De Moivre's theorem is one of the fundamental theorem of complex numbers which is used to solve various problems of complex numbers. This theorem is also widely used for solving trigonometric functions of multiple angles. DeMoivre’s Theorem is also called “De Moivre’s Identity” and “De Moivre’s Formula”. This theorem gets its name from the name of its founder the famous mathematician De Moivre. In this article, we will learn about De Moivre's Theorem, its proof, some examples based on the theorem, and others in detail. Table of Content De Moivre's Theorem StatementDe Moivre’s FormulaDe Moivre’s Theorem ProofUses of De Moivre’s TheoremFinding the Roots of Complex NumbersPower of Complex NumbersSolved Examples on De Moivre’s TheoremFAQsDe Moivre's Theorem StatementDe Moivre's Theorem is a special theorem of complex numbers which is used to expand the complex number raised to any integer. De Moivre's Formula states that for all real values of x, (cos x + i.sinx)n = cos(nx) + i.sin(nx) where, n is any integer De Moivre’s FormulaDe Moivre’s Formula for complex numbers is, for any real value of x, (cos x + i.sinx)n = cos(nx) + i.sin(nx) Also, we know that, eix = cos x + i.sinx Now, (eix)n = einx Note, "n" in the above formula is an integer, and "i" is an imaginary number iota. Such that, i = √(-1) De Moivre's Formula is shown in the image added below, De Moivre’s Theorem ProofDeMoivre's Theorem can be proved with the help of mathematical induction as follows: P(n) = (cos x + i.sinx)n = cos(nx) + i.sin(nx) ⇢ (1) Step 1: For n = 1, (cos x + i sin x)1 = cos(1x) + i sin(1x) = cos(x) + i sin(x), Which is true. thus, P(n) is true for n = 1. Step 2: Assume P(k) is true (cos x + i.sin x)k = cos(kx) + i.sin(kx) ⇢ (2) Step 3: Now we have to prove P(k+1) is true. (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 (1)] = cos (kx).cos x − sin (kx).sin x + i [sin (kx).cos x + cos (kx).sin x) = cos {(k+1)x} + i.sin {(k+1)x} = cos {(k+1)x} + i.sin {(k+1)x} Thus, P(k+1) is also true and by the principle of mathematical induction P(n) is true. Uses of De Moivre’s TheoremDe Moivre’s Theorem is used for various purposes. Some of its most important uses are, Finding the Roots of Complex Numbers.Finding the relationships between Powers of Trigonometric Functions and Trigonometric Angles.Solving the Power of Complex Numbers.Now, let's learn about them with the help of examples. Finding the Roots of Complex NumbersThe polar form of the complex number is, z = r(cos x + i sin x) Then for nth root of the complex number z1/n = r1/n(cos x + i sin x)1/n ⇒ z1/n = r1/n[cos (x + 2kπ)/n + i sin (x + 2kπ)/n] Where k = 0, 1, 2, 3, ... Power of Complex NumbersWe can easily solve the power of Complex numbers using De Moivre's Theorem. This can be understood using the example as follows, Example: Evaluate (√3 + i)200 Solution: Let, z = √3 + i comparing with z = x + iy x = √3, y = 1 Also, z = r(cos θ + i sin θ) r = √(x2 + y2) = √[(√3)2 + 12] r = 2 θ = tan-1(y/x) = tan-1(1/√3) = π/6 z = r(cos θ + i sin θ) ⇒ z = 2(cos π/6 +i.sin π/6) ⇒ z200 = [2(cos π/6 +i.sin π/6)]200 ⇒ z200 = [2]200[(cos π/6 +i.sin π/6)]200 Using DeMoivre’s Theorem z200 = [2]200[(cos 200π/6 +i.sin 200π/6)] ⇒ z200 = [2]200[-1/2 - i√3/2] ⇒ z200 = [2]200[1/2 + i√3/2] Read More, Complex Number Power FormulaPolar and Exponential Forms of Complex NumbersZ Bar in Complex NumbersIs Every Real Number a Complex Number? Solved Examples on De Moivre’s TheoremExample 1: Expand (1 + i)100. Solution: Let, z = 1 + i comparing with z = x + iy x = 1, y = 1 Also, z = r(cos θ + i sin θ) r = √(x2 + y2) = √[12 + 12] r = √2 θ = tan-1(y/x) = tan-1(1/1) = π/4 z = r(cos θ + i sin θ) ⇒ z = √2(cos π/4 +i.sin π/4) z100 = [2(cos π/4 +i.sin π/4)]100 ⇒ z100 = [2]50[(cos π/4 +i.sin π/4)]100 Using DeMoivre’s Theorem z100 = [2]50[(cos 100π/4 +i.sin 100π/4)] ⇒ z100 = [2]50[-1 + i.0] ⇒ z100 = [2]50 Example 2: Expand (3 + 3i)40. Solution: Let, z = 3 + 3i comparing with z = x + iy x = 3, y = 3 Also, z = r(cos θ + i sin θ) ⇒ r = √(x2 + y2) = √[32 + 32] ⇒ r = 2√3 θ = tan-1(y/x) = tan-1(3/3) = π/4 z = r(cos θ + i sin θ) ⇒ z = 3√2(cos π/4 +i.sin π/4) z40 = [3√2(cos π/4 +i.sin π/4)]140 ⇒ z40 = [3√2]40[(cos π/4 +i.sin π/4)]40 Using DeMoivre’s Theorem z40 = [3√2]40[(cos 40π/4 +i.sin 40π/4)] ⇒ z40 = [3√2]40[1 + i.0] ⇒ z40 = [3√2]40 Example 3: Simplify [(cos x + i.sin x)/(sin x + i.cos x)]8 Solution: [(cos x + i.sin x)/(sin x + i.cos x)]8 = (cos x + i.sin x)8/i8(cos x - i.sin x)8 Using De Movire's Theorem and taking the value of i8 = 1 = (cos 8x + i.sin 8x)/(cos 8x - i.sin 8x) Rationalising we get, = (cos 8x + i.sin 8x)2 / (cos2 8x – i2.sin2 8x) = cos 16x + i.sin 16x Since, [(cos2 8x – i2.sin2 8x) = (cos2 8x + sin2 8x) = 1] Practice Problems on DeMoivre's TheoremProblem 1: Use DeMoivre's Theorem to find (1 + i)6. Problem 2: Apply DeMoivre's Theorem to calculate (2 + 2i)4. Problem 3: Find the 5th power of (2−2i) using DeMoivre's Theorem. Problem 4: Determine (3+4i)3 using DeMoivre's Theorem. Problem 5: Calculate (2−3i)7 using DeMoivre's Theorem.

