Dirac delta function on the complex plane?

[pellman]

Supposedly,

∫ e^{z*(z - z₀)}f(z) dz*dz

is proportional to f(z₀) much in the same way that

(1/2π)∫ e^{iy(x - x₀)}f(x) dxdy
= ∫ δ(x - x₀)f(x) dx
= f(x₀)

Is this true? Could someone help convince me of it, or point me to a text?

I would say that even if true, it would be incorrect to say that

∫ e^{z*(z - z₀)}dz* = δ(z - z₀)

because the integration over dz and dz* cannot be done independently in the same way that a surface integral over dxdy in the plane can (sometimes) be separated into independent integrations over x and y. Or can it?

 

[fresh_42]

$z^*$ is only a function in $z$, too. So what you have is $\exp(F(z))$ with $F(z)=G(z)(z-z_0)$ where $G(z)$ is the conjugation.

(I used capital letters in order to avoid confusion with your function $f$.)