Let $c$ and $d$ be two complex-valued sequences that are absolutely summable, i.e. \begin{align*} c: \mathbb{Z} \to \mathbb{C}, \, \sum_{n \in \mathbb{Z}} |c_n| < \infty. \end{align*} We define the convolution of $c$ and $d$ to be \begin{align*} (c \ast d)(n) = \sum_{m \in \mathbb{Z}} c_m \, d_{n-m}. \end{align*} For $k \in \mathbb{Z}$, let the Kronecker delta be $\delta_k : \mathbb{Z} \to \{0, 1 \}, \; \delta_k (n) = 1 \iff k=n$.
Assuming that $c$ and $d$ both have at least two nonzero values, is it possible to have $c \ast d = \delta_0$?
If $c$ and $d$ were real-valued and positive, this is not possible. By rescaling $c$ and $d$ we can identify them with discrete independent random variables $X$ and $Y$. The sum $X+Y$ would now be given by the $c \ast d$. If $c \ast d = \delta_0$ holds, then $0 = \text{Var}(X+Y) = \text{Var}(X)+\text{Var}(Y)$. Thus $X$ and $Y$ have $0$ variance and must themselves be given by densities $\delta_n$ and $\delta_m$.
Intuitively, I'd think that a similar variance argument should still hold in $\mathbb{C}$. However, we can't just use their absolute values.
