Moments of random matrices - when are they finite I need to evaluate the moment
$$\mathbb{E} (AX)^n,$$ where A is an NxN Hermitian square matrix, and X is 
$$X=ZZ^{\ast},$$ where
$Z=\mu+Y$, where $\mu$ is mean of $Z$ and $Y$ is a zero-mean complex valued  matrix with independent elements, but they do not have equal variance.
Z is a NxK matrix where $K>N$.
Now, when are the moments finite as $n\to \infty$?
If it simplifies, let Y be complex Gaussian.
If it simplifies, let Y be IID (equal variances).
(I will use this to find the expected inverse of $AX$)
 A: For any $N\times N$ matrix $A$ we set $\DeclareMathOperator{\tr}{tr}$ $\newcommand{\bE}{\mathbb{E}}$
$$ |A|^2:=\sum_{i,j}|a_{ij}|^2 =\tr(A A^*). $$
There exists a constant $C=C(N)>0$ such that for any complex  $N\times N$   matrices $A, B$ we have
$$ |AB|\leq C |A|\cdot  |B|. $$
In particular, if $A_1,\dotsc, A_n$ are complex  $N\times N$ matrices we have
$$ |A_1\cdots A_n|\leq C^{n-1}|A_1|\cdots |A_n|. $$
We have
$$ \Bigl\vert \bE(AX)^n\Bigr\vert\leq \bE\bigl(\;|(A(X)^n|\;\bigr) \leq C^{2n-1}|A|^n \bE\bigl(\;|X|^n\;\bigr) =C^{2n-1}|A|^n \bE\bigl(\;|Z|^{2n}\;\bigr) $$
$$ = C^{2n-1}|A|^n\bE\Bigl( \;\Bigl(\; \sum_{i,j}|z_{ij}|^2\;\Bigr)^n\;\Bigr). $$
Now the question reduces to the following.  Suppose we are given independent, complex Gaussian variables $\zeta_1,\dotsc, \zeta_m$. Are the moments of $|\zeta_1|^2+\cdots +|\zeta_m|^2$ finite?  The answer is yes, because the probablity
$$ p(R)=P(|\zeta_1|^2+\cdots +|\zeta_m|^2 > R^2) $$
goes to zero exponentially fast since the random vector $(\zeta_1,\dotsc,\zeta_m)$ is Gaussian.
Next use the fact that for any nonnegative random variable $X$ we have the equality
$$ \bE(X^k) =\int_0^\infty kx^{k-1} P(X>x) dx,\;\;\forall k\geq 1. $$
