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# De Moivre's formula

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

(cosx + isinx)n = cos(nx) + isin(nx).

The formula is important because it connects complex numbers (i stands for the imaginary unit) and trigonometry. The expression "cos x + i sin x" is sometimes abbreviated to "cis x".

By expanding the left hand side and then comparing the real and imaginary parts, it is possible to derive useful expressions for cos(nx) and sin(nx) in terms of cos(x) and sin(x). Furthermore, one can use this formula to find explicit expressions for the n-th roots of unity, that is, complex numbers z such that zn = 1.

Abraham de Moivre was a good friend of Newton; in 1698 he wrote that the formula had been known to Newton as early as 1676. It can be derived from (but historically preceded) Euler's formula eix = cos x + i sin x and the exponential law (eix)n = einx (see exponential function).

De Moivre's formula is actually true in a more general setting than stated above: if z and w are complex numbers, then (cos z + i sin z)w is a multivalued function while cos (wz) + i sin (wz) is not, and one can state that

cos (wz) + i sin (wz) is one value of (cos z + i sin z)w.

## Proof

We consider three cases.

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

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

Now, considering the case n = k + 1:

$(\cos x+i\sin x)^{k+1}\,$
$= (\cos x+i\sin x)(\cos x+i\sin x)^{k}\,$
$= (\cos(kx)+i\sin(kx))(\cos x+i\sin x)\,$ (by the induction hypothesis)
$= \cos(kx)\cos x - \sin(kx)\sin x + i(\cos(kx)\sin x + \sin(kx)\cos x)\,$
$= \cos(k+1)x + i\sin(k+1)x\,$

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.

When n = 0 the formula is true since cos(0x) + isin(0x) = 1 + i0 = 1, and (by convention) z0 = 1.

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

$(\cos x + i\sin x)^{n}\, = (\cos x + i\sin x)^{-m}\,$
$=\frac{1}{(\cos x + i\sin x)^{m}} = \frac{1}{(\cos mx + i\sin mx)}\,$, from above
$=\cos(mx) - i\sin(mx)\,$, rationalizing the denominator
$=\cos(-mx) + i\sin(-mx)\, = \cos(nx) + i\sin(nx)\,$

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

03-10-2013 05:06:04
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