# fourier transform - why imaginary part represents the phase shift

I think this is still a topic not explained clearly and mathematicly enough in many sources (if mentioned at all).
Currently I'm working through Fourier again to fresh this up a little bit for an upcoming quantum mechanics lecture. There is one basic thing I still don't really understand.
When I transform a function $$f(t)$$ into the frequency domain, name it:

$$\tilde f(\omega) = Re[\tilde f(\omega)] + Im[\tilde f(\omega)]$$

our new function generally consists of a real and an imaginary part. The real part gives us information about the frequencies and their magnitude.
So all people say the imaginary part determines the phase shift of the corresponding frequencies, but they never go into detail. Why does the imaginary part represent the phase shift of my function $f(t)$? It obviously does, I tried it out myself. One can think of it in the way of Euler's formula (complex function in the polar form), and the imaginary part vanishes when there is no odd contribution, i.e. the sine becomes zero for a phase shift of multiples of $\pi$.

Is there any mathematicly clean way to show why and in which way this is the case?
And I'm also really interested in how I can interpret this imaginary part (quantitative and qualitative).
Like imagine you see the graph of $Im[\tilde f (\omega)]$, what can you say about the phase shift of different frequencies or about the original $f(t)$?

• The amplitude is $|\hat{f}(\omega)|$ and the phase is in $\text{arg}\ \hat{f}(\omega)$ where $f(\omega) = |\hat{f}(\omega)| e^{i\text{ arg}\ \hat{f}(\omega)}$
• The main thing you need to know is that a shift $t \mapsto t-a$ in the time domain is the same as a multiplication by $e^{-i a \omega}$ in the frequency domain
is inverse Fourier transform $\hat{g}(\omega)= \hat{f}(\omega) \phi(\omega)$ where $\phi$ rules out every frequencies except those in some interval $[a,b]$, to obtain $g(t)$ and look at $$\frac{1}{\|g\|^2} \int_{-\infty}^\infty t |g(t)|^2dt = \frac{\int_a^b \hat{g}'(\omega)\overline{\hat{g}(\omega)}d\omega}{\int_a^b |\hat{g}(\omega)|^2d\omega}$$ which indicates at which time $g(t)$ has the most energy.
• @Marc The local density of energy is $|f(t)|^2$, the energy on $[t_1,t_2]$ is $\int_{t_1}^{t_2} |f(t)|^2dt$ and Parseval theorem gives $\int_{-\infty}^\infty |f(t)|^2dt= \frac{1}{2\pi}\int_{-\infty}^\infty |\hat{f}(\omega)|^2d\omega$ (ie. the FT is an unitary operator : its inverse is its adjoint) so it makes sense to look at the energy in frequency bands $\int_a^b |\hat{f}(\omega)|^2d\omega$ Commented Oct 26, 2017 at 22:31