De Moivre's formula

From Wikipedia, the free encyclopedia

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 it holds that

Failed to parse (Missing texvc executable; please see math/README to configure.): \left(\cos x+i\sin x\right)^n=\cos\left(nx\right)+i\sin\left(nx\right).\,

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.

1 Derivation
2 Proof by induction
3 Generalization
4 Applications
5 See also
6 References

[edit] Derivation

Although historically proved earlier, de Moivre's formula can easily be derived from Euler's formula

Failed to parse (Missing texvc executable; please see math/README to configure.): e^{ix} = \cos x + i\sin x\,

and the exponential law

Failed to parse (Missing texvc executable; please see math/README to configure.): \left( e^{ix} \right)^n = e^{inx} .\,

Then, by Euler's formula,

Failed to parse (Missing texvc executable; please see math/README to configure.): e^{i(nx)} = \cos(nx) + i\sin(nx)\,

[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

Failed to parse (Missing texvc executable; please see math/README to configure.): \left(\cos x + i \sin x\right)^k = \cos\left(kx\right) + i \sin\left(kx\right). \,

Now, considering the case n = k + 1:

Failed to parse (Missing texvc executable; please see math/README to configure.): \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[\cos\left(kx\right) + i\sin\left(kx\right)\right] \left(\cos x+i\sin x\right) \qquad \mbox{by the induction hypothesis}\\ & = \cos \left(kx\right) \cos x - \sin \left(kx\right) \sin x + i \left[\cos \left(kx\right) \sin x + \sin \left(kx\right) \cos x\right]\\ & = \cos \left[ \left(k+1\right) x \right] + i\sin \left[ \left(k+1\right) x \right] \qquad \mbox{by the trigonometric identities} \end{alignat}

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≥1.

When n = 0 the formula is true since Failed to parse (Missing texvc executable; please see math/README to configure.): \cos (0x) + i\sin (0x) = 1 + i0 = 1 , and (by convention) Failed to parse (Missing texvc executable; please see math/README to configure.): z^0 = 1 .

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

Failed to parse (Missing texvc executable; please see math/README to configure.): \begin{alignat}{2} \left(\cos x + i\sin x\right)^{n} & = \left(\cos x + i\sin x\right)^{-m}\\ & = \frac{1}{\left(\cos x + i\sin x\right)^{m}}\\ & = \frac{1}{\left(\cos mx + i\sin mx\right)}\\ & = \cos\left(mx\right) - i\sin\left(mx\right)\\ & = \cos\left(-mx\right) + i\sin\left(-mx\right)\\ & = \cos\left(nx\right) + i\sin\left(nx\right). \end{alignat}

Hence, the theorem is true for all integer values of n.

[edit] Generalization

The formula is actually true in a more general setting than stated above: if z and w are complex numbers, then

Failed to parse (Missing texvc executable; please see math/README to configure.): \left(\cos z + i\sin z\right)^w

is a multivalued function while

Failed to parse (Missing texvc executable; please see math/README to configure.): \cos (wz) + i \sin (wz)\,

is not. Therefore one can state that

Failed to parse (Missing texvc executable; please see math/README to configure.): \cos (wz) + i \sin (wz) \,

is one value of Failed to parse (Missing texvc executable; please see math/README to configure.): \left(\cos z + i\sin z\right)^w\, .

Image:CubedRoot.png

A plot on the complex plane of the cubed roots of 1.

[edit] Applications

This formula can be used to find the Failed to parse (Missing texvc executable; please see math/README to configure.): n^{th}

roots of a complex number.  If Failed to parse (Missing texvc executable; please see math/README to configure.): z
is a complex number, written in polar form as

Failed to parse (Missing texvc executable; please see math/README to configure.): z=r\left(\cos x+i\sin x\right),\,

then

Failed to parse (Missing texvc executable; please see math/README to configure.): z^{1/n} = \left[ r\left( \cos x+i\sin x \right) \right]^{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 Failed to parse (Missing texvc executable; please see math/README to configure.): k

is an integer, to get the Failed to parse (Missing texvc executable; please see math/README to configure.): n
different roots of Failed to parse (Missing texvc executable; please see math/README to configure.): z
one only needs to consider values of Failed to parse (Missing texvc executable; please see math/README to configure.): k
from Failed to parse (Missing texvc executable; please see math/README to configure.): 0
to Failed to parse (Missing texvc executable; please see math/README to configure.): n-1