Prove by method of induction, for all n ∈ N:

(cos θ + i sin θ)^{n} = cos (nθ) + i sin (nθ)

#### Solution

Let P(n) ≡ (cos θ + i sin θ)^{n} = cos nθ + i sin nθ, for all n ∈ N.

**Step 1: **

For n = 1, L.H.S. = cos θ + i sin θ

R.H.S. = cos θ + i sin θ

∴ L.H.S. = R.H.S.

∴ P(1) is true.

**Step 2: **

Let us assume that for some k ∈ N, P(k) is true,

i.e., (cos θ + i sin θ)^{k} = cos kθ + i sin kθ ...(1)

**Step 3:** To prove that P(k + 1) is true,

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) ...[By (1)]

= cos kθ · cos θ + i cos kθ sin θ + i sin kθ cos θ + i^{2} sin kθ sin θ

= (cos kθ cos θ – sin kθ sin θ) + i(sin kθ cos θ + cos kθ sin θ) ......[as i^{2} = – 1]

= cos(kθ + 0) + i sin (kθ + θ)

= cos (k + 1)θ + i sin(k + 1)θ

= R.H.S.

∴ P(k + 1) is true.

**Step 4: **

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

i.e., (cos θ + i sin θ)^{n} = cos nθ + i sin nθ. for all n ∈ N