DeMoivre's Theorem

De Moivre's theorem states that when a complex number in trigonometric form is raised to a power, its magnitude is raised to that power, while its angle is multiplied by the same power.

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

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).

Theorem Proof

De Moivre'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

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 De Moivre’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]

Practice Questions on De Moivres Theorem