Free Books
   Free Books    Mathematics of the DFT  

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 (see §3.13 for details). 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 Section:
Conclusion
Previous Section:
Euler's Identity