# De Movier's Theorem

In the field of complex numbers, DeMoivre’s Theorem is one of the most important and useful theorems which connects complex numbers and trigonometry. Also helpful for obtaining relationships between trigonometric functions of multiple angles. DeMoivre’s Theorem also known as “De Moivre’s Identity” and “De Moivre’s Formula”. The name of the theorem is after the name of great mathematician De Moivre, who made many contributions to the field of mathematics, mainly in the areas of theory of probability and algebra.

Table of Content:

### De moivre’s Formula

Mathematical Statement: For any real number x, we have

(cos x + i sin x)^n = cos(nx) + i sin(nx)

OR $(e^{i \theta})^n = e^{in \theta}$

Where n is a positive integer and “ i “ is the imaginary part, and i = √(-1). Also assume i2 = -1.

Remark: Result can be shown true when n is a negative integer and even

n is a rational number.

Aspirants are advised, before starting this section should revise and get familiar with the argand diagram and polar form of complex numbers. This section introduces -De Moivre’s theorem statement, proof of theorem and some of its consequences.

## De Moivre’s Theorem Proof

Apply Mathematical Induction to prove De Moivre’s Theorem.

We know, (cos x + i sin x)n = cos(nx) + i sin(nx) …(i)

Step 1: For n = 1, we have

(cos x + i sin x)1 = cos(1x) + i sin(1x) = cos(x) + i sin(x)

Which is true.

Step 2: Assume that formula is true for n = k.

(cos x + i sin x)k = cos(kx) + i sin(kx) ….(ii)

Step 3: Prove that result is true for n = k + 1.

(cos x + i sin x)k+1 = (cos x + i sin x)k (cos x + i sin x)

= (cos (kx) + i sin (kx)) (cos x + i sin x) [Using (i)]

= cos (kx) cos x − sin(kx) sinx + i (sin(kx) cosx + cos(kx) sinx)

= cos {(k+1)x} + i sin {(k+1)x}

=> (cos x + i sin x)k+1 = cos {(k+1)x} + i sin {(k+1)x}

Hence the result is proved.

Since the theorem is true for n = 1 and n = k + 1, it is true ∀ n ≥ 1.

### Uses of De Moivre’s Theorem

To find the roots of complex numbers

If z is a complex number, and z = r(cos x + i sin x) [In polar form]

Then, the nth roots of z are:

$r^{\frac {1}{n}}\left (cos(\frac{x+2k \pi}{n}) + i\ sin (\frac{x+2k \pi}{n}) \right )$

where k = 0, 1, 2,….., (n − 1)

If k = 0, above formula reduce to

$r^{\frac {1}{n}}\left (cos(\frac{x}{n}) + i\ sin (\frac{x}{n}) \right )$

To obtain relationships between powers of trigonometric functions and trigonometric angles.

We use polar form of complex numbers to represent a complex number using trigonometry.

The parameters of the rectangular and polar form are as:

a = r cos x and b = r sin x

Where r = $\sqrt{a^2+b^2}$ and tan x = (b/a)

This implies, z = a + ib = r(cos x + i sin x)

Raising to a Power

Example: Evaluate ( 1 + i )1000.

Solution:

Let z = 1 + i

We have to represent z in the form of r(cos θ + i sin θ).

Here,

Argument = θ = arc(tan (1/1) = arc tan(1) = π/4

Absolute value = r = $\sqrt{(1)^2 + (1)^2}= \sqrt{2}$

Applying DeMoivre’s theorem, we get

z1000 = [√2{cos(π/4) + i sin(π/4)}]1000

= 21000 {cos(1000π/4) + i sin(1000π/4)}

= 21000 {1 + i (0)}

= 21000

## Problems on De Moivre’s Identity

Problem 1: Evaluate (2 + 2i)6

Solution: Let z = 2 + 2i

Here, r = 2√2 and θ = 45 degrees

Since z lies in the first quadrant, sinθ and cosθ functions are positive.

Applying De Moivre’s Theorem:

z6 = (2 + 2i)6 = (2√2)6 [cos 450 + i sin 450]6

= (2√2)6 [cos 2700 + i sin 2700]6

= – 512i

Problem 2: Express five fifth‐roots of (√3 + i) in trigonometric form.

Solution:

We know, z = a + ib = r(cos x + i sin x)

Where r = $\sqrt{a^2+b^2}$ and tan x = (b/a)

So,

Here r = 2 and θ = 30 degrees

Therefore, z = 2[cos(300 + 3600 k) + i sin cos(300 + 3600 k)]

Applying nth root theorem:

z1/5 = {2[cos(300 + 3600 k) + i sin cos(300 + 3600 k)]}1/5

= 21/5 [cos((300 + 3600 k)/5) + i sin cos((300 + 3600 k)/5)] …(1)

Where k = 0,1,2,3,4

At k = 0; (1)=> z1 = 21/5 [cos 60 + i sin 60]

At k = 1; (1)=> z1 = 21/5 [cos 780 + i sin 780]

At k = 2; (1)=> z1 = 21/5 [cos 1500 + i sin 1500]

At k = 3; (1)=> z1 = 21/5 [cos 2220 + i sin 2220]

At k = 4; (1)=> z1 = 21/5 [cos 2940 + i sin 2940]

Problem 3: Express $\left(\frac{cos \theta + i\ sin \theta}{sin \theta + i\ cos \theta}\right)^4$ in a+ib form.

Solution:

$\left(\frac{cos \theta + i\ sin \theta}{sin \theta + i\ cos \theta}\right)^4$

= $\frac{\left ( cos\ \theta + i\ sin\theta\right )^4}{i^4\left ( \ cos \theta – i sin \theta\right )^4}$

= (cos 4θ + i sin 4θ ) / (cos 4θ – i sin 4θ )

By rationalising the fraction, we have

= (cos 4θ + i sin 4θ )2 / (cos2 4θ – i sin2 4θ )

= cos 8θ + i sin 8θ