Complex numbers come into the game only because of mathematical convenience. All the quantities in AC circuit theory are, of course, real. The reason to use complex numbers is that it is easier to deal with exponential functions than trigonometric functions, since we consider an AC with a single frequency ##f=\omega/(2 \pi)##. E.g., a voltage is described as ##U(t)=U_0 \cos(\omega t)##, which is a real quantity. Now you can also describe this as ##U(t)=\text{Re}[U_0 \exp(\mathrm{i} \omega t)]##.
To describe a circuit you have to use differential equations. E.g., take a capacitor and a resistor in series with a voltage source. Then Faraday's Law, integrated along the circuit tells you that
$$Q/C+R \dot{Q}=U(t),$$
where ##Q## is the momentary charge on the capacitor. To solve this equation it's more simple to use the complex description with the exponential function for the voltage. Since the ##C## and ##R## are real you get then from any complex solution for ##Q(t)## the corresponding real solution by taking the real part.
To solve the equation note that you need the complete solution of the homogeneous equation (i.e., setting the right side to ##0##), which is
$$Q_{\text{hom}}(t)=A \exp[-t/(RC)],$$
where ##A## is an arbitray integration constant, and one special solution of the inhomogeneous equation. This we get with the ansatz
$$Q_{\text{inh}}(t)=B \exp(\mathrm{i} \omega t),$$
because the right-hand side is of this form, and taking derivatives always reproduces this exponential function. So plugging the ansatz into the equation, you get
$$\exp(-\mathrm{i} \omega t) B (1/C + \mathrm{i} \omega R)=U_0 \exp(-\mathrm{i} \omega t),$$
and this gives
$$B=\frac{U_0}{1/C+\mathrm{i} \omega R} = \frac{U_0 C}{1+\mathrm{i} \omega R C} = \frac{U_0 C (1-\mathrm{i} \omega RC)}{1+\omega^2 R^2C^2}.$$
So the complete solution is the sum of the homogeneous and the inhomogeneous solution
$$Q(t)=\frac{U_0 C (1-\mathrm{i} \omega R C)}{1+\omega^2 R^2 C^2} \exp(\mathrm{i} \omega t) + A \exp[-t/(R C)].$$
As you see, the homogeneous part goes exponentially to zero for ##t \rightarrow \infty##, i.e., after a sufficiently long time only the inhomogeneous solution is relevant.
The current for this stationary state is given by
$$I(t)=\dot{Q}_{\text{inh}}(t)=\frac{U_0 C \omega (\omega R C+\mathrm{i})}{1+\omega^2 R^2 C^2} \exp(\mathrm{i} \omega t).$$
To get the phase shift between current and voltage you have to write the prefactor in "polar form". The modulus is [EDIT: corrected in view of #8]
$$\left | \frac{U_0 C \omega (\omega R C+\mathrm{i})}{1+\omega^2 R^2 C^2} \right|=\frac{U_0 \omega C}{\sqrt{1+\omega^2 R^2 C^2}},$$
and the argument
$$\varphi=\arccos \left (\frac{\omega RC}{\sqrt{1+\omega^2 R^2 C^2}} \right).$$
So we can write
$$I(t)=\frac{U_0 \omega C}{\sqrt{1+\omega^2 R^2 C^2}} \exp[\mathrm{i}(\omega t+\varphi)],$$
i.e., ##\varphi## is the phase shift. The physical quantity is the real part of this expression, i.e., just the same but instead of the exp function the cos function,
$$I(t)=\frac{U_0 \omega C}{\sqrt{1+\omega^2 R^2 C^2}} \cos(\omega t+\varphi).$$
