Laplace Transforms and the imaginary input
I think you can easily see this from the \$i-v\$, characteristics of a capacitor,
$$i_c=C\frac{dv_c}{dt}$$
If voltage is of the form, \$v_c=Ve^{j\omega t}\$, then for LTI system current is of the form \$i_c = I_ce^{j\omega t}\$,
$$I_ce^{j\omega t} = C\frac{d}{dt}(Ve^{j\omega t})$$
$$I_ce^{j\omega t} = Cj\omega Ve^{j\omega t}$$
$$\frac{V_c}{I_c}=\frac{1}{j\omega C}$$
Thus, for sinusoidal inputs you can replace the capacitor by this equivalent impedance. Similar calculation can be performed for resistor or inductor to calculate their respective impedances.

In response to edit:

Intuitively, Fourier transform is decomposing a signal into its exponential components. Its amplitude at a given frequency implies how much a given frequency contributes to the total signal.
For a purely exponential signal, you just have a single frequency and so the amplitude of the exponential is itself the Fourier Transform.
Here is a more mathematical approach:
For a purely exponential input, \$v(t) = Ve^{j\omega_ot}\$, \$V\$ and \$\omega_o\$ are constants.
Taking the Fourier Transform, 
$$V(j\omega) = V\delta(\omega-\omega_o)$$
Thus, 
$$V(j\omega_o) = V$$
In other words, the amplitude of the exponential is equivalent to the Fourier Transform.