*θ *)n * *= cos *nθ *-
*i *sin *nθ*

(2) (cos*θ *+ *i *sin*θ *)-n * *= cos *nθ *-
*i *sin *nθ*

(3) (cos*θ *- *i *sin*θ *)-n* *= cos *nθ *+
*i *sin *nθ*

(4) sin*θ *+ *i *cos*θ *= *i *(cos*θ *-
*i *sin*θ *) .

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 z^{n} + 1/ z^{n} = 2 cos *nθ *and z^{n} – [1/ z^{n}] = 2*i* sin *nθ*
.

**Solution**

Let *z = (cosθ + i sinθ ) . *

By de Moivre’s theorem ,

*z*^{n}* = (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 + 3*i*)^{31} .

**Solution**

**(i) (1+ i)**^{18}

Let 1+ *i* = *r *(cos*θ* + *i *sin*θ* ) .
Then, we get

**(ii) (-√3 + 3 i)**

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

Raising power 31 on both sides,

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

*ω*^{n} = z
…………(1)

Let *ω *= *ρ *(cos*ϕ
*+ *i *sin*ϕ *) and

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

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

*ω*^{n}* *= *z*

⇒ *ρ*^{n}* *(cos*ϕ *+ *i *sin*ϕ*)^{n}_{ }= *r *(cos(*θ *+
2*kπ *) + *i *sin (*θ *+ 2*kπ *)) , *k* ∈ Z

By de Moivre’s theorem,

⇒ *ρ*^{n}* *(cos*nϕ *+ *i *sin*nϕ*)_{ }= *r* (cos (θ + 2*kπ*
) + *i* sin (*θ* + 2*kπ* )), 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 *(cos*θ
*+ *i *sin*θ *) 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 *x**θ** *+ *i *sin *x**θ *is one of the values
of (cos*θ** *+ *i *sin*θ** *)^{x}* *.

**(ii) Polar form of unit circle**

Let *z *= *e*^{i}^{θ}* *= cos*θ** *+ *i *sin*θ** *. Then, we get

|z|^{2} = |cosθ + *i* sinθ|2

⇒ | *x* + *iy*|^{2} = cos^{2}θ + sin^{2}*θ* = 1

⇒ x^{2} + y^{2} = 1.

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

The solutions of the equation *z*^{n}* *= 1 , for positive values
of integer *n *, are the *n *roots of the unity.

In polar form the equation *z*^{n}* *= 1 can be written as

z^{n} = 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 *z*^{n}* *= 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,ω^{2} ,... ...,ω^{n-1} are n^{th} roots of unity. The complex numbers 1,ω,ω^{2} ,... ...,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,ω^{2} ,... ...,ω^{n-1 }are in geometric progression with common ratio
ω

Therefore 1+ ω + ω^{2} +… + ω^{n-1} = 1- ω^{n} / 1- ω = 0 since ω^{n} = 1 and ω ≠ 1 .

The sum of all the nth roots of unity is

1+ ω + ω^{2} +… + ω^{n-1} = 0

The product of *n*, *n*^{th} roots of unit is

1ωω^{2} ... ...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

1ωω^{2} ... ...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 z^{3} = 1.

In polar form, the equation z^{3} = 1 can be written as

z^{3} = cos(0 + 2kπ) + *i* sin(0 + 2*k*π) = 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 z^{4} = 1 .

In polar form, the equation z^{4} = 1 can be written as

z^{4} = 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 z^{3} + 8*i* = 0 , where z ∈ **C**.

Let z^{3} + 8*i* = 0 . Then, we get

z^{3} = -8*i*

Find all cube roots of √3 + *i *

**Solution**

We have to find (√3 + *i*)^{1/3}. Let z^{3} = √3 + *i* = r (cosθ + *i* sinθ )

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

|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 z_{1} , z_{2} , and z_{3} 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 z_{2} and z_{3} by the rotation of z_{1} by 2π/3 and 4 π/3
respectively.

Given that

Therefore, z_{2} = -2, and z_{3} = 1- *i*√3.

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