note that 
$$e^{a\partial/\partial x}f(x)=f(x+a)$$
is the translation operator, so your exponent of the delta function gives $2\pi \delta(x-in)$, which is indeed consistent with

$$\int_{-\infty}^\infty e^{izy}dy=2\pi\delta(z)$$

for $z=x-in$.

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All of this is purely formal, but there have been attempts to put it on a more secure ground by defining the delta function of a complex argument as an "ultradistribution". See for example <A HREF="https://books.google.it/books?id=uOmiAgAAQBAJ&pg=121#v=onepage&q&f=false">Theories of Generalised Functions</A> (page 121) and <a href="https://books.google.nl/books?hl=en&lr=&id=6wPLAgAAQBAJ&oi=fnd&pg=204">Distribution Theory and Transform Analysis</a> (page 204). <I>Caveat emptor.</I>