# De Moivre's formula

In mathematics, de Moivre's formula (a.k.a. De Moivre's theorem and De Moivre's identity), 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

${\displaystyle (\cos x+i\sin x)^{n}=\cos(nx)+i\sin(nx),}$

where i is the imaginary unit (i2 = −1). While the formula was named after de Moivre, he never stated it in his works.[1]

The formula is important because it connects complex numbers 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 nth roots of unity, that is, complex numbers z such that zn = 1.

## Derivation from Euler's formula

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

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

and the exponential law for integer powers

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

Then, by Euler's formula,

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

A more elementary motivation of the theorem comes from calculating

{\displaystyle {\begin{aligned}(\cos x+i\sin x)^{2}&=\cos ^{2}x+2i\sin x\cos x-\sin ^{2}x=(\cos ^{2}x-\sin ^{2}x)+i(2\sin x\cos x)\\&=\cos(2x)+i\sin(2x)\end{aligned}}}

where the final equality follows from the trigonometric identities

${\displaystyle \cos ^{2}x-\sin ^{2}x=\cos(2x)}$
${\displaystyle 2\sin x\cos x=\sin(2x).}$

This proves the theorem for the case n = 2. Larger values of n correspond to trigonometric identities for the triple angle, quadruple angle, etc.

## Proof by induction (for integer n)

The truth of de Moivre's theorem can be established by mathematical induction for natural numbers, and extended to all integers from there. For an integer n, call the following statement S(n):

${\displaystyle (\cos x+i\sin x)^{n}=\cos(nx)+i\sin(nx).}$

For n > 0, we proceed by mathematical induction. S(1) is clearly true. For our hypothesis, we assume S(k) is true for some natural k. That is, we assume

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

Now, considering S(k+1):

{\displaystyle {\begin{alignedat}{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{alignedat}}}

We deduce that S(k) implies S(k+1). By the principle of mathematical induction it follows that the result is true for all natural numbers. Now, S(0) is clearly true since cos (0x) + i sin(0x) = 1 +i 0 = 1. Finally, for the negative integer cases, we consider an exponent of -n for natural n.

{\displaystyle {\begin{aligned}\left(\cos x+i\sin x\right)^{-n}&=\left[\left(\cos x+i\sin x\right)^{n}\right]^{-1}\\&=\left[\cos(nx)+i\sin(nx)\right]^{-1}\\&=\cos(-nx)+i\sin(-nx).\qquad (*)\\\end{aligned}}}

The equation (*) is a result of the identity ${\displaystyle z^{-1}={\frac {\bar {z}}{|z|^{2}}}}$, for z = cos nx + i sin nx. Hence, S(n) holds for all integers n.

## Formulas for cosine and sine individually

{{#invoke:see also|seealso}} 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 using binomial coefficients. This formula was given by 16th century French mathematician Franciscus Vieta:

${\displaystyle \sin nx=\sum _{k=0}^{n}{\binom {n}{k}}\cos ^{k}x\,\sin ^{n-k}x\,\sin \left({\frac {1}{2}}(n-k)\pi \right)}$
${\displaystyle \cos nx=\sum _{k=0}^{n}{\binom {n}{k}}\cos ^{k}x\,\sin ^{n-k}x\,\cos \left({\frac {1}{2}}(n-k)\pi \right).}$

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 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:

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

## 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 n. If a complex number is raised to a non-integer power, the result is multiple-valued (see failure of power and logarithm identities). For example, when n = ½, de Moivre's formula gives the following results:

for x = 0 the formula gives 1½ = 1, and
for x = 2π the formula gives 1½ = −1.

This assigns two different values for the same expression 1½, so the formula is not consistent in this case.

On the other hand, the values 1 and −1 are both square roots of 1. More generally, if z and w are complex numbers, then

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

is multi-valued while

${\displaystyle \cos(wz)+i\sin(wz)\,}$

is not. However, it is always the case that

${\displaystyle \cos(wz)+i\sin(wz)}$

is one value of

${\displaystyle \left(\cos z+i\sin z\right)^{w}.\,}$

### 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 nth roots of a complex number (equivalently, the (1/n)th power).

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

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

then the n nth roots of z are given by

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

where k varies over the integer values from 0 to n − 1.

This formula is also sometimes known as de Moivre's formula.[2]

## Analogues in other settings

### Hyperbolic trigonometry

Since ${\displaystyle \cosh x+\sinh x=e^{x}}$, an analog to de Moivre's formula also applies to the hyperbolic trigonometry. For all ${\displaystyle n\in {\mathbb {Z} }}$, ${\displaystyle (\cosh x+\sinh x)^{n}=\cosh nx+\sinh nx}$. Also, if ${\displaystyle n\in {\mathbb {Q} }}$, then one value of ${\displaystyle (\cosh x+\sinh x)^{n}}$ will be ${\displaystyle \cosh nx+\sinh nx}$.[3]

### Quaternions

To find the roots of a quaternion there is an analogous form of de Moivre's formula. A quaternion in the form ${\displaystyle d+a{\mathbf {\hat {i}} }+b{\mathbf {\hat {j}} }+c{\mathbf {\hat {k}} }}$ can be represented in the form ${\displaystyle q=k(\cos \theta +\epsilon \sin \theta )}$ for ${\displaystyle 0\leq \theta <2\pi }$. In this representation, ${\displaystyle k={\sqrt {d^{2}+a^{2}+b^{2}+c^{2}}}}$, and the trigonometric functions are defined as ${\displaystyle \cos \theta ={\frac {d}{k}}}$ and ${\displaystyle \sin \theta =\pm {\frac {\sqrt {a^{2}+b^{2}+c^{2}}}{k}}}$. In the case that ${\displaystyle a^{2}+b^{2}+c^{2}\neq 0}$, ${\displaystyle \epsilon =\pm {\frac {a{\mathbf {\hat {i}} }+b{\mathbf {\hat {j}} }+c{\mathbf {\hat {k}} }}{\sqrt {a^{2}+b^{2}+c^{2}}}}}$ (the unit vector). This leads to the variation of De Moivre's formula:

${\displaystyle q^{n}=k^{n}(\cos n\theta +\epsilon \sin n\theta )}$[4]

#### Example

To find the cubic roots of ${\displaystyle Q=1+{\mathbf {\hat {i}} }+{\mathbf {\hat {j}} }+{\mathbf {\hat {k}} }}$, write the quaternion in the form

${\displaystyle Q=2(\cos 60^{\circ }+\epsilon \sin 60^{\circ })\qquad {\mathrm {where} }\;\epsilon ={\frac {{\mathbf {\hat {i}} }+{\mathbf {\hat {j}} }+{\mathbf {\hat {k}} }}{\sqrt {3}}}}$

Then the cubic roots are given by ${\displaystyle q}$:

${\displaystyle q={\sqrt[{3}]{2}}(\cos \theta +\epsilon \sin \theta )\qquad \theta =20^{\circ },140^{\circ },260^{\circ }}$

## References

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