simple complex power: why is e^( i (2*Pi*n*t)/T ) not 1?


by Aziza
Tags: 2pint or t, complex, power, simple
Aziza
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#1
Dec23-12, 04:07 AM
P: 173
the complex form of Fourier series is:

f(t) = Ʃ c*e^[iωnt]
where c are the coefficients, the sum is from n= -inf to +inf; ω= 2*pi/T, where T is period....

but if you just look at e^[iωnt] = e^[ i (2*pi*n*t)/T] = {e^[ i (2*pi*n)] }^(t/T)

where I just took out the t/T....
well, e^[ i (2*pi*n)] = 1, since n is integer....and (1)^(t/T) is still equal to 1....so shouldnt the complex Fourier form just reduce to f(t) = Ʃ c ????

I feel i must be doing something stupid, if someone could just please point out what exactly....
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HallsofIvy
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#2
Dec23-12, 07:24 AM
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Every complex number, except 0, but including 1, has n distinct nth roots.
When dealing with complex numbers, 1 to a fractional power is not just 1.
mfb
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#3
Dec23-12, 07:38 AM
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In general, e^(a*b) != (e^a)^b with complex numbers a,b - unless you care about the phase of the expression in some other way.

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Dec23-12, 09:29 AM
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simple complex power: why is e^( i (2*Pi*n*t)/T ) not 1?


Quote Quote by HallsofIvy View Post
When dealing with complex numbers, 1 to a fractional power is not just 1.
Nice one!
I'm just realizing that ##1^\pi## is the complex unit circle! :)
HassanEE
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#5
Dec26-12, 01:54 PM
P: 11
Hello Aziza,
In case you're still skeptical, here's a couple of examples. If you had something like:

eiπ/3 = (e)1/3 = (ei/3) = 1/2+sqrt(3)/2.

Then the identity applies, but take a look here:

(e2πi)i = 1i =/= e-2π = e2∏ii

The identity does not hold, and you can't really guess when and where it does, or doesn't.
In your case, you know it doesn't work because you get such an odd result, 0 for all t,and 00 for t=0, when we know for a fact that eiωnt are n rotating vectors in the complex plain!


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