## DeMoivre's theorem

This theorem is named after the famous mathematician, Abraham De Moivre.

It is also known as DeMoivre's formula or DeMoivre's identity.

DeMoivre's formula:

Let  z= r(cosθ + i sinθ) and let 'n' be any real positive number, then zⁿ = rⁿ (cos nθ + i sin nθ), where r is the modulus and θ is the argument of the complex number.

Proof of DeMoivre's formula:

We can prove DeMoivre's theorem by mathematical induction.

When n=1, The formula is

(cosθ + i sinθ)¹ = cos 1θ + i sin 1θ

So it is true for n=1.

Let us assume this formula is true for n = k. That is

(cosθ + i sinθ)k = cos(kθ) + i sin(kθ)----(1)

We want to prove that it is true for n = k+1.

(cosθ + isinθ)k+1 = (cosθ + isinθ)k(cosθ + isinθ)

From the equation(1) we know that (cosθ + isinθ)k = cos(kθ) + i sin(kθ)

Substituting this for k+1

(cosθ + isinθ)k+1= [cos(kθ) + i sin(kθ)](cosθ + isinθ)

=  cos(kθ)cos(θ) +i² sin(kθ)sin(θ) +i[sin(θ)cos(kθ) +                                                 cos(θ)sin(kθ)]

= [cos(kθ)cos(θ) - sin(kθ)sin(θ)] + i[cos(kθ)sin(θ) +                                                 sin(kθ)cos(θ)]

=  cos(k+1)θ + i sin(k+1)θ

This is true for n = k+1,

So we proved by mathematical induction. 1.  Compute z¹⁰ when z = 1+i

Solution:

First let us write z in the polar form.

∣z∣  =   √(1²+1²)  = √2.

arg z = θ = tan ¹(1) = π/4

Applying DeMoivre's theorem

z¹⁰   =   (√2)¹⁰  [cos(10 x π/4) + i sin(10 x π/4)

=     2⁵ [ cos (5π/2) + i sin (5π/2)]

=    32 [cos (2π + π/2) + i sin (2π + π/2)]

=      32 [cos π/2 + i sin π/2]

=       32 [0 +i]

=        32i

2.   Compute (√3+ i)

Solution:

Let z = √(3+ 1)= 2

First let us write z in the polar form.

∣z∣     =   √( (√3)² + 1²) =√( 3+ 1) = 2.

arg z  =   θ = tan ¹(1/√3) = π/6

So   z   =   (2)  [cos(5 x π/6) + i sin(5 x π/6)

=     2⁵ [ cos (5π/6) + i sin (5π/6)]

=    32 [cos (π - π/6) + i sin (π - π/6)]

=      32 [-cos π/6 + i sin π/6]

=       32 [-√3/2 + 1/2]

=        -16√3 +16i

Students can go through the theorem and its proof step by step and become master in the theorem. If you are having any doubt you can contact us through mail, we will help you to clear the doubts.

We welcome your valuable suggestions for the improvement of our site. Please use the box given below to express your suggestions. 