Proof of the phase of the Fourier Transform

Question

From wikipdia:

The Fourier transform of a function of time is itself a complex-valued function of frequency, whose absolute value represents the amount of that frequency present in the original function, and whose complex argument is the phase offset of the basic sinusoid in that frequency.

Can anyone provide a proof of this?

EDIT

Allen suggested I should start by finding $\mathcal{F}[\sin(a(x-b))]$

Using the fact that

$$\sin(x) = \frac{e^{ix} - e^{-ix}}{2i}$$

I get

$$\mathcal{F}[\sin(a(x-b))] = \int_{-\infty}^{\infty} \sin(a(x-b) -2\pi xz) dx$$

if I haven't made a mistake.

EDIT

I'll try to restate what wikipedia is saying in mathematical terms.

The Fourier transform of a function of time is itself a complex-valued function of frequency

$ \mathcal{F}[f(x)] = z(t) = r(t) e^{i \theta (t)}$

For $z \in \mathbb{C}$

whose absolute value represents the amount of that frequency present in the original function

Amount of frequency present:

$ |z(t)| = r(t)$

and whose complex argument is the phase offset

Phase Offset:

$e^{i\theta(t)}$