Complex Numbers

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-De Moivre's Theorem-

De Moivre's theorem states that (cosθ + i.sinθ)ⁿ = cos(nθ) + i.sin(nθ) for any integer value of n.

Proof:

Induction is needed to prove De Moivre's theorem.
It is clearly true for n = 0 and n = 1. Now we suppose that it is true for n = k, then take

(cosθ + i.sinθ)^k+1 = (cosθ + i.sinθ)^k(cosθ + i.sinθ)
                           = (cos(kθ) + i.sin(kθ)).(cosθ + i.sinθ)           (as true for n=k)
                           = cos(kθ).cosθ - sin(kθ).sinθ + i.(cos(kθ).sinθ + sin(kθ).cosθ)
                           = cos(kθ + θ) + i.(sin(kθ + θ))
                           = cos((k+1)θ) + i.(sin((k+1)θ))
So by mathematical induction true for all n natural.

Also note that (cos(-nθ) + i.sin(-nθ)).(cos(nθ) + i.sin(nθ)) = cos²(nθ) + sin²(nθ) = 1.
Hence cos(-nθ) + i.sin(-nθ) = (cosθ + i.sinθ)^-n. So De Moivre's theorem holds for n is any integer