The differential equation behind the Lorentz Dispersion model is this one:
$$ \ddot x+\omega_S\dot x+\omega_0^2 x=\frac{qE_{0}}{m}\exp(i\omega_F t) $$
where $x(t)$ is the displacement of the electron from its equilibrium position.
Using the Fourier transform method the solution is easily found:
$$x(t)=\frac{qE_{0}}{m}\frac{\exp(i\omega_F t)}{\omega_0^2-\omega_F^2-i\omega_S\omega_F}$$
That is, of course, complex.
Since $$D(t)=\varepsilon_0E(t)+P(t)=\varepsilon_0E(t)+Nqx(t)=\varepsilon E(t) \tag{1}$$ this leads to the known Lorentz Model:
$$ \frac{\varepsilon}{\varepsilon_0}=1+\frac{Nq^2}{m\varepsilon_0}\frac{1}{\omega_0^2-\omega_F^2-i\omega_S\omega_F}$$
Since this is a complex quantity, it has an imaginary part that can be used to justify the fact that materials absorb energy from electromagnetic waves.
Indeed: $k=\omega\sqrt{\varepsilon\mu}=\beta +i\frac{\alpha}{2}$. So
$$E_0 e^{ikx}=E_0 e^{i\beta x}e^{-\alpha x/2} \tag{2}$$
But every single textbook says that
$E_0\exp(i\omega_F t)$ is just a short way to write $E(t)=\mathbb{Re}[E_0\exp(i\omega_F t)]=E_0\cos(\omega_F t)$. Writing again the equation:
$$ \ddot x+\omega_S\dot x+\omega_0^2 x=\frac{qE_{0}}{m}\cos(\omega_F t) $$
This time standard resolution methods lead to a real solution
$$ x(t)=Ae^{-{\omega_S\over 2}t}\cos(\omega_0 t+\phi)+B\cos(\omega_F t-\delta) $$
That will, of course, lead to a real dielectric function $\varepsilon(\omega)$. So the explanation of the absorption is now missing. How do I explain that?
Best Answer
If you phrase the equation of motion as explicitly real valued, $$ \ddot x+\omega_S\dot x+\omega_0^2 x=F(t)=\frac{qE_{0}}{m}\cos(\omega_F t), $$ then indeed you can write down the solution as $$ x(t)=Ae^{-{\omega_S\over 2}t}\cos(\omega_0 t+\phi)+B\cos(\omega_F t-\delta) $$ but because you are only interested in the steady-state behaviour you can reduce this to the form $$ x(t)=B\cos(\omega_F t-\delta).$$ This is completely correct and entirely equivalent to the complex-$x(t)$ phasor formulation. In particular, this solution does encode the absorption: it is hardwired into the phase term $\boldsymbol \delta$.
This is required by the correspondence with the phasor model, since it is easy to derive from its formulation that $\delta$ will vanish if and only if the damping $\omega_S$ is eliminated. However, it is probably more illuminating to tackle this directly, which you do by computing the instantaneous and average power delivered to the charge by the external forcing. This comes through the velocity of the particle, $$ \dot x(t) = -\omega_F B\sin(\omega_F t-\delta), $$ and it is essentially of the form $$ P(t) = \dot x(t) \cdot F(t) = -\frac{\omega_F qE_0}{m}B \sin(\omega_F t-\delta)\cos(\omega_F t). $$ However, in this form you get an awkward product, with a convoluted interplay of the field with the dephasing, so it's much cleaner to split things up a bit by cracking open the trigonometrics as \begin{align} P(t) &= -\frac{\omega_F qE_0}{m}\left[B \cos(\delta)\sin(\omega_F t)\cos(\omega_F t)-B\sin(\delta) \cos^2(\omega_F t)\right]. \end{align} This looks a bit messy, but you have now reduced the instantaneous power delivered to the charge into two very different contributions:
One is the first term, with coefficient $B \cos(\delta)$ and varying as $\sin(\omega_F t)\cos(\omega_F t)$, which takes both positive and negative values (i.e. power is delivered to, then taken from, the charge, alternatingly) and averages exactly to zero. This is known in circuit parlance as 'reactive power', and it reflects a back-and-forth exchange of energy between the field and the matter (much like capacitors and inductors bounce energy back and forth in an RLC circuit) without any net transfer of energy and hence without any absorption.
The second term, on the other hand, varies as $\cos^2(\omega_F t)$ and it is always positive, so it will average out to $\frac12$ over each cycle. This directly models the absorption, and not coincidentally it's proportional to $B\sin(\delta)$, which vanishes at $\delta=0$.
Alternatively, you could take a more macroscopic-focused approach, and try and see things through the electric polarization $P_\mathrm{dip}(t)$ generated by this motion of the charges, and here the absorption again shows up as intrinsically tied to the phase shift between the electric field and the electric polarization: waves can propagate without losses only so far as both vectors oscillate completely in sync; if the polarization lags behind the field then you can show that the amplitude will diminish.
With that in mind, you can indeed take the real-valued perspective and say that your forcing as $\cos(\omega_Ft)$ introduces a response in the form $$ F(t) = F_0 \cos(\omega_Ft) \ \rightarrow \ x(t) = B\cos(\omega_F t-\delta). $$ However, if you then simply define the "permittivity" $\varepsilon(\omega_F)$ as simply the amplitude ratio $B/F_0$, then yes, you lose information - but only because you're explicitly discarding it when you try to interpret the mathematics. If you were doing things correctly, you'd need to describe the response function as an ordered pair $(B/F_0,\delta)$ that included both the amplitude and the phase, but if you did this you'd discover that it is simply an awkward mapping of the complex number $\frac{B}{F_0}e^{i\delta}$ which directly corresponds with the complex permittivity from the phasor method. And, as usual, at that point the wise thing to do is to just give up and embrace the complex-valuedness of those quantities (while keeping mind that the physical part is still just the real part of the quantities involved.