Moments of complex random variables My question is that how much information we can get form integer moments of a complex random variable?
Let  $\mathcal{Z}$ be a complex value random variable, and assume that we can compute $$\int \mathcal{Z}^k d\mu,$$
For $k \in \mathbb{N}$ and $\mu$ be a measure.
I also am looking for an example of a complex random variable, that its moments positively approaches zero with
$$ 0< \int \mathcal{Z}^k d\mu \asymp \frac{1}{k!}.$$
If moments satisfy the above, can we occlude that $|\mathcal{Z}|$ is bounded almost surely?
One simple example: let $$ \mathcal{Z}= \sum_{m \in A} e^{imt} + \tfrac{1}{2},$$
where $A \subset \mathbb{Z}$ and $t$ is uniformly distributed in $(0, 2\pi].$ Then k-th moment of $\mathcal{Z}$ is 2^{-k}. Therefore moments decay but they fall short of decaying like $1/k!$.
 A: Let $u$ be uniformly distributed on the unit circle and let $X$ be a positive random variable with all moments finite. If $u$ and $X$ are independent then $Z:=uX$ has all moment equal to zero, because $\mathbb{E} u^{k} = 0$ for any $k\neq 0$. If you want an example with strictly positive moments then just add to $Z$ and independent positive random variable with moments decreasing very fast. You can easily cook up in this way examples that are not bounded.
EDIT: Here comes an actual example. Our random variable $Z$ will be of the form $Z=uX$, where $u$ is some distribution on the unit circle and $X$ is positive; we assume that $u$ and $X$ are independent. Then the moments are $\mathbb{E}Z^k = \mathbb{E} u^{k} \cdot \mathbb{E} X^{k}$. We want $X$ to be unbounded, so the moments of $X$ will grow to infinity at some rate, but it is not so important. We will use the oscillatory nature of $u$ to produce examples with an arbitrarily fast decay of $\mathbb{E} u^{k}$. The distribution of $u$, being supported in the unit circle, can be represented by the Fourier series $u \sim \sum_{n\in\mathbb{Z}} a_n e^{int}$. Note that $a_n= \mathbb{E} u^{-n}$. We have $a_0=1$ and we want the coefficients $(a_n)$ to decay very fast, while ensuring that the Fourier series above represents a probability measure. The moments are supposed to be real, so $a_n = a_{-n} = \overline{a_n}$. We can therefore rewrite the Fourier series as $1 + 2\sum_{n\geqslant 1} a_n \cos(nt)$. If we take $a_n$ small enough so that $|2\sum_{n\geqslant 1} a_n \cos(nt)|\leqslant 1$, then we have a probability measure. Now just pick $(a_n)$ in such a way that $\mathbb{E} u^{k} \simeq \frac{1}{k! \mathbb{E}X^{k}}$ and you get a counterexample.
A: This is to rewrite the excellent example by Mateusz Wasilewski in a more conventional form.
Let $Z:=XU$, where $X$ and $U$ are independent random variables (r.v.'s); $P(X>0)=1$; $X$ is unbounded; $EX^k<\infty$ for all natural $k$; $U=e^{iT}$; $T$ is a r.v. with values in the interval $[0,2\pi)$ and pdf $p$ given by the formula
$$p(t)=\frac1{2\pi}\,\Big(1+2\sum_{n=1}^\infty a_n\cos nt\Big)$$
for $t\in[0,2\pi)$;
$$0<a_n\sim\frac1{n!\,EX^n};$$
and $2\sum_{n=1}^\infty a_n<1$ (so that $p>0$).
Then for all natural $k$
$$EU^k=Ee^{ikT}=\int_0^{2\pi}e^{ikt}p(t)\,dt=a_k$$
and hence
$$EZ^k=EX^k\,EU^k=EX^k\,a_k\sim\frac1{k!}$$
and $EZ^k>0$, whereas the r.v. $|Z|=X$ is unbounded.
