View source for De Moivre's formula

{{Use American English|date = March 2019}}
{{Short description|1 = Theorem: (cos(x) + i sin(x))^n = cos(nx) + i sin(nx)}}
In [[mathematics]], '''de Moivre's formula ''' (also known as '''de Moivre's theorem''' and '''de Moivre's identity''') states that for any [[real number]] {{mvar|x}} and [[integer]] {{mvar|n}} it holds that

:<math>\big(\cos (x) + i \sin (x)\big)^n = \cos (nx) + i \sin (nx),</math>

where {{mvar|i}} is the [[imaginary unit]] ({{math|''i''<sup>2</sup> {{=}} −1}}). The formula is named after [[Abraham de Moivre]], although he never stated it in his works.<ref>{{cite book |first1=Margaret L. |last1=Lial |first2=John |last2=Hornsby |first3=David I. |last3=Schneider |first4=Daniels |last4=Callie J. |title=College Algebra and Trigonometry|edition=4th |year=2008 |location=Boston |publisher=Pearson/Addison Wesley |isbn=9780321497444 |page=792}}</ref> The expression {{math|cos(''x'') + ''i'' sin(''x'')}} is sometimes abbreviated to {{math|[[cis (mathematics)|cis]](''x'')}}.

The formula is important because it connects complex numbers and [[trigonometry]]. By expanding the left hand side and then comparing the real and imaginary parts under the assumption that {{mvar|x}} is real, it is possible to derive useful expressions for {{math|cos(''nx'')}} and {{math|sin(''nx'')}} in terms of {{math|cos(''x'')}} and {{math|sin(''x'')}}.

As written, the formula is not valid for non-integer powers {{mvar|n}}. However, there are generalizations of this formula valid for other exponents. These can be used to give explicit expressions for the {{mvar|n}}th [[Root of unity|roots of unity]], that is, complex numbers {{mvar|z}} such that {{math|''z<sup>n</sup>'' {{=}} 1}}.

==Example==
For <math> x = 30^\circ</math> and <math> n = 2</math>, de Moivre's formula asserts that
<math display=block>\left(\cos(30^\circ) + i \sin(30^\circ)\right)^2 = \cos(2 \cdot 30^\circ) + i \sin (2 \cdot 30^\circ),</math>
or equivalently that
<math display=block>\left(\frac{\sqrt{3}}{2} + \frac{i}{2}\right)^2 = \frac{1}{2} + \frac{i\sqrt{3}}{2}.</math>
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]]
<math display=block>e^{ix} = \cos x + i\sin x,</math>
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 [[exponentiation|exponential law]] for integer powers

:<math>\left( e^{ix} \right)^n = e^{inx}, </math>

since Euler's formula implies that the left side is equal to <math>\left(\cos x + i\sin x\right)^n</math> while the right side is equal to 

:<math>e^{inx} = \cos(nx) + i\sin(nx).</math>

== 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 {{mvar|n}}, call the following statement {{math|S(''n'')}}:

:<math>(\cos x + i \sin x)^n = \cos nx + i \sin nx.</math>

For {{math|''n'' > 0}}, we proceed by [[mathematical induction]]. {{math|S(1)}} is clearly true. For our hypothesis, we assume {{math|S(''k'')}} is true for some natural {{mvar|k}}. That is, we assume

:<math>\left(\cos x + i \sin x\right)^k = \cos kx + i \sin kx. </math>

Now, considering {{math|S(''k'' + 1)}}:

:<math>\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{by the 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{by the trigonometric identities}
\end{alignat}</math>

See [[Trigonometric identities#Angle sum and difference identities|angle sum and difference identities]].

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

:<math>\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 (*) \\
\end{align}</math>
The equation (*) is a result of the identity
:<math>z^{-1} = \frac{\bar z}{|z|^2},</math>
for {{nowrap|''z'' {{=}} cos (''nx'') + ''i'' sin (''nx'')}}. Hence, {{math|S(''n'')}} holds for all integers {{mvar|n}}.

== Formulae for cosine and sine individually ==
{{See also|List of trigonometric identities}}
Being an equality of [[complex number]]s, one necessarily has equality both of the [[real part]]s and of the [[imaginary part]]s of both members of the equation. If {{mvar|x}}, and therefore also {{math|cos ''x''}} and {{math|sin ''x''}}, are [[real number]]s, then the identity of these parts can be written using [[binomial coefficient]]s. This formula was given by 16th century French mathematician [[François Viète]]:

:<math>\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}</math>

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 {{mvar|x}}, because both sides are [[entire function|entire]] (that is, [[Holomorphic function|holomorphic]] on the whole [[complex plane]]) functions of {{mvar|x}}, and two such functions that coincide on the real axis necessarily coincide everywhere. Here are the concrete instances of these equations for {{math|''n'' {{=}} 2}} and {{math|''n'' {{=}} 3}}:

:<math>\begin{align}
 \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{align}</math>

