# How can the inverse Fourier transform return a real valued function

fourier analysis

I am relatively new to Fourier transforms, so I apologize for the rather basic nature of my question, but, despite much googling, I was not able to find a clear answer on-line, so I must be stuck with some misconception.

I do understand that the Fourier transform of a real valued function f: R => R is, in general, a complex valued function, which encodes both the amplitude and phase of the harmonics that make up the function itself. What I am having trouble with is understanding how the inverse transform of the Fourier transform of a real valued function f (which is complex valued) can return the original real valued function f. As far as I can tell, the inverse Fourier transform will, in general, return a complex valued function, not a real valued one. I am obviously missing something. Could anybody point me to the right direction and/or suggest some resources that would help me clarify this point?

Thank you so much for your help.

If $$f(t)$$ is a real-valued function, then its Fourier transform $$F(\omega)$$ may be a complex function, but $$F(\omega)$$ has the property that $$F(-\omega)$$ is the complex conjugate of $$F(+\omega)$$. The inverse Fourier transform of any function having this property is real.
A related fact is that any real function that is symmetric about the origin, so that $$f(t)=f(-t)$$, has a Fourier transform $$F(\omega)$$ that is a real function.