delta function on a complex number for a real number we know that
$$ f(a)= \int_{-\infty}^{\infty}dx \delta (x-a)f(x) $$
but what happens for $$  \int_{-\infty}^{\infty}dx \delta (x-2i)f(x) $$ ?
is this equal to $ f(2i) $ or equal to $0 $ , of course $ i= \sqrt -1 $ a complex number :)
if i use the generalize funtion approach $ \delta (ix)= \frac{sinh(nx)}{x \pi} $ but in the limit $ n \to \infty $ this limit makes no sense
 A: Depending on the application you can generalize the dirac delta distribution using its cauchy representation to get a reasonable analytic continuation on the complex plane.
$$
f(z)=\oint_{C}\frac{f(\zeta)}{z-\zeta}
$$
Here's a short paper that uses this perscription with a short description.look at equations 4.7 and 4.8 and the paragraph preceding them: http://arxiv.org/pdf/1212.2132v1.pdf
A: One precise way to define the $\delta$-function is as a unit point measure on $\mathbb{R}$, $${\delta}(x-a)dx=d{\mu_a}(x)$$ for real $a$. Given this, we want to extend it to have a meaning for non-real $a$. This looks problematic. Secondly, in your example, you have to require some analyticity to give a meaning to $f(2i)$
A: The integral $\int\limits_{-\infty}^{\infty}\delta(x-2i)\,f(x)\,dx$ doesn't make sense, but the integral illustrated in (1) below does make sense with the variable substitution $s=x+2\,i$ given certain assumptions on $f$ (e.g. complex analytic) .
(1) $\quad\int\limits_{-\infty+2\,i}^{\infty+2\,i}\delta(s-2i)\,f(s)\,ds=\int\limits_{-\infty}^{\infty}\delta(x)\,f(x+2\,i)\,dx=f(2\,i)$
An integral of $\delta(s)$ for $s\in\mathbb{C}$ makes sense if-and-only-if it can be mapped to an integral of $\delta(x)$ for $x\in\mathbb{R}$. This can be done for integrals along lines both parallel and perpendicular to the real axis which is illustrated in the following answer I posted to a similar question.
Delta function with imaginary argument
