De Moivre's formula

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

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

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 under the assumption that x is real, 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 a generalization of 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 Failure for non-integer powers
3 Proof by induction (for integer n)
4 Formulas for cosine and sine individually
5 Generalization
6 Applications
7 See also
8 References
9 External links

[ Derivation

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

$e^{ix} = \cos x + i\sin x\,$

and the exponential law for integer powers

$\left( e^{ix} \right)^n = e^{inx} .\,$

Then, by Euler's formula,

$e^{i(nx)} = \cos(nx) + i\sin(nx).\,$

[ Failure for non-integer powers

De Moivre's formula does not in general hold for non-integer powers. Non-integer powers of a complex number can have many different values, see failure of power and logarithm identities. However there is a generalization that the right hand side expression is one possible value of the power.

The derivation of de Moivre's formula above involves a complex number to the power n. When the power is not an integer, the result is multiple-valued, for example, when n = ½ then:

For x = 0 the formula gives 1^½ = 1

For x = 2π the formula gives 1^½ = −1

Since the angles 0 and 2π are the same this would give two different values for the same expression. The values 1 and −1 are however both square roots of 1 as the generalization asserts.

No such problem occurs with Euler's formula since there is no identification of different values of its exponent. Euler's formula involves a complex power of a positive real number and this always has a preferred value. The corresponding expressions are:

eⁱ⁰ = 1

e^iπ = −1

[ Proof by induction (for integer n)

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

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

Now, considering the case n = k + 1:

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

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 cos (0x) + i sin(0x) = 1 +i 0 = 1.

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

$\begin{align} \left(\cos x + i\sin x\right)^{n} & = \left(\cos x + i\sin x\right)^{-m}\\ & = \frac{1}{\left(\cos x + i\sin x\right)^{m}}\\ & = \frac{1}{\cos mx + i\sin mx}\\ & = \frac{\cos\left(mx\right) - i\sin\left(mx\right)}{\left(\cos mx + i\sin mx\right)\left(\cos\left(mx\right) - i\sin\left(mx\right)\right)}\\ & = \frac{\cos\left(mx\right) - i\sin\left(mx\right)}{\cos^2 mx + \sin^2 mx}\\ & = \cos\left(mx\right) - i\sin\left(mx\right)\\ & = \cos\left(-mx\right) + i\sin\left(-mx\right)\\ & = \cos\left(nx\right) + i\sin\left(nx\right). \end{align}$

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

[ Formulas for cosine and sine individually

Being an equality of complex numbers, one necessarily has equality both of the real parts and of the imaginary parts of both members of the equation. If x, and therefore also cos x and sin x, are real numbers, then the identity of these parts can be written (interchanging sides) as

$\begin{alignat}2 \cos(nx)&=\sum_{k=0}^{\lfloor n/2\rfloor}{\tbinom{n}{2k}}(-1)^k(\cos{x})^{n-2k}(\sin{x})^{2k}& &=\sum_{k=0}^{\lfloor n/2\rfloor}{\tbinom{n}{2k}}(\cos{x})^{n-2k}((\cos{x})^2-1)^k\\ \sin(nx)&=\sum_{k=0}^{\lfloor (n-1)/2\rfloor}{\tbinom{n}{2k+1}}(-1)^k(\cos{x})^{n-2k-1}(\sin{x})^{2k+1}& &=(\sin{x})\sum_{k=0}^{\lfloor(n-1)/2\rfloor}{\tbinom{n}{2k+1}}(\cos{x})^{n-2k-1}((\cos{x})^2-1)^k. \end{alignat}$

These equations are in fact even valid for complex values of x, because both sides are entire (that is, holomorphic on the whole complex plane) functions of x, and two such functions that coincide on the real axis necessarily coincide everywhere. Here are the concrete instances of these equations for n = 2 and n = 3:

$\begin{alignat}2 \cos(2x) &= (\cos{x})^2 +((\cos{x})^2-1) &&= 2(\cos{x})^2-1\\ \sin(2x) &= 2(\sin{x})(\cos{x})\\ \cos(3x) &= (\cos{x})^3 +3\cos{x}((\cos{x})^2-1) &&= 4(\cos{x})^3-3\cos{x}\\ \sin(3x) &= 3(\cos{x})^2(\sin{x})-(\sin{x})^3 &&= 3\sin{x}-4(\sin{x})^3.\\ \end{alignat}$

The right hand side of the formula for cos(nx) is in fact the value T_n(cos x) of the Chebyshev polynomial T_n at cos x.

[ Generalization

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

$\left(\cos z + i\sin z\right)^w$

is a multi-valued function while

$\cos (wz) + i \sin (wz)\,$

is not. Therefore one can state that

$\cos (wz) + i \sin (wz) \text{ is one value of } \left(\cos z + i\sin z\right)^w.\,$

[ Applications

This formula can be used to find the n^th roots of a complex number. This application does not strictly use de Moivre's formula as the power isn't an integer. However considering the right hand side to the power of n will in each case give the same value left hand side.

If z is a complex number, written in polar form as

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

then

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