De Moivre's formula

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De Moivre's formula, named after Abraham de Moivre, states that for any complex number (and, in particular, for any real number) x and any integer n,

$(\cos x+i\sin x)^n=\cos(nx)+i\sin(nx).\,$

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 easily derived from (but historically preceded) Euler's formula e^ix = cos x + i sin x and the exponential law (e^ix)ⁿ = e^inx (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.

[edit] Applications

This formula can be used to find the $n t h$ roots of a complex number. If $z$ is a complex number, written in polar form as

$z=r(\cos x+i\sin x),\,$

then

$z^{1/n}=(r(\cos x+i\sin x))^{1/n}=r^{1/n}\left\{\cos\left(\frac{x+2k\pi}{n}\right)+i\sin\left(\frac{x+2k\pi}{n}\right)\right\}$

where $k$ varies from $0$ to $n - 1$ to give the $n$ roots of the complex number.

[edit] Proof by induction

We consider three cases.

For n > 0, we proceed by mathematical 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:

$(\cos x+i\sin x)^{k+1}\,$

$= (\cos x+i\sin x)^{k}(\cos x+i\sin x)\,$

$= (\cos(kx)+i\sin(kx))(\cos x+i\sin x)\,$ (by the induction hypothesis)

$= \cos(kx)\cos x - \sin(kx)\sin x + i(\cos(kx)\sin x + \sin(kx)\cos x)\,$

$= \cos(k+1)x + i\sin(k+1)x\,$ , by using trigonometric identities

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 $cos(0 x) + i sin(0 x) = 1 + i 0 = 1$ , and (by convention) $z 0 = 1$ .

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

$(\cos x + i\sin x)^{n}\, = (\cos x + i\sin x)^{-m}\,$