# Why fourier transformation use complex number?

I know that the Fourier transform is as follows:$$\hat{f}(\xi)= \int_{-\infty}^{\infty}\exp(-\mathrm ix\xi)f(x)\mathrm{d}x$$ but I couldent understand why should use complex number $i$ in the integration. Is that means I have a real number function and after fourier transformation I get a complex function? I know that $\hat{f}(\xi)$ stand for the amplitude of each frequency. But how to understand when the amplitude is a complex number?

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I'm not sure if there's any good geometric intuition behind the Fourier transform of a complex function, but if you write $f(x) = g(x) + i h(x)$, with $g,h$ real, then the $\hat{f} = \hat{g} + i \hat{h}$, and I guess the Fourier transform of a real-valued function is easier to visualize. – Christopher A. Wong Jan 10 '13 at 12:31
@ChristopherA.Wong. I think maple is asking about the Fourier transform of a real function, but wonders why the operation itself involves the complex number $i$. – T. Eskin Jan 10 '13 at 12:35
maybe $\exp(-iz)=\cos(z) - i\sin(z)$ helps... – draks ... Jan 10 '13 at 12:43
They don't have to use complex numbers. See the Hartley transform: en.wikipedia.org/wiki/Hartley_transform . It has the same information in the sense that H[g] = Re[F[g]] - Im[F[g]], where g is a function and H and F are the Hartley and Fourier transform. But it never uses complex anything. – Jess Riedel Mar 25 '14 at 21:56
in my opinion, the reason why the Fourier transform is the most natural transform (more than the Hartley transform or the cosine transform) is that when solving the differential equation $f'(x) = a f(x)$ we need the complex exponentials, in the same way, $(e^{i wx})' = i \omega e^{i \omega x}$ i.e. to get the Fourier transform of the derivative we simply multiply by $i \omega$. finally, there is also the polynomials in $e^{i \omega x}$ point of view close to the power series. all these properties make the Fourier transform the most natural 'Hilbert space' transform. – user1952009 Jan 4 at 18:46

You need to ask yourself why we use Fourier transforms. We want to transfer the signal from the space or time domain to another domain - the frequency domain. In this domain, the signal has two "properties" - magnitude and phase. If we want to get only the signal's "power" in a specific frequency bin, we indeed need only to take the absolute value of the Fourier transform, which is real. But, the Fourier transform gives as the phase of each frequency as well. While the first's (magnitude) importance is immediate, the phase is sometimes just as important. For example, for images, most of the information is contained in the phase and NOT in the amplitude. Also, frequency responses (fourier transform) are used in digital and analog filters, and the phase plays a major role here as well, especially for audio filters where a linear phase is required: This is what enables an audio filter to process all frequencies and output them without a different delay for each frequency (which will distort the sound - imagine a filter that makes your bass sound come a little before your treble...).

So I hope I convinced you the phase is important as well as the magnitude. And in order to get these two properties, we need something other than just real numbers, we need something with magnitude and phase. Something like a complex number.

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thank you. It's really help – maple Jan 10 '13 at 13:00
You explained it very clearly, +1 – Cloverr May 21 '15 at 18:41

You can use the Fourier sine and cosine transforms instead, and these only depend on real functions. This is how Fourier originally worked. \begin{align} f & \sim \frac{1}{\pi}\int_{-\infty}^{\infty}\cos(sx)\int_{-\infty}^{\infty}f(y)\cos(sy)dy \\ & +\frac{1}{\pi}\int_{-\infty}^{\infty}\sin(sx)\int_{-\infty}^{\infty}f(y)\sin(sy)dy. \end{align} The real transforms are analogous to the real Fourier series expansions. This form is cumbersome because of requiring two separate transforms to reconstruct $f$, just as for the real Fourier series. It was realized that the sum of these could be written as $$\lim_{R\rightarrow\infty}\frac{1}{\pi}\int_{-\infty}^{\infty}\int_{-R}^{R}f(y)\cos(s(x-y))ds dy \\ =\lim_{R\rightarrow\infty}\frac{1}{\pi}\int_{-\infty}^{\infty}f(y)\frac{\sin(s(x-y))}{x-y}dy \\ =\lim_{R\rightarrow\infty}\frac{1}{2\pi}\int_{-\infty}^{\infty}f(y)\int_{-R}^{R}e^{is(x-y)}dy \\ = \lim_{R\rightarrow\infty}\frac{1}{2\pi}\int_{-R}^{R}e^{isx}\int_{-\infty}^{\infty}e^{-isy}f(y)dyds$$ This was an advanced because it simplified the expression to one transform pair instead of two. But you can work with real transforms. The sin transform is $0$ for an odd function $f$ and the cos transform is $0$ for an even function $f$, which corresponds to the traditional Fourier series.

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Just remember like this, fourier series is collection of waves with each wave having set of amplitude and frequency, when added as per specifications all the net crests and net troughs get canceled or added out each other at that time instant leaving, the desired shape on whole resultant wave

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## Actually when you expand the series all the Complex terms gets canceled out each other leaving you the Real terms.

Here frequency is in the considered periods of the cosine or sin terms. And the Amplitude of wave is in the constants before trigonometric sin and cosine terms.

## - Just to memorize easily, the expression is shortened to complex number notation :)

For understanding the derivation and cancellation of those complex terms, follow this link. http://lpsa.swarthmore.edu/Fourier/Series/DerFS.html

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$\widehat f$ needs not be real, if $f$ is not an even function. For instance the Fourier trasform of $x\mapsto f(x-\tau)$ is $$\xi\mapsto e^{i\tau\xi}\widehat f(\xi)$$ so, whenever $\widehat f$ is real, you can find a function whose Fourier transform is not real. – G. Sassatelli Dec 16 '15 at 18:00
Well, it shoud actually be $x\mapsto f(x+\tau)$, but the point stands. – G. Sassatelli Dec 16 '15 at 18:07
While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review – Alex M. Dec 16 '15 at 18:22
I didn't understood why every one is downvoting and saying that explanation is wrong. As per my understanding Fourier transform may be complex but not the Fourier Series. Am i wrong ? Hmm :/ ( I just want to extend the explanation even about Fourier series ) – Srinadh kch Dec 17 '15 at 15:18

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