Q:

Plane wave complex notation

As far as I know, the function:
$$
\vec{E}(\vec{r},t)=\vec{E_0}\cdot e^{i(\vec{k}\cdot \vec{r}-\omega t)} \hspace{2cm}(1)
$$
is a mathematical solution of the wave equation:
$$
\nabla^2 \vec{E}=\mu\varepsilon\frac{\partial^2 \vec{E}}{\partial t^2}
$$
if and only if $\omega$ satisfies the dispersion relation:
$$\omega(k)=\frac{k}{\sqrt{\mu\varepsilon}}$$
Previously I wrote "mathematical" because the complex fuction $(1)$ has no physical meaning, if we want to have a physically significative function we have to take the real part of (1) 
$$
\vec{E}(\vec{r},t)=\vec{E_0}\cdot \cos(\vec{k}\cdot \vec{r}-\omega t)
$$
Up to now there should be no problem. 
The problem arises when I consider a 1D wave packet. 
Using the complex notation:
the initial wave packet is given by:
$$
E(x,t=0)=e^{-\frac{(x-x_c)^2}{2\sigma^2}}\cdot e^{i k_{\text{c}}x}
$$
where the derm $e^{i k_{\text{c}}x}$ derives from the fact that the initial wave packet is a moving wave packet and not a static one. 
Its temporal evolution can be determined doing a Fourier transform. 
Let's indicate with $g(k)$ the Fourier transform of the initial wave packet:
$$ 
  g(k)=\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty}E(x,t=0)\cdot e^{-ikx}dx
$$
So, using the fact that every component of the spectra evolves according to a specific $\omega(k)$, I can determine the temporal evolution:
$$
E(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}g(k)e^{i(kx-\omega(k) t)}dk
\hspace{20mm}(2)
$$
Finally, according to what I said previouisly, if I am interested in physical quantities like the temporal evoultion of the amplitude, I have to take the real part of (2):
$$
E_{\text{phys}}(x,t)=\Re \left[\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}g(k)e^{i(kx-\omega(k) t)}dk\right]\hspace{20mm}(3)
$$
Reasoning on the physical meaning step by step
The initial condition is a real value wave packet:
$$
E_r(x,t=0)=\Re\left[e^{-\frac{(x-x_c)^2}{2\sigma^2}}\cdot e^{i k_{\text{c}}x}\right]=e^{-\frac{(x-x_c)^2}{2\sigma^2}}\cdot\cos(k_{\text{c}}x)
$$
so, because of I am interested in determining the temporal evolution of the wave packet, I perform a Fourier transform of the initial wave packet:
$$
 g_r(k)=\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty}E_r(x,t=0)\cdot e^{-ikx}dx
$$
where now $g_r(k)$ is could be a complex value function but this does not hurt me because this function lives in the k-space and I am interested in having real function only in the x-space. 
The problem is thai if I evolve $g_r(k)$ in the function in the Fourier space multiplying it for $e^{-i\omega(k) t}$ and then I come back in the x space antitrasforming:
$$
E_r(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}g_r(k)e^{i(kx-\omega(k) t)}dk
\hspace{20mm}(4)
$$
I obtain that (4) is a function whose amplitude becomes zero in a very rapid time, a very different behaviour from (3). 
Could anyone give me an explanation of that, or any reference in which the complex notation is well explained? 

A:

Take the wave equation
$$\nabla^2\vec{E} = \frac{1}{c^2} \frac{\partial^2 \vec{E}}{\partial t^2},$$
and let $\vec{E}(\vec{r},t)$ be a solution.  Indeed taking the real part $\Re(\vec{E}(\vec{r},t))$ yields the physical significant values. 
The initial values at $t = 0$ are $\Re(\vec{E}(\vec{r},0))$ and the problem arises here: this does not give you enough information to predict the future evolution of the field! 
For example, consider the 1D case with:
$$\begin{align}
f_1(x,t) &= e^{i(kx-\omega t)}, \\
f_2(x,t) &= \cos(kx - \omega t). 
\end{align}$$
Note that $\Re(f_1(x,0)) = \Re(f_2(x,0)) = \cos(kx)$. Therefore the two functions have the same initial values at $t = 0$, yet they evolve differently as $t$ progresses. 
To fully specify the initial state of a solution, you would need $\Re(\vec{E}(\vec{r},0))$ and one of in addition:

$\Im(\vec{E}(\vec{r},0))$,
$\Re(\frac{\partial\vec{E}}{\partial t}(\vec{r},0))$,
$\Im(\frac{\partial\vec{E}}{\partial t}(\vec{r},0))$.  [Added in Edit 2]. 

In your calculation, the moment you took $\Re(\vec{E}(\vec{r},0))$ and applied Fourier Transform, you've implicitly stipulated $\Im(\vec{E}(\vec{r},0)) = 0$. Therefore you changed the initial state to something else, so of course you'd end up with a different solution.