De Moivre's formula

For x = \frac{\pi}{6} and n = 2, de Moivre's formula asserts that \left(\cos\bigg(\frac{\pi}{6}\bigg) + i \sin\bigg(\frac{\pi}{6}\bigg)\right)^2 = \cos\bigg(2 \cdot \frac{\pi}{6}\bigg) + i \sin \bigg(2 \cdot \frac{\pi}{6}\bigg), or equivalently that \left(\frac{\sqrt{3}}{2} + \frac{i}{2}\right)^2 = \frac{1}{2} + \frac{i\sqrt{3}}{2}. In this example, it is easy to check the validity of the equation by multiplying out the left side. ==Relation to Euler's formula==

De Moivre's formula is a precursor to Euler's formula e^{ix} = \cos x + i\sin x, with expressed in radians rather than degrees, which establishes the fundamental relationship between the trigonometric functions and the complex exponential function. One can derive de Moivre's formula using Euler's formula and the exponential law for integer powers :\left( e^{ix} \right)^n = e^{inx}, since Euler's formula implies that the left side is equal to \left(\cos x + i\sin x\right)^n while the right side is equal to \cos nx + i\sin nx. == Proof by induction ==

The truth of de Moivre's theorem can be established by using mathematical induction for natural numbers, and extended to all integers from there. For an integer , call the following statement : :(\cos x + i \sin x)^n = \cos nx + i \sin nx. For , we proceed by mathematical induction. is clearly true. For our hypothesis, we assume is true for some natural . That is, we assume :\left(\cos x + i \sin x\right)^k = \cos kx + i \sin kx. Now, considering : :\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 kx + i\sin kx \right) \left(\cos x+i\sin x\right) &&\qquad \text{via induction hypothesis}\\ & = \cos kx \cos x - \sin kx \sin x + i \left(\cos kx \sin x + \sin kx \cos x\right)\\ & = \cos ((k+1)x) + i\sin ((k+1)x) &&\qquad \text{via trigonometric identities} \end{alignat} See angle sum and difference identities. We deduce that implies . By the principle of mathematical induction it follows that the result is true for all natural numbers. Now, is clearly true since . Finally, for the negative integer cases, we consider an exponent of for natural . :\begin{align} \left(\cos x + i\sin x\right)^{-n} & = \big( \left(\cos x + i\sin x\right)^n \big)^{-1} \\ & = \left(\cos nx + i\sin nx\right)^{-1} \\ & = \cos nx - i\sin nx \qquad\qquad(*)\\ & = \cos(-nx) + i\sin (-nx).\\ \end{align} The equation (*) is a result of the identity :z^{-1} = \frac{\bar z}{|z|^2}, for . Hence, holds for all integers . == Formulae for cosine and sine individually ==

For 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 , and therefore also and , are real numbers, then the identity of these parts can be written using binomial coefficients. This formula was given by 16th century French mathematician François Viète: :\begin{align} \sin nx &= \sum_{k=0}^n \binom{n}{k} (\cos x)^k\,(\sin x)^{n-k}\,\sin\frac{(n-k)\pi}{2} \\ \cos nx &= \sum_{k=0}^n \binom{n}{k} (\cos x)^k\,(\sin x)^{n-k}\,\cos\frac{(n-k)\pi}{2}. \end{align} In each of these two equations, the final trigonometric function equals one or minus one or zero, thus removing half the entries in each of the sums. These equations are in fact valid even for complex values of , because both sides are entire (that is, holomorphic on the whole complex plane) functions of , and two such functions that coincide on the real axis necessarily coincide everywhere. Here are the concrete instances of these equations for and : :\begin{alignat}{2} \cos 2x &= \left(\cos x\right)^2 +\left(\left(\cos x\right)^2-1\right) &{}={}& 2\left(\cos x\right)^2-1 \\ \sin 2x &= 2\left(\sin x\right)\left(\cos x\right) & & \\ \cos 3x &= \left(\cos x\right)^3 +3\cos x\left(\left(\cos x\right)^2-1\right) &{}={}& 4\left(\cos x\right)^3-3\cos x \\ \sin 3x &= 3\left(\cos x\right)^2\left(\sin x\right)-\left(\sin x\right)^3 &{}={}& 3\sin x-4\left(\sin x\right)^3. \end{alignat} The right-hand side of the formula for is in fact the value of the Chebyshev polynomial at . == Failure for non-integer powers, and generalization ==

