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

1. Dec 23, 2012

### Aziza

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. Dec 23, 2012

### HallsofIvy

Staff Emeritus
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. Dec 23, 2012

### Staff: Mentor

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. Dec 23, 2012

### I like Serena

Nice one!
I'm just realizing that $1^\pi$ is the complex unit circle! :)

5. Dec 26, 2012

### HassanEE

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!

Last edited: Dec 26, 2012