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* [[Answer Here]]
 
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On the other hand, if n is an integer, it is true that
  
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<math> e^{j 2 \pi n}  = \left( e^{j 2 \pi} \right)^n= \left( cos{2 \pi } + j sin{2 \pi } \right)^n= 1^n =1 </math>
  
 
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Latest revision as of 09:24, 23 July 2009

Adam Frey's Euler Identity Summary

The identity is a special case of Euler's formula from complex analysis, which states that where j = i = $ \sqrt{-1} $

$ e^{jx} = \cos x + j \sin x \,\! $

for any real number x. (Note that sine and cosine should be in radians)

In particular,

$ e^{j \pi} = \cos \pi + j \sin \pi.\,\! $

We know from trig identities that:

$ \cos \pi = -1 \, \! $

and

$ \sin \pi = 0,\,\! $


which results in

$ e^{j \pi} = -1,\,\! $

which gives the identity

$ e^{j \pi} +1 = 0.\,\! $


Also useful is the relationship in splitting sine and cosine is where

$ \cos wt = \frac{e^{j w}}{2} + \frac{e^{-j w}}{2} \,\! $

and

$ \sin wt = \frac{e^{j w}}{2j} - \frac{e^{-j w}}{2j} \,\! $

--Freya 15:04, 20 July 2009 (UTC)

A related mystery: why is the following wrong? Can somebody please explain?

$  e^{j 2 \pi t}  = \left( e^{j 2 \pi} \right)^t= \left( cos{2 \pi } + j sin{2 \pi } \right)^t= 1^t =1  $

On the other hand, if n is an integer, it is true that

$ e^{j 2 \pi n} = \left( e^{j 2 \pi} \right)^n= \left( cos{2 \pi } + j sin{2 \pi } \right)^n= 1^n =1 $

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