Question

Prove by the method of induction, for all n ∈ N.

(cos θ + i sin θ)ⁿ = cos (nθ) + i sin (nθ)

Answer 1

Let P(n) ≡ (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ,

for all n ∈ N.

Step I: Put n = 1

L.H.S. = (cos θ + i sin θ)¹ = cos θ + i sin θ

R.H.S. = cos[(1)θ] + i sin[(1)θ] = cos θ + i sin θ

∴ L.H.S. = R.H.S.

∴ P(n) is true for n = 1.

Step II: Let us assume that P(n) is true for n = k

 ∴ (cos θ + i sin θ)^k = cos kθ + i sin kθ …….(i)

Step III: We have to prove that P(n) is true for n = k + 1,

i.e., to prove that (cos θ + i sin θ)^k+1 = cos (k + 1)θ + i sin (k + 1)θ

L.H.S. = (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) …… [From (i)]

= cos kθ cos θ + i sin θ cos kθ + i sin kθ cosθ – sin kθ sin θ ……[∵ i² = -1]

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

= cos(kθ + θ) + i sin(kθ + θ)

= cos(k + 1) θ + i sin (k + 1) θ

= R.H.S.

∴ P(n) is true for n = k + 1.

Step IV:

From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.

∴ (cos θ + i sin θ)ⁿ = cos (nθ) + i sin (nθ), for all n ∈ N.