DeMoivre's Theorem

Last Updated : 11 Jun, 2024

De Moivre's theorem is one of the fundamental theorem of complex numbers which is used to solve various problems of complex numbers. This theorem is also widely used for solving trigonometric functions of multiple angles. DeMoivre’s Theorem is also called “De Moivre’s Identity” and “De Moivre’s Formula”. This theorem gets its name from the name of its founder the famous mathematician De Moivre.

In this article, we will learn about De Moivre's Theorem, its proof, some examples based on the theorem, and others in detail.

Table of Content

De Moivre's Theorem Statement

De Moivre's Theorem is a special theorem of complex numbers which is used to expand the complex number raised to any integer. De Moivre's Formula states that for all real values of x,

(cos x + i.sinx)ⁿ = cos(nx) + i.sin(nx)

where, n is any integer

De Moivre’s Formula

De Moivre’s Formula for complex numbers is, for any real value of x,

(cos x + i.sinx)ⁿ = cos(nx) + i.sin(nx)

Also, we know that,

e^ix = cos x + i.sinx

Now,

(e^ix)ⁿ = e^inx

Note, "n" in the above formula is an integer, and "i" is an imaginary number iota. Such that, i = √(-1)

De Moivre's Formula is shown in the image added below,

De Moivre’s Theorem Proof

DeMoivre's Theorem can be proved with the help of mathematical induction as follows:

P(n) = (cos x + i.sinx)ⁿ = cos(nx) + i.sin(nx) ⇢ (1)

Step 1: For n = 1,

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

Which is true. thus, P(n) is true for n = 1.

Step 2: Assume P(k) is true

(cos x + i.sin x)^k = cos(kx) + i.sin(kx) ⇢ (2)

Step 3: Now we have to prove P(k+1) is true.

(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 (1)]

= cos (kx).cos x − sin (kx).sin x + i [sin (kx).cos x + cos (kx).sin x)

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

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

Thus, P(k+1) is also true and by the principle of mathematical induction P(n) is true.

Uses of De Moivre’s Theorem

De Moivre’s Theorem is used for various purposes. Some of its most important uses are,

Finding the Roots of Complex Numbers.
Finding the relationships between Powers of Trigonometric Functions and Trigonometric Angles.
Solving the Power of Complex Numbers.

Now, let's learn about them with the help of examples.

