simple complex power: why is e^( i (2*Pi*n*t)/T ) not 1?by Aziza Tags: 2pint or t, complex, power, simple 

#1
Dec2312, 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.... 



#2
Dec2312, 07:24 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,882

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. 



#3
Dec2312, 07:38 AM

Mentor
P: 10,809

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.




#4
Dec2312, 09:29 AM

HW Helper
P: 6,189

simple complex power: why is e^( i (2*Pi*n*t)/T ) not 1?I'm just realizing that ##1^\pi## is the complex unit circle! :) 



#5
Dec2612, 01:54 PM

P: 11

Hello Aziza,
In case you're still skeptical, here's a couple of examples. If you had something like: e^{iπ/3} = (e^{iπ})^{1/3} = (e^{i/3}) = 1/2+sqrt(3)/2. Then the identity applies, but take a look here: (e^{2πi})^{i} = 1^{i} =/= e^{2π} = e^{2∏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 0^{0} for t=0, when we know for a fact that e^{iωnt} are n rotating vectors in the complex plain! 


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