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$ 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 $

It is correct that:

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


However

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

So there error would be that (2 Pi t) is the theta, so it must stay in the theta when converted into sin and cos.

Therefore the "t" may not be separated into the exponent in that case.

But if "t" starts off in the exponent, then the result will equal 1.


A great source would be this web site:

http://www00.wolframalpha.com/input/?i=(e^(2+i+pi+))^t

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva