"If the signal spectrum is asymmetric, the original signal spectrum will be corrupted"
If your mixer is a simple multiplication with a single oscillator, then yes, that will be the case.
As Dilip explained, real (RF) signals are always symmetrical in spectrum to the 0Hz axis. That means your original doesn't only exist around 3kHz, it also exists around -3kHz, only mirrored!
I think your image captures that pretty fine:
Now, if you shift your 3kHz signal (black) to 0Hz, the higher spectral peak ends up on the positive frequency side, and the lower on the negative frequency side.
Problem is that for real-valued signals, there's no physical difference between positive and negative frequencies – and thus, your mirrored signal also gets overlayed with this. So you have the small and the big spectral peak overlay - and since they were originally two different parts of your signal, this overlay is not actually just a "shifted copy" of your original signal, but a "corrupted" version.
The only way this wouldn't happen is if the "upper" half of your signal (3-4 kHz) and the "lower" half (2-3 kHz) were already symmetrical in spectrum to its center frequency – because then, the your mirrored spectrum would be identical to the original, and nothing was lost.
But: simply read on. The fact that your text uses negative frequencies definitely works towards showing you how a complex baseband mixer works. In two or three pages, you'll understand what needs to be done to still be