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=1

Since

Assume that de Moivre's Theorme is true for

Z

Now. let's try it for n = k+1

[r(cos θ + i sinθ)]

= [r

= [r

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

Since

n is negative =>n<0

Let n=-m, where m is a positive number.Since the theorem is true for positive integers,

= 1 / [r(cos θ + i sinθ)]

= 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

= [cos mθ - i sin mθ] / r

= [cos mθ - i sin mθ] / r

= [cos mθ - i sin mθ] / r

=r

=r

Since -m = n

r

For fractions

Let n=p/q, where p and q are integers.[r(cos θ + i sinθ)]

= r

= r

Now take the q

=r

=r

= r

Therefore, deMoivre's Theorem is valid for fractions as well.

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

(a + x)^{n} = Σ ^{n}C_{r} a^{n-r} x^{r}

**E.g.1**

Find the identities for sin2θ and cos2θ.

[cosθ + i sinθ]^{2} = Σ ^{2}C_{r} cosθ^{2-r} (i sinθ)^{r}

= cos^{2}θ + 2i cosθsinθ + (isinθ)^{2}

= cos^{2}θ + 2i cosθsinθ - sin^{2}θ

From de Moivre's Theorem,

[cosθ + i sinθ]^{2} = cos2θ + i sin2θ

cos2θ + i sin2θ = cos^{2}θ + 2i cosθsinθ - sin^{2}θ

cos2θ + i sin2θ = cos^{2}θ - sin^{2}θ + 2i sinθcosθ

Comparing the real and imaginary parts on both sides,

cos2θ = cos^{2}θ - sin^{2}θ

sin2θ = 2 sinθ cosθ

** cos2θ = cos ^{2}θ - sin^{2}θ **

sin2θ = 2 sinθ cosθ

**E.g.2**

Find an identity for cos5θ.

[cosθ + i sinθ]^{5} = Σ ^{5}C_{r} cosθ^{5-r} (i sinθ)^{r}

= cos^{5}θ + 5 cos^{4}θ(i sinθ) + 10 cos^{3}θ(i sinθ)^{2} + 10 cos^{2}θ(i sinθ)^{3} + 5 cosθ(i sinθ)^{4} + (i sinθ)^{5}

= cos^{5}θ + 5 cos^{4}θ(i sinθ) + 10 cos^{3}θ(-sin^{2}θ) + 10 cos^{2}θ(-i sin^{3}θ) + 5 cosθ(sin^{4}θ) + (i sin^{5}θ)

From de Moivre't Theorem,

[cosθ + sinθ]^{5} = [cos5θ + i sin5θ]

So, [cos5θ + i sin5θ] = cos^{5}θ + 5 cos^{4}θ(i sinθ) + 10 cos^{3}θ(-sin^{2}θ) + 10 cos^{2}θ(-i sin^{3}θ) + 5 cosθ(sin^{4}θ) + (i sin^{5}θ)

Now equate the **real** parts on both sides:

cos5θ = cos^{5}θ + 10 cos^{3}θ(-sin^{2}θ) + 5 cosθsin^{4}θ

= cos^{5}θ + 10 cos^{3}θ(-sin^{2}θ) + 5 cosθ(sin^{2}θ)^{2}

= cos^{5}θ -10 cos^{3}θsin^{2}θ + 5 cosθ(1 - cos^{2}θ)^{2}

= cos^{5}θ -10 cos^{3}θ(1 - cos^{2}θ) + 5 cosθ(1 -2cos^{2}θ + cos^{4}θ)

= cos^{5}θ -10 cos^{3}θ + 10 cos^{5}θ + 5 cosθ -10cos^{3}θ +5cos^{5}θ

= 16cos^{5}θ -20 cos^{3}θ + 5 cosθ

**cos 5θ = 16cos ^{5}θ -20 cos^{3}θ + 5 cosθ**

**E.g.3**

Find an identity for sin4θ.

[cosθ + i sinθ]^{4} = Σ ^{4}C_{r} cosθ^{4-r} (i sinθ)^{r}

= cos^{4}θ + 4 cos^{3}θ(i sinθ) + 6 cos^{2}θ(i sinθ)^{2} + 4 cosθ(i sinθ)^{3} + (i sinθ)^{4}

= cos^{4}θ + 4 cos^{3}θ(i sinθ) + 6 cos^{2}θ(-sin^{2}θ) + 4 cosθ(-i sin^{3}θ) + (i sin^{4}θ)

From de Moivre't Theorem,

[cosθ + sinθ]^{4} = [cos4θ + i sin4θ]

So, [cos4θ + i sin4θ] = = cos^{4}θ + 4 cos^{3}θ(i sinθ) + 6 cos^{2}θ(-sin^{2}θ) + 4 cosθ(-i sin^{3}θ) + (i sin^{4}θ)

Now equate the **imaginary** parts on both sides:

sin4θ = 4cos^{3}θsinθ - 4 cosθ sin^{3}θ + sin^{4}θ

**sin 4θ = 4cos ^{3}θsinθ - 4 cosθ sin^{3}θ + sin^{4}θ**

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 z^{3} = 27 = 27(1 + 0 x i) = 27(cos 0 + i sin 0)

In more generalized form,

z^{3} = 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.

z_{0} = 3(cos 0 + i sin 0) = 3 - the real root

z_{1} = 3(cos 2π/3 + i sin 2π/3) = 3(-1/2 + i√3/2) = -3/2 + i√3/2

z_{2} = 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

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 z^{3} = 1 = 1(1 + 0 x i) = 1(cos 0 + i sin 0)

In more generalized form,

z^{3} = 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.

z_{0} = 1(cos 0 + i sin 0) = 1 - the real root

z_{1} = 1(cos 2π/3 + i sin 2π/3) = (-1/2 + i√3/2) = -1/2 + i√3/2

z_{2} = 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:

z_{0} + z_{1} + z_{2} = 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.

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 - 7^{th} edition in print.