# Mathematics

### 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,
Zk = rk[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)]
= [rk(cos kθ + i sin kθ)][r(cosθ + i sinθ)]
= [rk+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)rm(cos mθ + i msinθ)] =>
= [cos mθ - i sin mθ] / rm[cos2mθ - i2 sin 2mθ]
= [cos mθ - i sin mθ] / rm[cos2mθ + sin 2mθ]
= [cos mθ - i sin mθ] / rm 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 = rp[cos pθ + i sin pθ]
= rp[cos pq(1/q)θ + i sin pq(1/q)θ]
= rp[cos p(1/q)θ + i sin p(1/q)θ]q
Now take the qth root of both sides
=rp/q[cos p(1/q)θ + i sin p(1/q)θ]
=rp/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 nCr an-r xr

E.g.1

Find the identities for sin2θ and cos2θ.

[cosθ + i sinθ]2 = Σ 2Cr cosθ2-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 5Cr cosθ5-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 4Cr cosθ4-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 nth 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.  Maths is challenging; so is finding the right book. K A Stroud, in this book, cleverly managed to make all the major topics crystal clear with plenty of examples; popularity of the book speak for itself - 7th edition in print.

### Recommended - GCSE & iGCSE  This is the best book available for the new GCSE(9-1) specification and iGCSE: there are plenty of worked examples; a really good collection of problems for practising; every single topic is adequately covered; the topics are organized in a logical order.

### Recommended for A Level  This is the best book that can be recommended for the new A Level - Edexcel board: it covers every single topic in detail;lots of worked examples; ample problems for practising; beautifully and clearly presented.