Fourier transform in the complex plane

I see the following formula when reading a textbook on quantum optics:

$$g(u)=\int f(\alpha)\, e^{\alpha^*u-\alpha u^*} \, \mathrm d^2\alpha,$$

$$f(\alpha)=\frac{1}{\pi^2}\int g(u)\, e^{\alpha u^*-\alpha^*u}\, \mathrm d^2u.$$

The book defines them as 'Fourier transforms in the complex plane.' Are there any proofs or justifications of them?

• Might this question be better suited for Mathematics as there is no relevant physical context given? – ACuriousMind Jan 9 '17 at 14:29

You have $i\Im (\alpha^*u) = \frac{1}{2}(\alpha^*u - u^*\alpha)$ from complex calculus; remember $$z-z^* = (a+ib)-(a-ib) = 2ib.$$ Therefore, the exponent is purely imaginary. The integration element $\mathrm d^2\alpha$ means that you have to integrate over the real part of $\alpha$ and over its imaginary part.
Substitute $\alpha = \alpha_R+i\alpha_I,u=u_R+iu_I$ and compute $\Im(\alpha^*u)$. This will give you a two-fold Fourier integral, one is over $\alpha_R$ and another is over $\alpha_I$. The second Fourier integral is the inverse transformation; it can be proven analogously.
• Normally when we do a 2D fourier transform, we only deal with a function of, say, $x$ and $y$. But now $f(\alpha)=f(\alpha_R+i\alpha_i)$. There is an $i$ hanging around. Also is $d^2\alpha=d\alpha_R+i d\alpha_i$? How to change this back to the usual $dxdy$? I don't know how to make these two changes from the original definition of Fourier transform involving only real numbers. – Physicist Jan 9 '17 at 15:18