DeMoivre’s Theorem

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.

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 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,

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.

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^th root 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,

• Complex Number Power Formula
• Polar and Exponential Forms of Complex Numbers
• Z Bar in Complex Numbers
  
• Is Every Real Number a Complex Number? 

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]

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.

What is De Moivre’s Theorem?

The De Moivre’s theorem is the basic theorem used in complex numbers for solving various problems. The De Moivre’s theorem states that,

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

What are the uses of De Moivre’s Theorem?

The various uses of De Moivre’s theorem include the solving of complex roots, finding the power of the complex number and others.

Is De Moivre’s Theorem work for Non-Integer Powers?

No, De Moivre’s formula does not work for non-integer powers. The result for non-negative integers is the multiple-value different from the original results.

Who invented De Moivre’s Theorem?

The De Moivre’s theorem was first introduced by the French mathematician Abraham De Moivre.