Prove De Moivre's theorem.

De Moivre's theorem states that for any complex number z and any positive integer n, (cos θ + i sin θ)^n = cos nθ + i sin nθ.

To prove De Moivre's theorem, we can use mathematical induction. First, we will prove the base case for n = 1:

(cos θ + i sin θ)^1 = cos 1θ + i sin 1θ
(cos θ + i sin θ) = cos θ + i sin θ

This is true, so the base case holds.

Next, we assume that the theorem holds for some positive integer k:

(cos θ + i sin θ)^k = cos kθ + i sin kθ

Now we need to prove that it also holds for k + 1:

(cos θ + i sin θ)^(k+1) = (cos θ + i sin θ)^k * (cos θ + i sin θ)
= (cos kθ + i sin kθ) * (cos θ + i sin θ)
= cos kθ cos θ - sin kθ sin θ + i(sin kθ cos θ + cos kθ sin θ)
= cos (k+1)θ + i sin (k+1)θ

Therefore, the theorem holds for all positive integers n by mathematical induction. Understanding the trigonometric identities involved is crucial, as demonstrated in the Trigonometric Identities notes. Additionally, the application of these identities in complex numbers can be further explored in Trigonometric Form of Complex Numbers. The method of proof, mathematical induction, is detailed further in Proof by Mathematical Induction.