next Conclusion
previous Euler's Identity
up Complex Numbers   Index   Search


De Moivre's Theorem

As a more complicated example of the value of the polar form, we'll prove De Moivre's theorem:

$\displaystyle \left[\cos(\theta) + j \sin(\theta)\right] ^n =
\cos(n\theta) + j \sin(n\theta)
$

Working this out using sum-of-angle identities from trigonometry is laborious. However, using Euler's identity, De Moivre's theorem simply ``falls out'':

$\displaystyle \left[\cos(\theta) + j \sin(\theta)\right] ^n =
\left[e^{j\theta}\right] ^n = e^{j\theta n} =
\cos(n\theta) + j \sin(n\theta)
$

Moreover, by the power of the method used to show the result, $ n$ can be any real number, not just an integer.


next Conclusion
previous Euler's Identity
up Complex Numbers   Index   Search

``Mathematics of the Discrete Fourier Transform (DFT), with Music and Audio Applications'', by Julius O. Smith III, W3K Publishing, 2003, ISBN 0-9745607-0-7.

(Browser settings for best viewing results)
(How to cite this work)
(Order a printed hardcopy)

Copyright © 2004-09-24 by Julius O. Smith III
W3K Publishing,
World Wide Web of Knowledge