When is a convolution of two functions equal to their product, i.e. when is $f(t) \star g(t) \equiv \int_{-\infty}^\infty \mathrm{d}\tau f(\tau) g(t-\tau)=f(t) g(t)$ ?
Or equivalently, when is a Fourier transform of a product equal to the product of the relevant Fourier transforms, $\mathcal{F}[h(\omega) k(\omega)]=\mathcal{F}[h(\omega)] \mathcal{F}[k(\omega)]$ ?
On
Let $\omega=\tfrac{1}{2}+\tfrac{\sqrt{3}}{2}\mathrm{i}$, and let $\bar{\omega}$ be the complex conjugate of $\omega$. Let $$\begin{align} f(t)&=\mathrm{e}^{-\pi \omega t^2}\\ g(t)&=\mathrm{e}^{-\pi \bar{\omega} t^2}\text{.} \end{align}$$
Then from $$\begin{align} \omega+\bar{\omega}&=1\\ \omega\bar{\omega}&=1 \end{align}$$ we get $$\begin{align} f(t)g(t)&=\mathrm{e}^{-\pi t^2}\\ (f\star g)(t)&=\mathrm{e}^{-\pi t^2}\text{.} \end{align}$$
Suppose $f(t)=1$ and $g(t)=\delta(t)$. Then$$(f\ast g)(t)=\int_{-\infty}^\infty (1)\delta(t-\tau)\,d\tau=\delta(t)=(1)\delta(t)=f(t)g(t)$$ as was desired. Additionally, note that their Fourier transforms are $$\mathcal{F}[f](\omega)=\delta(\omega),\qquad \mathcal{F}[g](\omega)=1$$ (up to an overall multiplicative constant depending on your convention for the FT). So the Fourier transform just swaps which function is the constant function and which is the delta function.