Finding the Roots of Complex Numbers

The polar form of the complex number is,

z = r(cos x + i sin x)

Then for n^throot of the complex number

z^1/n = r^1/n(cos x + i sin x)^1/n

⇒ z^1/n= r^1/n[cos (x + 2kπ)/n + i sin (x + 2kπ)/n]

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

Power of Complex Numbers

We can easily solve the power of Complex numbers using De Moivre's Theorem. This can be understood using the example as follows,

Example: Evaluate (√3 + i)²⁰⁰

Solution:

Let, z = √3 + i comparing with z = x + iy

x = √3, y = 1

Also, z = r(cos θ + i sin θ)

r = √(x² + y²) = √[(√3)² + 1²]

r = 2

θ = tan^-1(y/x) = tan^-1(1/√3) = π/6

z = r(cos θ + i sin θ)

⇒ z = 2(cos π/6 +i.sin π/6)

⇒ z²⁰⁰ = [2(cos π/6 +i.sin π/6)]²⁰⁰

⇒ z²⁰⁰ = [2]²⁰⁰[(cos π/6 +i.sin π/6)]²⁰⁰

Using DeMoivre’s Theorem

z²⁰⁰ = [2]²⁰⁰[(cos 200π/6 +i.sin 200π/6)]

⇒ z²⁰⁰ = [2]²⁰⁰[-1/2 - i√3/2]

⇒ z²⁰⁰ = [2]²⁰⁰[1/2 + i√3/2]

Read More,

Solved Examples on De Moivre’s Theorem

Example 1: Expand (1 + i)¹⁰⁰.

Solution:

Let, z = 1 + i comparing with z = x + iy

x = 1, y = 1

Also, z = r(cos θ + i sin θ)

r = √(x² + y²) = √[1² + 1²]

r = √2

θ = tan^-1(y/x) = tan^-1(1/1) = π/4

z = r(cos θ + i sin θ)

⇒ z = √2(cos π/4 +i.sin π/4)

z¹⁰⁰ = [2(cos π/4 +i.sin π/4)]¹⁰⁰

⇒ z¹⁰⁰ = [2]⁵⁰[(cos π/4 +i.sin π/4)]¹⁰⁰

Using DeMoivre’s Theorem

z¹⁰⁰ = [2]⁵⁰[(cos 100π/4 +i.sin 100π/4)]

⇒ z¹⁰⁰ = [2]⁵⁰[-1 + i.0]

⇒ z¹⁰⁰ = [2]⁵⁰

Example 2: Expand (3 + 3i)⁴⁰.

Solution:

Let, z = 3 + 3i comparing with z = x + iy

x = 3, y = 3

Also, z = r(cos θ + i sin θ)

⇒ r = √(x² + y²) = √[3² + 3²]

⇒ r = 2√3

θ = tan^-1(y/x) = tan^-1(3/3) = π/4

z = r(cos θ + i sin θ)

⇒ z = 3√2(cos π/4 +i.sin π/4)

z⁴⁰ = [3√2(cos π/4 +i.sin π/4)]¹⁴⁰

⇒ z⁴⁰ = [3√2]⁴⁰[(cos π/4 +i.sin π/4)]⁴⁰

Using DeMoivre’s Theorem

z⁴⁰ = [3√2]⁴⁰[(cos 40π/4 +i.sin 40π/4)]

⇒ z⁴⁰ = [3√2]⁴⁰[1 + i.0]

⇒ z⁴⁰ = [3√2]⁴⁰

Example 3: Simplify [(cos x + i.sin x)/(sin x + i.cos x)]⁸

Solution:

[(cos x + i.sin x)/(sin x + i.cos x)]⁸

= (cos x + i.sin x)⁸/i⁸(cos x - i.sin x)⁸

Using De Movire's Theorem and taking the value of i⁸= 1

= (cos 8x + i.sin 8x)/(cos 8x - i.sin 8x)

Rationalising we get,

= (cos 8x + i.sin 8x)² / (cos² 8x – i².sin² 8x)

= cos 16x + i.sin 16x Since, [(cos² 8x – i².sin² 8x) = (cos² 8x + sin² 8x) = 1]

Practice Problems on DeMoivre's Theorem

Problem 1: Use DeMoivre's Theorem to find (1 + i)⁶.

Problem 2: Apply DeMoivre's Theorem to calculate (2 + 2i)⁴.

Problem 3: Find the 5^th power of (2−2i) using DeMoivre's Theorem.

Problem 4: Determine (3+4i)³ using DeMoivre's Theorem.

Problem 5: Calculate (2−3i)⁷ using DeMoivre's Theorem.

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Article Tags :

DeMoivre's Theorem

De Moivre's Theorem Statement

De Moivre’s Formula

De Moivre’s Theorem Proof

Uses of De Moivre’s Theorem

Finding the Roots of Complex Numbers

Power of Complex Numbers

Solved Examples on De Moivre’s Theorem

Practice Problems on DeMoivre's Theorem

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