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Question
Prove by method of induction, for all n ∈ N:
(cos θ + i sin θ)n = cos (nθ) + i sin (nθ)
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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 + i2 sin kθ sin θ
= (cos kθ cos θ – sin kθ sin θ) + i(sin kθ cos θ + cos kθ sin θ) ......[as i2 = – 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
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