I know that $N = (N_1, N_2)$, where $N_1, N_2$ follow normal distributions on $\mathbb{R}$ is characterized by its moments. I am also aware of Carleman condition for the moments problem for real random variable.
I have a complex random variable $X \in \mathbb{C}$ that satisfies the following properties : $\operatorname{E}[X] = 0; \operatorname{E}[X \overline{X}] = 1$. Moreover $C_m(X) = 0$ for all $m \geq 1$ where $C_m$ is the complex cumulant $C_m(X) = \tilde{C}_m(X, \dots, X)$ with $\tilde{C}_m$ defined by : for all complex random variables $Y_1, \dots, Y_m$ : $$\tilde{C}_m(Y_1, \dots, Y_n)= \sum_{ \pi \in \Pi_m} (-1)^{|\pi|}(|\pi|-1)!\prod_{I \in \pi} \operatorname{E}\Big[\prod_{i \in I} Y_i\Big]\quad (1)$$ where $\Pi_m$ denote the set of partitions of $\{1, \dots, m\}$. The size $|\pi|$ of a partition $\pi = \{I_1, \dots, I_n\}$ is $n$, where $I_i = \{a_{i, 1}, \dots, a_{i, n_i}\}$ with $\coprod_{i = 1}^n I_i = \{1, \dots, m\}$ (disjoint union). In fact $(1)$ can be inverted by : $$\operatorname{E}[Y_1 \times \dots Y_m] = \sum_{ \pi \in \Pi_m} \prod_{I \in \pi} \tilde{C}_{|I|}((Y_i)_{i \in I})$$ Thus, if $Y$ and $Z$ have the same cumulant, i.e. for all $m \geq 1$, $C_m(Y) = C_m(Z)$, then for all $k \geq 1$, $\operatorname{E}(Y^k) = \operatorname{E}(Z^k)$.
To come back to my problem, the random variable $Z = N_1 + i N_2 \in \mathbb{C}$ whith $N_1, N_2 \sim \mathcal{N}(0, 1/2)$ and $N_1$ and $N_2$ independent satifies the same condition as $X$. Indeed, by independence, $C_m(N_1 + i N_2) = C_m(N_1) + i^m C_m(N_2) = 0$ for $m \geq 3$ because the cumulants of order greater than 3 of the normal distribution are zero. Moreover $C_2(N_1 + i N_2) = C_2(N_1) - C_2(N_2) = \operatorname{E}[N_1^2] - \operatorname{E}[N_2^2] = 0$. Finally, $\operatorname{E}[Z \overline{Z}] = \operatorname{E}[N_1^2] + \operatorname{E}[N_2^2] = 1$.
My question is : is $X$ equal in distribution to $Z$ ? Or, do you have reference that can help for solving this problem ?
Context : I know that if $\tilde{C}_m(X^{\sigma_1}, \dots, X^{\sigma_m}) = \tilde{C}_m(Z^{\sigma_1}, \dots, Z^{\sigma_m})$ for all $m \geq 1$ and $\sigma_i \in \{\pm 1\}$ (notation $X^1 = X$ and $X^{-1} = \overline{X}$), and $Z$ is caracterized by its moment, then $X$ is equal to $Z$ in distribution. But, the question come from the method of moments with a rather complicated random variable $X$. I was able to prove that $\operatorname{E}[X] = 0; \operatorname{E}[X \overline{X}] = 1$, $C_m(X) = 0$ for all $m \geq 1$.
In fact, if it helps, I can also prove that $\tilde{C}_{2m+1}(X^{\sigma_1}, \dots, X^{\sigma_m}) = 0$ for all $\sigma_i \in \{\pm 1\}$ and that, $\tilde{C}_{2m}(X^{\sigma_1}, \dots, X^{\sigma_{2m}}) = 0$ if $\#\{i = 1, \dots, 2m~| \sigma_i = 1\} \neq m$. Thus, theses last two hypothesis can be assumed.