De Moivre's Theorem

De Moivre's Theorem is an important element in complex numbers. It helps raise complex numbers to higher powers and prove famous trigonometric identities.

(cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ)

de Moivre's Theorem is valid for any rational number - negative and positive integers as well as fractions.
Normally, it is proved by the mathematical induction method.

   It is true for any positive integer, n

[r(cos θ + i sinθ)]ⁿ = rⁿ[cos nθ + i sin nθ]
n=1
LHS: [r(cos θ + i sinθ)]¹ = r[cosθ + i sinθ]
RHS: r¹[(cos1θ + i sin1θ)] = r[cosθ + i sinθ]
Since LHS: = RHS:, de Moivre's Theorem is true for n=1.   
Assume that de Moivre's Theorme is true for n=k. So,
Z^k = r^k[cos kθ + i sin kθ]
Now. let's try it for n = k+1
[r(cos θ + i sinθ)]^k+1 = [r(cosθ + i sinθ)]^k[r(cos θ + i sin θ)]
= [r^k(cos kθ + i sin kθ)][r(cosθ + i sinθ)]
= [r^k+1(cos (k+1)θ + i sin (k+1)θ)]   

Therefore, if de Moivre's Theorem is true for n=k, it is true for n=k+1; Since it is true for n=1 too, de Moivre's Theorem is true for any integer, n>=1.

   n=0

LHS: [r(cos θ + i sinθ)]⁰ = 1
RHS: r⁰[(cos0θ + i sin 0θ)] = 1[1 + 0] = 1
Since LHS: = RHS:, de Moivre's Theorem is true when n=0.

   n is negative =>n<0

Let n=-m, where m is a positive number.
Since the theorem is true for positive integers,
LHS: [r(cos θ + i sinθ)]ⁿ = [r(cos θ + i sinθ)]^-m
= 1 / [r(cos θ + i sinθ)]^m
= 1 / [r(cos mθ + i msinθ)]   (since it is valid for positive integers)
X [cos mθ - i sin mθ] / [cos mθ - i sin mθ] =>
= [cos mθ - i sin mθ] / [(cos mθ - i sin mθ)r^m(cos mθ + i msinθ)] =>
= [cos mθ - i sin mθ] / r^m[cos²mθ - i² sin ²mθ]
= [cos mθ - i sin mθ] / r^m[cos²mθ + sin ²mθ]
= [cos mθ - i sin mθ] / r^m X 1
=r^-m[cos mθ - i sin mθ]
=r^-m[cos (-mθ) + i sin (-mθ)],    since cos-θ= cosθ and sin-θ=-sinθ
Since -m = n
rⁿ[cos nθ) + i sin nθ]
RHS:

   For fractions

Let n=p/q, where p and q are integers.
[r(cos θ + i sinθ)]^p = r^p[cos pθ + i sin pθ]
= r^p[cos pq(1/q)θ + i sin pq(1/q)θ]
= r^p[cos p(1/q)θ + i sin p(1/q)θ]^q
Now take the q^th root of both sides
=r^p/q[cos p(1/q)θ + i sin p(1/q)θ]
=r^p/q[cos (p/q)θ + i sin (p/q)θ]
= rⁿ[cos nθ + i sin nθ]
Therefore, deMoivre's Theorem is valid for fractions as well.

Derivation of Trigonometric Identities by de Moivre's Theorem

In order to derive the trigonometric identities, we are going to use the binomial expansion along with de Moivre Theorem.

(a + x)ⁿ = Σ ⁿC_r a^n-r x^r

E.g.1

Find the identities for sin2θ and cos2θ.

[cosθ + i sinθ]² = Σ ²C_r cosθ^2-r (i sinθ)^r
= cos²θ + 2i cosθsinθ + (isinθ)²
= cos²θ + 2i cosθsinθ - sin²θ
From de Moivre's Theorem,
[cosθ + i sinθ]² = cos2θ + i sin2θ
cos2θ + i sin2θ = cos²θ + 2i cosθsinθ - sin²θ
cos2θ + i sin2θ = cos²θ - sin²θ + 2i sinθcosθ
Comparing the real and imaginary parts on both sides,
cos2θ = cos²θ - sin²θ
sin2θ = 2 sinθ cosθ

cos2θ = cos²θ - sin²θ
sin2θ = 2 sinθ cosθ

E.g.2

Find an identity for cos5θ.

