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) ........ (cosθn + i sinθn)
then z = cos ( θ1 + θ2 + ........ + θn) + i sin (θ1 + θ2 + ...... + θn)
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, .... αn-1 . Which are in G.P.
Sum of nth roots of unity = 0