Simplification with exponentials and imaginary numbers

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SUMMARY

The discussion centers on the application of Euler's identity, specifically exp(iπ) = exp(-iπ) = -1, to simplify the integration of the equation exp(0.5it) + exp(-0.5it). The simplification leads to a cosine function, cos(0.5t), but the user encounters a problem when the numerator becomes zero due to the misapplication of the identity. The key takeaway is that while Euler's identity holds true, it does not apply universally for all values of x, particularly when x is not an odd integer.

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JieXian
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Hi, I'm using euler's identity : exp(i∏) = exp (-i∏) = -1 to simplify the equation after integrating it.

[PLAIN]http://img443.imageshack.us/img443/5504/captureikm.jpg
Note: the equation to be integrated is exp(0.5it) + exp(-0.5it) and they have simplified it, it was actually a cos(0.5t) function and the 1/2 from cos(t) = 0.5[ exp(it) + exp(-it) ] isn't included.

However, after trying I got the numerator to be 0, because of the exp(i∏) = exp (-i∏) problem.

Letting x and y be any variable,

exp(i∏ x) - exp(i∏ y) - [ exp(i(-∏) x) - exp(i(-∏) y) ]

= exp(i∏ x) - exp(i(-∏) x) - exp(i∏ y) + exp(i(-∏) y) --------- (rearranging)

= exp(i∏ x) - exp(i(∏) x) - exp(i∏ y) + exp(i(∏) y) ----------- (since exp(i∏) = exp (-i∏))

= 0

Please help me, thank you very much.
 
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whilst it is true
[tex] e^{i \pi} = -1 = e^{-i \pi}[/tex]

unless x=2n+1 for some integer n, then
[tex] e^{i \pi x} \neq e^{-i \pi x}[/tex]

this makes your last step invalid, though I have not checked the rest
 

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