When dealing with Fourier transforms, it is often useful to take advantage of the following property in order to simplify work:
$$\mathcal{F}(e^{i\omega_0t}f(t))=G(\omega-\omega_0)$$
where $G(\omega)=\mathcal{F}(f(t))$, and $\omega_0 \in \mathbb{R}$. Is there a more general relation where $\omega_0 \in \mathbb{C}$? I know that, physically speaking, having a complex frequency shift (or complex time shift) does not have a meaning, but mathematically it could transform a sinusoidal function, for example, into a hyperbolic one. In short, is there a quick mathematical trick to doing a complex shift? If so, what is it? If not, why?
My guess is that a complex shift would cause the residues to move around the plane relatively to the real line, either keeping them on the same side, or moving them across. This would require a case by case study to determine the result. So there might not be a general relation of the form shown above. And come to think of it, it might be related to the stability of the system the function might model.
Any input is appreciated…
Best Answer
The answer by Hans is good and explains under what conditions this makes sense. The OP asked me to post my comment as an answer. To justify this, I'll try to expand it with a bit more detail.
There are various conventions for defining the Fourier transform. I will assume we are working with the following: $$\hat{f}(\omega) = \int_{-\infty}^{\infty}f(t) e^{-i \omega t} dt$$ Then, applying the definition to the function $g(t) = e^{i \omega_0 t}f(t)$, we can manipulate formally as follows: $$\begin{aligned} \hat{g}(\omega) &= \int_{-\infty}^{\infty}f(t)e^{i\omega_0 t}e^{-i \omega t} dt \\ & = \int_{-\infty}^{\infty}f(t) e^{-i(\omega - \omega_0)t} dt \\ &= \hat{f}(\omega - \omega_0) \end{aligned}$$ Now, under what conditions does that above formal manipulation make sense?
Note that the integral defining the Fourier transform, namely $$\hat{f}(\omega) = \int_{-\infty}^{\infty}f(t)e^{-i \omega t} dt$$ converges whenever $\int_{-\infty}^{\infty}|f(t)| dt$ conveges, in other words, whenever $f$ is absolutely integrable.
Now, if $\omega_0$ is real, then $|e^{i\omega_0 t}| = 1$, so $$\int_{-\infty}^{\infty}f(t)e^{i \omega_0 t}e^{-i \omega t} dt$$ also converges whenever $f$ is absolutely integrable.
However, if $\omega_0$ is not real, then we can write $\omega_0 = a_0 + i b_0$, where $a_0$ and $b_0$ are both real, and $a_0$ is nonzero. Then $e^{i\omega_0 t} = e^{(a_0 + ib_0)t} = e^{a_0 t} e^{ib_0 t}$, so $$|e^{i\omega_0 t}| = |e^{a_0 t}||e^{i b_0 t}| = |e^{a_0 t}| = e^{a_0 t}$$ Therefore, convergence of the integral $$\int_{-\infty}^{\infty}f(t)e^{i \omega_0 t}e^{-i \omega t} dt$$ requires convergence of $$\int_{-\infty}^{\infty}|f(t)|e^{a_0 t} dt$$ Now, $e^{a_0 t}$ grows exponentially in either the positive $t$ or negative $t$ direction, depending on the sign of $a_0$. Consequently, if the integral is to converge, we require $f$ to decay exponentially in that direction in order to compensate.
The conclusion is that when $\omega_0$ is real, the integral defining $\mathcal F(f(t) e^{i\omega t})$ always makes sense for any absolutely integrable function $f$. But when $\omega_0$ is not real, this is no longer the case. So one has to be careful about what conditions on $f$ and/or the real part of $\omega_0$ are required so that the integral will converge. The answer by Hans summarizes these conditions nicely.