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# De Moivre's Theorem

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

### 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

$$\begin{array}{l}(e^{i \theta})^n = e^{in \theta}\end{array}$$

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

Remark: The 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, to revise and get familiar with the argand diagram and polar form of complex numbers. This section introduces De Moivre’s theorem statement, proof of the 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 the 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 proven.

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:

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

Where k = 0, 1, 2,….., (n − 1)

If k = 0, the above formula reduces to

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

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

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

The parameters of the rectangular and polar forms are as follows:

a = r cos x and b = r sin x

Where

$$\begin{array}{l}r = \sqrt{a^2+b^2}\end{array}$$
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

$$\begin{array}{l}=\sqrt{(1)^2 + (1)^2}= \sqrt{2}\end{array}$$

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

$$\begin{array}{l}r= \sqrt{a^2+b^2}\end{array}$$
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 the 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

$$\begin{array}{l}\left(\frac{cos \theta + i\ sin \theta}{sin \theta + i\ cos \theta}\right)^4\end{array}$$
in a+ib form.

Solution:

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

= (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

Problem 4: If

$$\begin{array}{l}{{x}_{r}}=\cos \left( \frac{\pi }{{{2}^{r}}} \right)+i\sin \left( \frac{\pi }{{{2}^{r}}} \right), then \ {{x}_{1}}\ .\ {{x}_{2}}……\infty \ is\end{array}$$

Solution:

$$\begin{array}{l}{{x}_{1}},{{x}_{2}},{{x}_{3}}…..upto\infty =\left( \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} \right)\,\,\left( \cos \frac{\pi }{{{2}^{2}}}+i\sin \frac{\pi }{{{2}^{2}}} \right) upto …..\infty \\=\cos \left( \frac{\pi }{2}+\frac{\pi }{{{2}^{2}}}+….. \right)+i\sin \left( \frac{\pi }{2}+\frac{\pi }{{{2}^{2}}}+….. \right) \\=\cos \left( \frac{\frac{\pi }{2}}{1-\frac{1}{2}} \right)+i\sin \left( \frac{\frac{\pi }{2}}{1-\frac{1}{2}} \right)\\=\cos \pi +i\sin \pi \\=-1\end{array}$$

Problem 5: The value of

$$\begin{array}{l}\frac{4(\cos {{75}^{o}}+i\sin {{75}^{o}})}{0.4(\cos {{30}^{o}}+i\sin {{30}^{o}})} \ is\end{array}$$

Solution:

$$\begin{array}{l}\frac{4(\cos {{75}^{o}}+i\sin {{75}^{o}})}{0.4(\cos {{30}^{o}}+i\sin {{30}^{o}})} \\=10(\cos {{75}^{o}}+i\sin {{75}^{o}})(\cos {{30}^{o}}-i\sin {{30}^{o}}) \\=10(\cos {{45}^{o}}+i\sin {{45}^{o}})\\=\frac{10}{\sqrt{2}}(1+i)\end{array}$$

Problem 6: If

$$\begin{array}{l}(\cos \theta +i\sin \theta )(\cos 2\theta +i\sin 2\theta )…….. (\cos n\theta +i\sin n\theta )=1,\end{array}$$
then what is the value of θ?

Solution:

$$\begin{array}{l}(\cos \theta +i\sin \theta )(\cos 2\theta +i\sin 2\theta ) ……(\cos n\theta +i\sin n\theta )=1 \\ \cos (\theta +2\theta +3\theta +…+n\theta )+i\sin (\theta +2\theta +.+n\theta )=1 \\ \cos \left( \frac{n(n+1)}{2}\theta \right)+i\sin \left( \frac{n(n+1)}{2}\theta \right)=1\\ \cos \left( \frac{n\,(n+1)}{2}\theta \right)\,=\,1\text{ and }\sin \left( \frac{n(n+1)}{2}\theta \right)\,=0 \\ \frac{n(n+1)}{2}\theta =2m\pi \\\Rightarrow \theta =\frac{4m\pi }{n(n+1)},\text where \ m\in I.\end{array}$$

#### De Moivre’s Theorem – Video Lesson Q1

### Give De Moivre’s Formula.

De Moivre’s formula is given by (cos x + i sin x)n = cos(nx) + i sin(nx)

Q2

### Give two uses of De Moivre’s theorem.

De Moivre’s theorem is used to find the roots of complex numbers.

It is also used to solve complex numbers raised to powers.

Q3

### Does De Moivre theorem work for non-integer powers?

De Moivre’s formula does not work for non-integer powers.

Q4

### Who invented De Moivre’s theorem?

The French mathematician Abraham De Moivre invented De Moivre’s theorem.

Test your knowledge on De Moviers Theorem