De Moivre's formula

De Moivre's formula states that for any real number x and any integer n,

Proof

We consider three cases.

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For n > 0, we proceed by induction. When n=1, the result is clearly true. For our hypothesis, we assume the result is true for some positive integer k. That is, we assume

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:(cos x + i sin x)^k = cos(kx) + i sin(kx). ,

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Now, considering the case n = k + 1:

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:(cos x+isin x)^{k+1},

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:= (cos x+isin x)(cos x+isin x)^{k},

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:= (cos(kx)+isin(kx))(cos x+isin x), (by the induction hypothesis)

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:= cos(kx)cos x - sin(kx)sin x + i(cos(kx)sin x + sin(kx)cos x),

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:= cos(k+1)x + isin(k+1)x,

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We deduce that the result is true for n = k + 1 when it is true for n = k. By the Principle of Mathematical Induction it follows that the result is true for all positive integers n.

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When n = 0 the formula is true since cos(0x) + isin(0x) = 1 + i0 = 1, and (by convention) z^{0} = 1.

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When n < 0, we consider a positive integer m such that n = −m. So

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:(cos x + isin x)^{n}, = (cos x + isin x)^{-m},

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:=rac{1}{(cos x + isin x)^{m}} = rac{1}{(cos mx + isin mx)},, from above

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:=cos(mx) - isin(mx),, rationalizing the denominator

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:=cos(-mx) + isin(-mx), = cos(nx) + isin(nx).,

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Hence, the theorem is true for all integral values of n. Q.E.D.

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