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de Moivre identity (Theorem)

From the Euler relation

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

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,
$\displaystyle \cos 2 \theta + i \sin 2 \theta = (\cos \theta + i \sin \theta)^2 = \cos^2 \theta + 2 i \sin \theta \cos \theta - \sin^2 \theta.$    

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

$\displaystyle \cos 2 \theta = \cos^2 \theta - \sin^2 \theta$    

and
$\displaystyle \sin 2 \theta = 2 \sin \theta \cos \theta.$    



"de Moivre identity" is owned by Daume. [ full author list (3) | owner history (2) ]
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See Also: Euler relation, double angle identity, argument of product and quotient

Other names:  de Moivre's theorem, de Moivre's formula

Attachments:
proof of de Moivre identity (Proof) by rspuzio
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Cross-references: side, real parts, imaginary parts, angle, Euler relation
There are 7 references to this entry.

This is version 8 of de Moivre identity, born on 2002-02-16, modified 2005-06-15.
Object id is 1994, canonical name is DeMoivreIdentity.
Accessed 14345 times total.

Classification:
AMS MSC12E10 (Field theory and polynomials :: General field theory :: Special polynomials)

Pending Errata and Addenda
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Discussion
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Is this already proven? by NeuRet on 2002-05-27 11:23:18
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?
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