Find the cumulants of the power level of a given gaussian process I have given a zero-mean, circular symmetric, complex gaussian stochastic process $x[n] = a[n] + jb[n]$, where $a[n]$, $b[n]$, are jointly independent and $\sim N (0, \sigma^2)$. 
And I want to find the cumulants of the function:
$z[n] = \sum_{l=1}^{L} |x[l]|^2$
Whereby all samples of $x[n]$ are i.i.d. $\sim \mathcal{N}(0, \sigma ^2)$.
It is clear I can write:
$|x[l]|^2 = a[l]^2 + b[l]^2$
However, I am stuck at the further steps of the derivation.
 A: I think this is correct, but I changed the notation to something I was more comfortable with. I may have misunderstood the question. 
You gave that:
\begin{equation}
|x_l|^2 = a_l^2 + b_l^2
\end{equation}
and we know that $a_l$ and $b_l$ are independent and distributed $\textrm{N}(0,\sigma^2)$. This gives us that:
\begin{equation}
\begin{split}
a_l^2 + b_l^2 & = \sigma^2 (\alpha_l^2 +\beta_l^2)\\
& =  \sigma^2 (\mathcal{Z}_1^2 +\mathcal{X}_1^2)
\end{split}
\end{equation}
Where $\alpha,\beta \sim \textrm{N}(0,1)$ are independent, giving us that $\mathcal{Z}_1^2$ and $\mathcal{X}_1^2$ are two independent chi squared distributions. Because they are independent, we can sum them together to get:
\begin{equation}
\begin{split}
\sigma^2(a_l^2 + b_l^2) \overset{d}{=} \sigma^2\chi_2^2 \overset{d}{=} \textrm{Gamma}(1,2\sigma^2)
\end{split}
\end{equation}
Where the final gamma is in the shape-scale parameterization.
Summing these Gamma distributions from $1$ to $L$ gives us:
\begin{equation}
\sum_{l=1}^L|x_l|^2 \overset{d}{=} \sum_{l=1}^L\textrm{Gamma}(1,2\sigma^2) \overset{d}{=}L\textrm{Gamma}(1,2\sigma^2)\overset{d}{=}\textrm{Gamma}(1,L2\sigma^2)
\end{equation}
Finding the characteristic function of a Gamma gives us:
\begin{equation}
K(\textrm{Gamma}(1,L2\sigma^2)) = (1-L2\sigma^2it)^{-1} = (1-\omega it)^{-1} 
\end{equation}
Where $\omega = 2L\sigma^2$.
Taking the Taylor series of the log of this gives us:
\begin{equation}
log K(\textrm{Gamma}(1,L2\sigma^2)) = \sum_{n=1}^\infty \frac{(n-1)!(\omega it)^n}{n!} 
\end{equation}
Giving us that the $n$'th cumulant is $(n-1)!\omega^ni^n$.
