# 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 $i$. –  Thomas E. 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

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.