Can one compute the Fourier transform of a function whose argument is complex?

I ask because, for an engineering purpose, I have derived several Fourier properties. That is, I have derived several Fourier transforms of compositions of unspecified functions. Moreover, I have worked out the duals of my properties.

The unspecified functions have complex arguments.

When I derived my properties, the results looked okay to me; but now, months later, reviewing the work, it occurs to me to ask: can I do that? That is, can I transform a function whose argument is complex?

Or, insofar as Fourier's integral runs from $-\infty$ to $\infty$, usually along the real number line (none of my work requiring a complex contour integration), is it just meaningless to transform a function whose argument is complex?


All my calculations would be too long to burden StackExchange readers with. However, an example of one of my results is that

$$\mathcal{F}_{\omega t}\left\{\overline{h(t)} f(t)\right\} = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \overline{H\left(\tau-\frac{1}{2}\overline \omega\right)} F\left(\tau+\frac{1}{2}\omega\right) \,d\tau;$$

or, if you prefer engineering notation, that

$$\mathcal{F}_{\omega t}\left\{h^{*}(t) f(t)\right\} = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} H^{*}\bigg(\tau-\frac{\omega^{*}}{2}\bigg) F\bigg(\tau+\frac{\omega}{2}\bigg) \,d\tau.$$

Besides Fourier's integral, another integral of convolutional/correlational type appears, as you see (which may or may not be relevant to the question). In either notation, $H(\cdot)$ and $F(\cdot)$ are respectively the transforms of unspecified functions $h(\cdot)$ and $f(\cdot)$.

I do not ask you to check my result in detail (too tedious), but look at that $\overline\omega$ or $\omega^{*}$ on the right side of my transform pair. Can I do that?

Or does the logic of Fourier require the argument of a transformed function to be real?


Besides engineering books, I have on my bookshelf one book, dedicated to Fourier mathematics, dated 1963, whose author was professor of applied mathematics at the Univ. of Waterloo, Ontario. The author is Harry F. Davis. The book is Fourier Series and Orthogonal Functions.

Davis gets to the present point in his book's sect. 6.7, where, curiously, he does not seem to consider the case of a complex argument at all, despite that complex quantities otherwise appear in the section.

See also this question and answer.


1 Answer 1


Of course you can take the Fourier transform of complex function. There is nothing mathematically that prevents you from doing that. Moreover, consider that when you take the Fourier transform of a real function you get a complex function. In taking the inverse Fourier transform are not actually taking a 'Fourier' transform of a complex function. The bottom line is that it many times in my career as an engineer doing mathematical modeling. Here are some examples:

Consider the functions

$$ \psi(\tau;n)=\frac{i^n\Gamma(n+1)}{(1-i\tau)^{n+1}}\\ \gamma(\tau;n)=\tau^n e^{-\tau}u(\tau) $$

where $u(\tau)$ is the Heaviside step function.

I have shown, for example, that

$$ \mathscr{F}\{\psi(\tau;n) \}=2\pi i^n \gamma(\varpi;n)\\ \mathscr{F}\{\gamma(\tau;n) \}=i^n \psi^*(\varpi;n) $$

Here, $\tau$ is a dimensionless time and $\varpi$ is the corresponding frequency in similarity space. These is just meant as examples. $\psi$ arose from a static two-dimensional electric field , so it's real and ia=maginary parts are the $x$ and $y$ field components respectively. You can see that when $n=1$ this is a classic dipole.

  • $\begingroup$ And if $\tau$ or $\varpi$ is complex? $\endgroup$
    – thb
    Commented Jan 21, 2018 at 18:41
  • $\begingroup$ Then you have a different problem, but $\psi$ or $\gamma$ is still just a complex function. Consider a logarithmic spiral, say $z=e^{(b+i)\theta}$, with a flair coefficient of $b$. If $\theta$ was complex, then the argument would be $\Re(\theta)$ and the flair coefficient would become $b-\Im(\theta)$. $\endgroup$ Commented Jan 21, 2018 at 22:55
  • $\begingroup$ Your point is well taken. Maybe your point reveals another, clearer way I can ask an equivalent question to get at the point I wish to address, because your answer is right as far as it goes. Indeed, your answer does not so much miss the mark as my mark misses your answer. Perhaps I will return after a couple of days to ask a briefer, modified question. My problem arises when $\tau$ and $\tau^{*}$ are both simultaneously in play, which might be making my calculations nonanalytic; but I need to explain this better before asking again. $\endgroup$
    – thb
    Commented Jan 21, 2018 at 23:04

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