Why can one ignore the real part of a complex number because of Fourier transform “only using the rotational part”?

Why can one ignore the real part of a complex number because of Fourier transform "only using the rotational part"? Particularly, what property of Fourier transform is this?

• Quoting the accepted answer by Kaz: "What's happening is that $\sigma$ is being ignored because it is assumed to be zero. The reason for it is that we are looking at the response of the system to periodic (and thus non-decaying) sinusoidal signals, whereby Laplace conveniently reduces to Fourier along the imaginary axis. The real axis in the Laplace domain represents exponential decay/growth factors that pure signals do not have, and which Fourier does not model." – Saucy O'Path Aug 17 '18 at 16:16
• Physically $\omega$ encodes frequency of pure oscillation while $\sigma$ encodes dissipation. A common simplifying assumption in electronics is that the dissipation is negligible so you can deal with pure oscillation. – Ian Aug 17 '18 at 16:16
• @mavavilj More or less what you are doing in both cases is writing a function as a "linear combination" of a continuum of exponentials with different coefficients. The Laplace transform with a positive real part of $s$ corresponds to finding the coefficients of exponentials which are decaying (and oscillating) as you go forward in time. The Fourier transform with real $\omega$ corresponds to exponentials with no real part, which are purely oscillating. – Ian Aug 17 '18 at 16:20