Hi
is pure complex if k is odd and real if k is even
Set k=2r
If Z = cos θ + i sin θ, (θ belongs to set of real numbers, i = √-1), then according to DeMoivre's theorem, Z^n = cos nθ + i sin nθ, (n belongs to set of natural numbers). Using this theorem alongwith binomial expansion, prove that:
(i) cos 2nθ = ∑ ( r = 0 to n) (2n C 2r) cos^(2(n - r)) θ sin^(2r) θ
(ii) cos 2nθ = ∑ (r = 0 to n) (2n C (2r + 1))cos^(2(n-r) - 1) θ sin^(2r + 1) θ
NOTE: 2n C 2r is 2n choose 2r.