EM wave reflection on a perfect conductor Suppose the region $x>0$ of 3D space is a perfect conductor, and the region $x<0$ is vacuum. You send a monochromatic plane wave $\vec{E_i}=\vec{E_0}e^{i(\omega t-kx)}$ from the left to the conductor. When calculating the total field, people usually suppose that the reflected wave will be of the form $\vec{E_r}=\vec{E_0'}e^{i(\omega't+k'x)}$, then they show that $\omega=\omega'$ and $k=k'$ by saying that $\vec{E_i}+\vec{E_r}=\vec{0}$ for $x=0$ and for all $t$.
I get that, and also that we take an incident wave that is monochromatic and plane, but how can we show that the reflected wave will be a monochromatic plane wave too?
 A: The reflection coefficient is a constant independent of time and the amplitude of the reflected wave is basically -1.  Since all frequencies are reflected exactly the same way with the reflection coefficient $\Gamma=-1$, the form of the pulse is not distorted. 
The reflected plane wave only has a single frequency component so it can only remain a plane wave.  If it shifted in frequency the boundary condition on the electric field at the interface would not be true for all times.
When there is transmission, i.e when the transmission coefficient $\tau\ne 0$ and $\Gamma\ne -1$; because $\sigma$ and $\epsilon$ are frequency-dependent (often slowly varying functions of $\omega$) not all frequency components of the wave-packet will be equally transmitted or reflected, so the actual shape of the pulse can change (usually slightly) upon reflection or transmission.
An alternative approach is to compute the length of the wave vector $\vec k$: this depends on the properties of the medium in which the wave propagates: basically for vacuum $k=\omega/c$ and for air one might as well take the velocity to be $c$ as well.  To match phases in a time-independent way requires $\omega_r=\omega_i$, which in turn implies that the lengths of $\vec k_r$ and $\vec k_i$ are the same.  By translational invariance, the component of $\vec k$ parallel to the interface cannot change, which means the component that is normal must reverse its sign: in other words, assuming the interface is the $z=0$ plane, the boundary conditions show that
$$
k_{rx}=k_{ix}\, ,\qquad k_{ry}=k_{ir}\, , \tag{1}
$$
and we know that $k_r=\sqrt{k_{rx}^2+k_{ry}^2+k_{rz}^2}=
\sqrt{k_{ix}^2+k_{iy}^2+k_{iz}^2}$.  Of course as a vector $\vec k_r\ne \vec k_i$ so this leaves $k_{rz}=-k_{iz}$ as the only possible solution, i.e. the component of $\vec k$ normal to the interface changes sign without changing magnitude, since the reflected and incident waves are in the same medium.
A: The reflected wave has to have a physical origin. A source.  In your case the source is induced currents at the surface.  The incident wave induces a sinusoidal current distribution. That current distribution generates a plane wave. 
A: Let $\vec{E_r}= E_0^{'}\exp[i(\omega't - \vec{k'}\cdot \vec{r})]$
If the sum of the incident and reflected waves is the same for all $t$ at the plane $x=0$, 
$$E_0^{'}\exp[i(\omega't - k_y^{'}y +k_z^{\prime}z)] + E_0\exp[i\omega t] =0$$
To be true for all $t$ then $\omega = \omega'$. To be true for all points on the plane, $k_y^{'}y +k_z^{\prime}z =0$.
One way to arrange this is if $k_y^{'} = k_z^{'} =0$. In which case, $k_x^{'}$ is non-zero, but must have a magnitude of $\omega/c=k$, and results in a plane wave heading normally away from the plane with $E_0^{'}=-E_0$.
Any other solution for $k_y^{'}y+ k_z^{'}z=0$ would require the wavevector to be position dependent, but there is nothing about this situation that depends on position on the plane.
Note that this argument isn't any different when there is also a transmitted component, it just implies a different relationship between the incident, reflected and transmitted amplitudes.
