Don't forget the context you're working in. You're solving for the phasor voltage across the resistor. When you measure the actual time domain voltage with, say, an oscilloscope, you'll see a sinusoid with an amplitude and a phase (referenced to the source $U_e$).
The magnitude of $U_a$ is the amplitude you will measure. The angle (phase) of $U_a$ is the phase you will measure.
UPDATE: In response to the last question: First, we limit $\omega$ to being a real number. Now, if you work out $\frac{U_a}{U_e}$ by hand (something I highly recommend if you haven't), you should get (assuming I've not made an error):
$\frac{U_a}{U_e} = \dfrac{j \omega RC}{1 - (\omega RC)^2 + j3 \omega RC}$
Now, you can "see" where this is maximum without calculation. As $\omega$ increases from zero, the real part of the denominator is decreasing and becomes zero when $\omega = \frac{1}{RC}$. From that point on, the magnitude of the denominator increases faster than the magnitude of the numerator.
In general, the voltage across and current through a capacitor or inductor do not have the same form:
$$i_C(t) = C \dfrac{dv_C}{dt} $$
$$v_L(t) = L \dfrac{di_L}{dt} $$
Thus, in general, the ratio of the voltage across to the current through is not a constant.
However, recalling that:
$$\dfrac{d e^{st}}{dt} = s e^{st} $$
where s is a complex constant $s = \sigma + j \omega$, we find that, for these excitations only:
$$i_C(t) = (sC) \cdot v_C(t)$$
$$v_L(t) = (sL) \cdot i_L(t)$$
In words, for complex exponential excitation, the voltage across and current through are proportional.
Now, there are no true complex exponential excitations but since:
$$e^{j \omega t} = \cos(\omega t) + j \sin(\omega t) $$
we can pretend that a circuit with sinusoidal excitation has complex exponential excitation, do the math, and take the real part of the solution at the end and it works.
This is called phasor analysis. The relationship between a sinusoidal voltage its phasor representation is:
$$ v_A(t) = V_m \cos (\omega t + \phi) \rightarrow \mathbf{V_a} = V_m e^{j \phi}$$
This is because:
$$v_A(t) = \Re \{V_me^{j(\omega t + \phi)}\} = \Re\{V_m e^{j \phi}e^{j \omega t}\} = \Re\{\mathbf{V_a} e^{j \omega t}\} $$
Since all of the voltages and currents in a circuit will have the same time dependent part, in phasor analysis, we just "keep track" of the complex constant part which contains the amplitude and phase information.
Thus, the ratio of the phasor voltage and current, a complex constant, is called the impedance:
$$\dfrac{\mathbf{V_c}}{\mathbf{I_c}} = \dfrac{1}{j\omega C} = Z_C$$
$$\dfrac{\mathbf{V_l}}{\mathbf{I_l}} = j \omega L = Z_L$$
$$\dfrac{\mathbf{V_r}}{\mathbf{I_r}} = R = Z_R$$
(Carefully note that though the impedance is the ratio of two phasors, the impedance is not itself a phasor, i.e., it is not associated with a time domain sinusoid).
Now, we can use the standard techniques to solve DC circuits for AC circuits where, by AC circuit, we mean: linear circuits with sinusoidal excitation (all sources must have the same frequency!) and in AC steady state (the sinusoidal amplitudes are constant with time!).
So my question is why does the magnitude of the ratio of the complex
voltage to the complex current now suddenly carries a physical meaning
(if my understanding is correct, it is the resistance which can be
measured in ohms, just like in the resistor).
Remember, the complex sources are a convenient fiction; if there were actually physical complex sources to excite the circuit, the phasor representation would by physical.
The physical sources are sinusoidal, not complex but, remarkably, we can mathematically replace the sinusoidal sources with complex sources, solve the circuit in the phasor domain using impedances, and then find the actual, physical sinusoidal solution as the real part of the complex time dependent solution.
Here's an example of the physical content of impedance:
Let the time domain inductor current be:
$i_L(t) = I_m \cos (\omega t + \phi)$
Find the time domain inductor voltage using phasors and impedance. The phasor inductor current is:
$\mathbf{I_l} = I_m e^{j\phi}$
and the impedance of the inductor is:
$Z_L = j \omega L = e^{j\frac{\pi}{2}}\omega L$
Thus, the phasor inductor voltage is:
$\mathbf{V_l} = \mathbf{I_l} Z_L = I_m e^{j\phi}e^{j\frac{\pi}{2}}\omega L = \omega L I_m e^{j(\phi + \frac{\pi}{2})}$
Converting to the time domain:
$v_L(t) = \omega L I_m \cos (\omega t + \phi + \frac{\pi}{2})$
Note that the magnitude of the impedance shows up in the amplitude of the sinusoid and the phase angle of the impedance shows up in the phase of the sinusoid.
Best Answer
The physical 'meaning' of the imaginary part of the impedance is that it represents the energy storage part of the circuit element.
To see this, let the sinusoidal current $i = I\cos(\omega t)$ be the current through a series RL circuit.
The voltage across the combination is
$$v = Ri + L\frac{di}{dt} = RI\cos(\omega t) - \omega LI\sin(\omega t)$$
The instantaneous power is the product of the voltage and current
$$p(t) = v \cdot i = RI^2\cos^2(\omega t) - \omega LI^2\sin(\omega t)\cos(\omega t) $$
Using the well known trigonometric formulas, the power is
$$p(t) = \frac{RI^2}{2}[1 + \cos(2\omega t)] - \frac{\omega LI^2}{2}\sin(2\omega t) $$
Note that the first term on the RHS is never less than zero - power is always delivered to the resistor.
However, the power for the second term has zero average value and alternates symmetrically positive and negative - the inductor stores energy half the time and releases the energy the other half.
But note that $\omega L$ is the imaginary part of the impedance of the series RL circuit:
$$Z = R + j\omega L$$
Indeed, via the complex power S, we see that the imaginary part of the impedance is related the reactive power Q
$$S = P + jQ = \tilde I^2Z = \frac{I^2}{2}Z = \frac{RI^2}{2} + j\frac{\omega L I^2}{2} $$
Thus, as promised, the imaginary part of the impedance is the energy storage part while the real part of the impedance is the dissipative part.