Clarification regarding physical fields from Fourier amp's

[Max Karlsson]

My professor in Classical Electrodynamics is great and all, but sometimes he has trouble understanding what it is that I don't understand. So here I am.

Let's say we have the some sort of (monochromatic) radiating system generating a electric field with Fourier amplitude E_ω(x) and want to retrieve the physical electric field from this. One then takes the inverse Fourier transform which in this case should look something like E(t,x)=Re{E_ω(x)e^-iωt'} where t'=t-kr/ω is the retarded time (this is at least how my professor does it, and I think this makes sense).

But let's say E_ω(x) =(constant) * (e^(ikr)/r)e₁, then he will say that

E(t,x)=Re{E_ω(x)e^-iωt'}=Re{E_ω(x)e^i(kr-ωt)}=(constant) * cos(kr-ωt).

Here is where I get confused, what happens to the e^(ikr) in E_ω(x)? If this in included, my brain tells me that E(t,x)=(constant) * cos(2kr-ωt), but this is never the case.

I am probably misunderstanding something and any kind of clarification would be highly appreciated. I am including a picture of an example he gave to further explain my question. In the attached picture, I think the answer should have sin(2kr-ωt) instead of sin(kr-ωt).

http://i.imgur.com/RJuHwMK.png

 

[blue_leaf77]

Eω(x) =(constant) * (e^(ikr)/r)e1,

You can't make such assumption since the complex exponential has been included in the retarded time. If you do this, the E field expression won't satisfy the Helmholtz equation.