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Homede Moivre identity

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# de Moivre identity

From the Euler relation

$e^{{i\theta}}=\cos\theta+i\sin\theta$ |

it follows that

$\displaystyle e^{{i\theta\cdot n}}$ | $\displaystyle=(e^{{i\theta}})^{n}$ | ||

$\displaystyle\cos n\theta+i\sin n\theta$ | $\displaystyle=(\cos\theta+i\sin\theta)^{n}$ |

where $n\in\mathbb{Z}$.
This is called *de Moivre’s formula*, and besides being generally useful, it’s a convenient way to remember double- (and higher-multiple-) angle formulas. For example,

$\cos 2\theta+i\sin 2\theta=(\cos\theta+i\sin\theta)^{2}=\cos^{2}\theta+2i\sin% \theta\cos\theta-\sin^{2}\theta.$ |

Since the imaginary parts and real parts on each side must be equal, we must have

$\cos 2\theta=\cos^{2}\theta-\sin^{2}\theta$ |

and

$\sin 2\theta=2\sin\theta\cos\theta.$ |

Related:

EulerRelation, DoubleAngleIdentity, ArgumentOfProductAndSum, ArgumentOfProductAndQuotient

Synonym:

de Moivre's theorem, de Moivre's formula

Type of Math Object:

Theorem

Major Section:

Reference

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12E10*no label found*

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## Attached Articles

## Corrections

classification? by bwebste ✓

centering of equations by mathcam ✓

n = ? by pahio ✓

negative n by pahio ✓

centering of equations by mathcam ✓

n = ? by pahio ✓

negative n by pahio ✓

## Comments

## Is this already proven?

This entry is classified as unproven, yet the

entry itself already contains the short proof, in my opinion, for it follows easily from Euler's identity. Is there a way to change the classification of this object?

## Re: Is this already proven?

It's not really proven. Ok, so we have shown that if (exp(ix))^n = exp(nix) then the result follows, but we need to show that (exp(ix))^n does actually equal exp(nix). Recall that exp(ix) is DEFINED as isin(x) + cos(x). So we need to show the above identity; the proof for real numbers does not carry across to here.

It's like saying "dy/dx = dy/du * du/dx does not need proof, its just obvious". We are dealing with different objects than usual fractions there, and we are dealing here with different objects than real numbers.

Proof for integer n is trivial, by the way. Simply show that it is true for all n >= 0 by induction on n, and then use exp(-ix) = 1/exp(ix) (which IS provable from the definition) to show that it extends to negative numbers.

## Re: Is this already proven?

Nag the owner into check the little "contains own proof" box. They often forget to. Many entries on the "Unproven" list actually have proofs in the entries.

## Re: Is this already proven?

While I agree that the entry does contain a proof, I nevertheless added an elementary proof based on angle sum identities. I figured thatthe people mot likely to benefit from this article were newcomers to complex analysis who would find the proof based on Euler's identity a bit too slick and would feel more comfortable with a proof based on familiar trigonometric identities.