The right hand side of the formula for {{math|cos ''nx''}} is in fact the value {{math|''T''<sub>''n''</sub>(cos ''x'')}} of the [[Chebyshev polynomial]] {{math|''T''<sub>''n''</sub>}} at {{math|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 {{mvar|n}}. If a complex number is raised to a non-integer power, the result is [[multiple-valued]] (see [[Exponentiation#Failure of power and logarithm identities|failure of power and logarithm identities]]). For example, when {{math|''n'' {{=}} {{sfrac|1|2}}}}, de Moivre's formula gives the following results:

:for {{math|''x'' {{=}} 0}} the formula gives 1<sup>{{frac|1|2}}</sup>&nbsp;=&nbsp;1, and
:for {{math|''x'' {{=}} 2{{pi}}}} the formula gives 1<sup>{{frac|1|2}}</sup>&nbsp;=&nbsp;−1.

This assigns two different values for the same expression 1<sup>{{frac|1|2}}</sup>, 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 {{mvar|z}} and {{mvar|w}} are complex numbers, then

:<math>\left(\cos z + i\sin z\right)^w</math>

is multi-valued while

:<math>\cos wz + i \sin wz</math>

is not. However, it is always the case that

:<math>\cos wz + i \sin wz</math>

is one of the values of

:<math>\left(\cos z + i\sin z\right)^w.</math>

===Roots of complex numbers===
A modest extension of the version of de Moivre's formula given in this article can be used to find [[nth root|the {{mvar|n}}th roots]] of a complex number (equivalently, the power of {{math|{{sfrac|1|''n''}}}}).

If {{mvar|z}} is a complex number, written in [[polar form]] as

: <math>z=r\left(\cos x+i\sin x\right),</math>

then the {{mvar|n}} {{mvar|n}}th roots of {{mvar|z}} are given by

: <math>r^\frac1n \left( \cos \frac{x+2\pi k}{n} + i\sin \frac{x+2\pi k}{n} \right)</math>

where {{mvar|k}} varies over the integer values from 0 to {{math|''n'' − 1}}.

This formula is also sometimes known as de Moivre's formula.<ref>{{springer|title=De Moivre formula|id=p/d030300}}</ref>

== Analogues in other settings ==

===Hyperbolic trigonometry===

Since {{math|cosh ''x'' + sinh ''x'' {{=}} ''e<sup>x</sup>''}}, an analog to de Moivre's formula also applies to the [[Hyperbolic function|hyperbolic trigonometry]]. For all {{math|''n'' ∈ ℤ}},
:<math>(\cosh x + \sinh x)^n = \cosh nx + \sinh nx.</math>
Also, if {{math|''n'' ∈ ℚ}}, then one value of {{math|(cosh ''x'' + sinh ''x'')<sup>''n''</sup>}} will be {{math|cosh ''nx'' + sinh ''nx''}}.<ref>{{cite journal|last=Mukhopadhyay|first=Utpal|title=Some interesting features of hyperbolic functions|journal=Resonance|date=August 2006|volume=11|issue=8|pages=81–85|doi=10.1007/BF02855783}}</ref>

===Extension to complex numbers===

The formula holds for any complex number <math>z=x+iy</math>
:<math>( \cos z + i \sin z)^n = \cos {nz} + i \sin {nz}.</math>
where
: <math>\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}</math>

===Quaternions===

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

:<math>q^n = k^n(\cos n \theta + \varepsilon \sin n \theta).</math><ref>{{cite journal|last=Brand|first=Louis|title=The roots of a quaternion|journal=The American Mathematical Monthly|date=October 1942|volume=49|issue=8|pages=519–520|jstor=2302858|doi=10.2307/2302858}}</ref>

====Example====
To find the [[cube root]]s of
:<math>Q = 1 + \mathbf{\hat i} + \mathbf{\hat j}+ \mathbf{\hat k},</math>
write the quaternion in the form
:<math>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}.</math>
Then the cube roots are given by:
:<math>\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}.</math>
=== {{gaps|2|×|2}} matrices ===
Consider the following matrix
<math>A=\begin{pmatrix}\cos\phi & \sin\phi \\ -\sin\phi & \cos\phi \end{pmatrix}</math>. Then <math>\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}</math>. This fact (although it can be proven in the very same way as for complex numbers) is a direct consequence of the fact that the space of matrices of type <math>\begin{pmatrix}a & b \\ -b & a \end{pmatrix}</math> is isomorphic to the space of complex numbers.

== References ==
*{{cite book|first=Milton |last=Abramowitz |first2=Irene A. |last2=Stegun |title=[[Abramowitz and Stegun|Handbook of Mathematical Functions]] |year=1964 |publisher=Dover Publications |location=New York |isbn=0-486-61272-4 |page=74 }}.

{{Reflist}}

== External links ==
* [http://demonstrations.wolfram.com/DeMoivresTheoremForTrigIdentities/ De Moivre's Theorem for Trig Identities] by Michael Croucher, [[Wolfram Demonstrations Project]].

[[Category:Theorems in complex analysis]]
[[Category:Articles containing proofs]]