The domain of the Fourier transform

  • #1
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Summary:

How does the Fourier transform handle functions of complex variables when only integrating over the reals?
Given the domain of the integral for the Fourier transform is over the real numbers, how does the Fourier transform transform functions whose independent variable is complex?

For example, given
\begin{equation}
\begin{split}
\hat{f}(k_{\mathbb{C}}) &= \int_{\mathbb{R}} f(z_{\mathbb{C}}) e^{2 \pi i k_{\mathbb{C}} z_{\mathbb{C}}} d z_{\mathbb{C}}
\end{split}
\end{equation}
where ##z_{\mathbb{C}} = z_1 + z_2 i## and ##k_{\mathbb{C}} = k_1 + k_2 i##.

It seems that evaluating ##f(z_{\mathbb{C}}) e^{2 \pi i k_{\mathbb{C}} z_{\mathbb{C}}}## over the reals includes only ##z_1## and ignores ##z_2 i## in ##f(z_{\mathbb{C}}) ##.
 
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Answers and Replies

  • #2
Delta2
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The way I 've been taught fourier transform is that it applies to complex valued functions of a real variable and it gives as result also a complex valued function of a real variable.

That is, it is ##f:\mathbb{R}\to\mathbb{C}## and its fourier transform ##\hat f:\mathbb{R}\to\mathbb{C}## so the integrals of the Fourier and the inverse Fourier transform are inevitably over the real line.

The integral you are proposing here seems to be an interesting generalization of the Fourier transform so that it can include functions ##f:\mathbb{C}\to\mathbb{C}##, that is complex valued functions of a complex variable. Only thing is that you have to make it a contour integral in order to make it integrate over the complex plane. Or simply a double integral over the real line like for example
$$\int_\mathbb{R}\int_\mathbb{R}f(z_1+z_2i)e^{2\pi i k_{\mathbb{C}}(z_1+z_2i)}dz_1dz_2$$

I cant tell if such an integral has interesting theoretical or practical applications.
 
  • #3
RPinPA
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There is such a thing as a two-dimensional Fourier transform, for instance the spatial Fourier transform of an image. It consists of Fourier transforms applied independently in the x and y directions.

Perhaps that's what you need to do here, transform Re(z) and Im(z) to Re(k) and Im(k).
 
  • #4
Svein
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... and remember that [itex]i=e^{\frac{i\pi}{2}} [/itex], which means that the transformation of a purely imaginary variable is the same as the transformation of the real version of the variable rotated π/2. This leads to the Laplace transformation.
 

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