[cosθ + i sinθ]⁵ = Σ ⁵C_r cosθ^5-r (i sinθ)^r
= cos⁵θ + 5 cos⁴θ(i sinθ) + 10 cos³θ(i sinθ)² + 10 cos²θ(i sinθ)³ + 5 cosθ(i sinθ)⁴ + (i sinθ)⁵
= cos⁵θ + 5 cos⁴θ(i sinθ) + 10 cos³θ(-sin²θ) + 10 cos²θ(-i sin³θ) + 5 cosθ(sin⁴θ) + (i sin⁵θ)
From de Moivre't Theorem,
[cosθ + sinθ]⁵ = [cos5θ + i sin5θ]
So, [cos5θ + i sin5θ] = cos⁵θ + 5 cos⁴θ(i sinθ) + 10 cos³θ(-sin²θ) + 10 cos²θ(-i sin³θ) + 5 cosθ(sin⁴θ) + (i sin⁵θ)
Now equate the real parts on both sides:
cos5θ = cos⁵θ + 10 cos³θ(-sin²θ) + 5 cosθsin⁴θ
= cos⁵θ + 10 cos³θ(-sin²θ) + 5 cosθ(sin²θ)²
= cos⁵θ -10 cos³θsin²θ + 5 cosθ(1 - cos²θ)²
= cos⁵θ -10 cos³θ(1 - cos²θ) + 5 cosθ(1 -2cos²θ + cos⁴θ)
= cos⁵θ -10 cos³θ + 10 cos⁵θ + 5 cosθ -10cos³θ +5cos⁵θ
= 16cos⁵θ -20 cos³θ + 5 cosθ

cos 5θ = 16cos⁵θ -20 cos³θ + 5 cosθ

E.g.3

Find an identity for sin4θ.

[cosθ + i sinθ]⁴ = Σ ⁴C_r cosθ^4-r (i sinθ)^r
= cos⁴θ + 4 cos³θ(i sinθ) + 6 cos²θ(i sinθ)² + 4 cosθ(i sinθ)³ + (i sinθ)⁴
= cos⁴θ + 4 cos³θ(i sinθ) + 6 cos²θ(-sin²θ) + 4 cosθ(-i sin³θ) + (i sin⁴θ)
From de Moivre't Theorem,
[cosθ + sinθ]⁴ = [cos4θ + i sin4θ]
So, [cos4θ + i sin4θ] = = cos⁴θ + 4 cos³θ(i sinθ) + 6 cos²θ(-sin²θ) + 4 cosθ(-i sin³θ) + (i sin⁴θ)
Now equate the imaginary parts on both sides:
sin4θ = 4cos³θsinθ - 4 cosθ sin³θ + sin⁴θ

sin 4θ = 4cos³θsinθ - 4 cosθ sin³θ + sin⁴θ

Finding the roots of a number by de Moivre's Theorem

If you want to find ∛27, the calculator gives just one value - 3. There are, however, two more roots to it, which are complex numbers in conjugate form - a ± ib. Since a normal calculator does not provide us with these two values, we have to turn to an alternative way to find them out. de Moivre's Theorem comes to our rescue in this situation.

Let z³ = 27 = 27(1 + 0 x i) = 27(cos 0 + i sin 0)
In more generalized form,
z³ = 27(cos 2nπ + i sin 2nπ), where n can be 0, 1, 2 etc.
z = ∛27(cos 2nπ+ i sin 2nπ)^1/3
By de Moivre's Theorem,
z = ∛27(cos 2nπ/3 + i sin 2nπ/3)
z = 3(cos 2nπ/3 + i sin 2nπ/3)
Since there are three roots, let's substitute 0, 1 and 2 for n.
z₀ = 3(cos 0 + i sin 0) = 3 - the real root
z₁ = 3(cos 2π/3 + i sin 2π/3) = 3(-1/2 + i√3/2) = -3/2 + i√3/2
z₂ = 3(cos 4π/3 + i sin 4π/3) = 3(-1/2 + i√3/2) = -3/2 - i√3/2
∛27 = 3 or -3/2 ± i√3/2

Finding Cube Roots of Unity by de Moivre's Theorem

If you want to find ∛1, the calculator gives just one value - 1. There are, however, two more roots to it, which are complex numbers in conjugate form - a ± ib. Since a normal calculator does not provide us with these two values, we have to turn to an alternative way to find them out. de Moivre's Theorem comes to our rescue in this situation.

The roots are called cube roots of unity.
Let z³ = 1 = 1(1 + 0 x i) = 1(cos 0 + i sin 0)
In more generalized form,
z³ = 1(cos 2nπ + i sin 2nπ), where n can be 0, 1, 2 etc.
z = ∛1(cos 2nπ+ i sin 2nπ)^1/3
By de Moivre's Theorem,
z = ∛1(cos 2nπ/3 + i sin 2nπ/3)
z = 1(cos 2nπ/3 + i sin 2nπ/3)
Since there are three roots, let's substitute 0, 1 and 2 for n.
z₀ = 1(cos 0 + i sin 0) = 1 - the real root
z₁ = 1(cos 2π/3 + i sin 2π/3) = (-1/2 + i√3/2) = -1/2 + i√3/2
z₂ = 1(cos 4π/3 + i sin 4π/3) = (-1/2 + i√3/2) = -1/2 - i√3/2
∛1 = 1 or -1/2 ± i√3/2
You can see that the sum of cube roots of unity is zero:
z₀ + z₁ + z₂ = 0
The sum of n^th roots of unity is always zero, if n >= 2.

In the following diagram, the cube roots of unity are shown:

As you can see, they all lie in a circle of radius of one unit.