De Moivre's formula does not hold for non-integer powers. The derivation of de Moivre's formula above involves a complex number raised to the integer power . If a complex number is raised to a non-integer power, the result is multiple-valued (see failure of power and logarithm identities). Roots of complex numbers A modest extension of the version of de Moivre's formula given in this article can be used to find the -th roots of a complex number for a non-zero integer . If is a complex number, written in polar form as z=r\left(\cos x+i\sin x\right), then the -th roots of are given by r^\frac1n \left( \cos \frac{x+2\pi k}{n} + i\sin \frac{x+2\pi k}{n} \right) where varies over the integer values from 0 to . This formula is also sometimes known as de Moivre's formula. Complex numbers raised to an arbitrary power Generally, if z=r\left(\cos x+i\sin x\right) (in polar form) and are arbitrary complex numbers, then the set of possible values is z^w = r^w \left(\cos x + i\sin x\right)^w = \lbrace r^w \cos(xw + 2\pi kw) + i r^w \sin(xw + 2\pi kw) | k \in \mathbb{Z}\rbrace\,. (Note that if is a rational number that equals in lowest terms then this set will have exactly distinct values rather than infinitely many. In particular, if is an integer then the set will have exactly one value, as previously discussed.) In contrast, de Moivre's formula gives r^w (\cos xw + i\sin xw)\,, which is just the single value from this set corresponding to . == Analogues in other settings ==

Hyperbolic trigonometry Since , an analog to de Moivre's formula also applies to the hyperbolic trigonometry. For all integers , : (\cosh x + \sinh x)^n = \cosh nx + \sinh nx. If is a rational number (but not necessarily an integer), then will be one of the values of . Extension to complex numbers For any integer , the formula holds for any complex number z=x+iy :( \cos z + i \sin z)^n = \cos {nz} + i \sin {nz}. where : \begin{align} \cos z = \cos(x + iy) &= \cos x \cosh y - i \sin x \sinh y\, , \\ \sin z = \sin(x + iy) &= \sin x \cosh y + i \cos x \sinh y\, . \end{align} Quaternions To find the roots of a quaternion there is an analogous form of de Moivre's formula. A quaternion in the form :q = d + a\mathbf{\hat i} + b\mathbf{\hat j} + c\mathbf{\hat k} can be represented in the form :q = k(\cos \theta + \varepsilon \sin \theta) \qquad \mbox{for } 0 \leq \theta In this representation, :k = \sqrt{d^2 + a^2 + b^2 + c^2}, and the trigonometric functions are defined as :\cos \theta = \frac{d}{k} \quad \mbox{and} \quad \sin \theta = \pm \frac{\sqrt{a^2 + b^2 + c^2}}{k}. In the case that , :\varepsilon = \pm \frac{a\mathbf{\hat i} + b\mathbf{\hat j} + c\mathbf{\hat k}}{\sqrt{a^2 + b^2 + c^2}}, that is, the unit vector. This leads to the variation of De Moivre's formula: :q^n = k^n(\cos n \theta + \varepsilon \sin n \theta). Example To find the cube roots of :Q = 1 + \mathbf{\hat i} + \mathbf{\hat j}+ \mathbf{\hat k}, write the quaternion in the form :Q = 2\left(\cos \frac{\pi}{3} + \varepsilon \sin \frac{\pi}{3}\right) \qquad \mbox{where } \varepsilon = \frac{\mathbf{\hat i} + \mathbf{\hat j}+ \mathbf{\hat k}}{\sqrt 3}. Then the cube roots are given by: :\sqrt[3]{Q} = \sqrt[3]{2}(\cos \theta + \varepsilon \sin \theta) \qquad \mbox{for } \theta = \frac{\pi}{9}, \frac{7\pi}{9}, \frac{13\pi}{9}. 2 × 2 matrices With matrices, \begin{pmatrix}\cos\phi & -\sin\phi \\ \sin\phi & \cos\phi \end{pmatrix}^n=\begin{pmatrix}\cos n\phi & -\sin n\phi \\ \sin n\phi & \cos n\phi \end{pmatrix} when is an integer. This is a direct consequence of the isomorphism between the matrices of type \begin{pmatrix}a & -b \\ b & a \end{pmatrix} and the complex plane. == References ==