Question

State and prove de-moivre's theorem,

Answer 1

De-moivre’s theorem:

(cosθ + i sinθ)n = (cos nθ + i sin nθ) for all n ϵ N

Proof:

We have to prove this theorem using mathematical induction.

Let p(n): (cosθ + i sinθ)n = (cos nθ + i sin nθ) for

all n є N

Put n = 1

We get p(1): (cosθ + i sinθ)1 = (cos θ + i sin θ)

That is (cosθ + i sinθ) = cosθ + i sinθ

Thus p(n) is true for n = 1 → (1)

Let us assume that p(n) is true for n = 2, 3, …k

⇒ (cosθ + i sinθ)k = (cos kθ + i sin kθ) → (2)

Now we have to prove that p(n) is also true for

n = (k + 1)

That is we have to prove that (cosθ + i sinθ)k +
1 = [cos (k + 1)θ + i sin (k + 1)θ]

Consider, (cosθ + i sinθ)k + 1 = (cosθ + i sinθ)k × (cosθ + i sinθ)
= (cos kθ + i sin kθ) × (cosθ + i sinθ)

= [cos kθ × cosθ + i cos kθ × sinθ + i sin kθ × cosθ + i2sin kθ × sinθ]

= [cos kθ × cosθ + i cos kθ × sinθ + i sin kθ × cosθ – sin kθ × sinθ]

= [(cos kθ cosθ – sin kθ sinθ) + i (cos kθ sinθ + sin kθ cosθ)]

= [cos (k + 1)θ + i sin(k + 1)θ]

Thus p(n) is also true for n = n + 1 → (3)

From (1), (2) and (3) by the principle of mathematical induction, we get

(cosθ + i sinθ)n = (cos nθ + i sin nθ)

for all n ϵ N

hope it will help you
.