De Moivre's Formula - Proof By Induction (for Integer n)

Proof By Induction (for Integer n)

The truth of de Moivre's theorem can be established by mathematical induction for natural numbers, and extended to all integers from there. Consider S(n):

For n > 0, we proceed by mathematical induction. S(1) is clearly true. For our hypothesis, we assume S(k) is true for some natural k. That is, we assume

Now, considering S(k+1):

\begin{alignat}{2} \left(\cos x+i\sin x\right)^{k+1} & = \left(\cos x+i\sin x\right)^{k} \left(\cos x+i\sin x\right)\\ & = \left \left(\cos x+i\sin x\right) &&\qquad \text{by the induction hypothesis}\\ & = \cos \left(kx\right) \cos x - \sin \left(kx\right) \sin x + i \left\\ & = \cos \left + i\sin \left &&\qquad \text{by the trigonometric identities} \end{alignat}

We deduce that S(k) implies S(k+1). By the principle of mathematical induction it follows that the result is true for all natural numbers. Now, S(0) is clearly true since cos (0x) + i sin(0x) = 1 +i 0 = 1. Finally, for the negative integer cases, we consider an exponent of -n for natural n.

\begin{align} \left(\cos x + i\sin x\right)^{-n} & = \left^{-1} \\ & = \left^{-1} \\ & = \cos(-nx) + i\sin (-nx). \qquad (*) \\ \end{align}

The equation (*) is a result of the identity, for z = cos nx + i sin nx. Hence, S(n) holds for all integers n.

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