Moment generating function of non-central Chi-squared distribution with complex mean? I have random variables $(X_1, \dots, X_k)$ distributed independently according to normal distributions with complex means, i.e. $j\mu_i, i=1\dots k, j^2=-1$, with unit variances.
I want to study the random variable
$$
Z = \sum_{i=1}^k X_i^2, 
$$
Can I use directly the result of the moment-generating function of Chi-squared distribution for $Z$?
$$
M(t) = \frac{\exp(\frac{\lambda t }{ (1 - 2t)})}{(1-2t)^{k/2}}.
$$
with $$\lambda = \sum_{i=1}^k j^2\mu_i^2 = - \sum_{i=1}^k \mu_i^2$$
 A: According to a comment, it looks like your problem is to compute
$$E\left[\exp(t(X-j\mu)^2)\right]$$
where $X$ has a standard Normal distribution and, quite possibly, $j^2=-1.$  But that detail scarcely matters, because in any case $(X-j\mu)^2$ has (by definition) a non-central Chi-squared distribution and this expectation is its moment generating function,
$$\psi_{(X-j\mu)^2}(t) = \exp\left(i j\mu t\right)\left(1-2it\right)^{-1/2}$$
where $i^2=-1$ is the Complex unit.  When $j=i$ this reduces to
$$\psi_{(X-i\mu)^2}(t) = \exp\left(-\mu t\right)\left(1-2it\right)^{-1/2}.$$
The justification for just plugging a complex value into the non-centrality parameter is that every step in the derivation of $\psi$ (by integrating against the standard Normal density function) goes through even with complex values: it just comes down to completing the square and noting that the resulting (exponential) integrand (a Gaussian with imaginary mean) is holomorphic in the region between the x-axis and any line parallel to it, thereby assuring the integral over that parallel line equals the corresponding real integral.
