Definition:
               If n is any integer then (cosθ + i sinθ)ⁿ = cos nθ + i sin nθ is called De Moiver's theorem.

Conceptual theorem
1. State and prove De Moiver's theorem
Statement: If n is any integer then (cosθ + i sinθ)ⁿ = cos nθ + i sin nθ.
proof: Case (i) : Let 'n' be a positive integer. The proof is by mathematical induction.
p(n) = {n  N / (cosθ + i sinθ)ⁿ = cos nθ + i sin nθ}
Put n = 1
   (cosθ + i sinθ)¹ = cos (1)θ + i sin (1)θ
    cosθ + i sinθ = cosθ + i sinθ
When n = 1, the given stament is true.
Put n = k
 (cosθ + i sinθ)^k = cos kθ + i sin kθ
Let the given stament be true for n = k .

Now, we shall prove that the given stament is true for n = k + 1.
  (cosθ + i sinθ)^k+1 = (cosθ + i sinθ)^k (cosθ + i sinθ)
  (cosθ + i sinθ)^k+1 = (cos kθ + i sin kθ) (cosθ + i sinθ)
  (cosθ + i sinθ)^k+1 = cos kθ cosθ + i cos kθ sinθ + i sin kθ cosθ − sin kθ sin θ
  (cosθ + i sinθ)^k+1 = (cos kθ cos θ − sin kθ sin θ) + i (sin kθ cos θ + cos kθ sin θ)
  (cos θ + i sin θ)^k+1 = cos (kθ + θ) + i sin(kθ + θ)
  (cos θ + i sin θ)^k+1 = cos (k+1)θ + i sin(k+1)θ
   The given stament is true for n = k+1.
Hence P(n) is true for all n   N.
   (cosθ + i sinθ)ⁿ = cos nθ + i sin nθ
Case (ii) : Let 'n' be a negative integer .
                 Let n = −m, where 'm' is positive integer.
                 (cosθ  +  i sinθ)n  =  (cosθ  +  i sinθ)^−m
                  

   (cosθ + i sinθ)ⁿ   =   cos mθ − i sin mθ
   (cosθ + i sinθ)ⁿ   =   cos (−m)θ + i sin(−m)θ
   (cosθ + i sinθ)ⁿ   =   cos nθ + i sin nθ

Case (iii) : Let n = 0.
(cosθ + i sinθ)ⁿ    =  (cosθ + i sinθ)ⁿ 
                                = 1
                                =  cos 0 + i sin 0
                                =  cos (0)θ + i sin (0)θ
  (cosθ + i sinθ)n = cos nθ + i sin nθ
From cases (i), (ii) and (iii)
  (cosθ + i sinθ)n = cos nθ + i sin nθ

Note (1) : We know that (cos nθ + i sin nθ) (cos nθ − i sin nθ) = 1.
                
Note (2) : (cosθ + i sinθ)⁻ⁿ   =   cos (−n)θ + i sin (−n)θ
                   (cosθ + i sinθ)⁻ⁿ  =   cos nθ − i sin nθ
Note (3) : (cosθ − i sinθ)n = [cosθ + i (−sinθ)]n
                = [cos (−θ)  +  i sin (−θ)]ⁿ
                = cos (−nθ)  +  i sin (−nθ)
                  (cosθ − i sinθ)ⁿ  =  cos nθ − i sin nθ
Note (4) : If z = (cosθ₁ + i sinθ1) (cosθ₂ + i sinθ₂) ........ (cosθ_n + i sinθ_n)
                 then z = cos ( θ₁ + θ₂ + ........ + θ_n) + i sin (θ₁ + θ₂ + ...... + θ_n)
Note (5) : z = r (cosθ + i sinθ) and 'n' is a positive integer, then
         
                where k = 0, 1, 2 ....... (n−1)

Cube Roots Of Unity
           The equation x³ =  1 has three roots. Which are called Cube Roots Of Unity.
                                   x³ =  1
                               x³ − 1  =  0
                               (x − 1) (x² + x + 1)  =  0
                            
If the second root is represented by ω, then third root is ω² .
                             Cube Roots Of Unity: 1, ω, ω²

Properties :
(1)  1  +  ω  + ω²  =  0
(2)  ω³   =  1
(3) ω³ⁿ  =  1
      ω³ⁿ⁺¹   =  ω
      ω³ⁿ⁺²  = ω²
eg: ω²⁰¹¹  = ω³ × 670 +1  =   ω
  ω²⁰¹²   =   ω³ × 670 +2  =  ω²
       ω²⁰¹³   =   ω³ × 671  =  1
(4)   = ω², (  )2  =  ω

nth Roots Of Unity
     The equation xn = 1 has n roots. Which are called the nth Roots Of Unity. xn  =  1
    xⁿ  =  cos 0 + i sin 0
    xⁿ  =   cos (2kπ + 0) + i sin (2kπ + 0)

Then nth roots of unity are
αs (s = 0, 1, 2, ........ n−1)
i.e. the nth roots of unity are α⁰, α¹, α², .... α^n-1 . Which are in G.P.

  Sum of nth roots of unity  =  0