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? 
 A: 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.
A: This is analogous to ordinary Fourier transformation. Bring the expression for $f\left(\alpha\right)$ into the r.h.s. of the equation for $g\left(u\right)$, it goes like
$$
\text{r.h.s.} = \frac{1}{\pi^2}\int\mathrm{d}^2\alpha\ \mathrm{d}^2v\  g\left(v\right) \mathrm{e}^{v^*\alpha - v\alpha^*}\mathrm{e}^{\alpha^*u-\alpha u^*}\\
=\frac{1}{\pi^2}\int\mathrm{d}^2\alpha\ \mathrm{d}^2v\ g\left(v\right)\mathrm{e}^{\alpha^*\left(u-v\right)-\alpha\left(u-v\right)^*}\\
=\int\mathrm{d}^2v\ g\left(v\right)\delta^2\left(u-v\right)\\
=g\left(u\right)
$$
where the integration of the delta-function can be proved in the following way. Denote $\alpha = x+\mathrm{i}y$ and $w=u+\mathrm{i}v$, we have $\mathrm{d}^2\alpha=\mathrm{d}x\ \mathrm{d}y$ and $\alpha^*w - \alpha w^* = 2\mathrm{i}\left(vx-uy\right)$.
$$
\int\mathrm{d}^2\alpha\ \mathrm{e}^{\alpha^*w-\alpha w^*} = \int\mathrm{d}x\ \mathrm{e}^{2\mathrm{i}vx}\cdot\int\mathrm{d}y\ \mathrm{e}^{-2\mathrm{i}uy} = \pi\delta\left(v\right)\cdot\pi\delta\left(u\right) = \pi^2\delta^2\left(w\right).
$$
(which is eqn 3.99 in this textbook.)
