> 0$. The $n$th complex moment is given by $m_n(\mu_0) = \sqrt{2}i^n \cos(\frac{n-1}{4}\pi)$. In particular, $m_1(\mu_0) = \sqrt{2}i$, $m_2(\mu_0) = -1$.\
(3) More generally, $\mu_a = \frac{\sqrt{2}(x-a)^2}{\pi(1+(x-a)^4)}dx$ on $\operatorname*{\mathbb{R}}$, $a \in \operatorname*{\mathbb{R}}$. This is the convolution of $\mu_0$ and $\delta_a$. Since the binomial expansion holds for the moments of the tensor convolution, the $n$th complex moment is $m_n(\mu_a) = \sum_{k=0}^n \frac{n!}{k!(n-k)!}\sqrt{2}i^k \cos(\frac{k-1}{4}\pi) a^{n-k}$. In particular, $m_1(\mu_a) = a + \sqrt{2}i$, $m_2(\mu_a) = -1+a^2 + 2\sqrt{2}ai$.

In general, if $\mu \in \mathcal{P}_1$ is symmetric, there exist $r_n \in \operatorname*{\mathbb{R}}$ such that $m_n(\mu) = r_n i^n$. This is because the Fourier transform takes real values on $\operatorname*{\mathbb{R}}$ and therefore the coefficient of $z^n$ ($z > 0$) in the Taylor expansion is a real number for any $n$.

In addition to moments, we can extend cumulants. We need the following fact whose proof is identical to that in [@H-S].

For any $\mu \in \mathcal{P}_1$, $m_n(\mu^{\ast N})$, $m_n(\mu^{\boxplus N})$, $m_n(\mu^{\uplus N})$ and $m_n(\mu^{\rhd N})$ are all polynomials of $N$ and $m_k(\mu)$ ($1 \leq k \leq n$).

(Complex cumulants). We define the $n$th *complex tensor cumulant* $K_n^T(\mu)$, *complex free cumulant* $K_n^F(\mu)$, *complex Boolean cumulant* $K_n^B(\mu)$ and *complex monotone cumulant* $K_n^M(\mu)$ of $\mu \in \mathcal{P}_1$ respectively 