how would you prove the equality written beneath please? I beg for as much detail as possible please. I never seen that equality and I want to master it.
$$ \sum_{j \in \mathbb Z} \sum_{k \in \mathbb Z} \psi_j \psi_{j + |k|} z^k = (\sum_{j \in \mathbb Z} \psi_j z^j )(\sum_{k \in \mathbb Z} \psi_k z^{-k} )$$ where $z \in \mathbb C$, the $\psi \in \mathbb R $ such that everything converges smoothly.
This answer is partially incorrect, but it should lead in the right direction
I don't seem to be getting the absolute value, but usually it's actually just a matter of renaming indices after applying the distributive law: \begin{align}(\sum_{j \in \mathbb Z} \psi_j z^j )(\sum_{k \in \mathbb Z} \psi_k z^{-k} )&= \sum_{j \in \mathbb Z} \left(\psi_j z^j \cdot \sum_{k \in \mathbb Z}\psi_k z^{-k} \right)\\ &=\sum_{j \in \mathbb Z}\psi_j\sum_{k \in \mathbb Z} \psi_k z^j z^{-k} \\ &=\sum_{j \in \mathbb Z}\psi_j\sum_{k \in \mathbb Z} \psi_k z^{j-k}\\ &=\sum_{j \in \mathbb Z}\psi_j\sum_{k \in \mathbb Z} \psi_{j-(j-k)} z^{j-k}\\ &=\sum_{j \in \mathbb Z}\psi_j\sum_{k \in \mathbb Z} \psi_{j-k} z^{k} \\ &=\sum_{j \in \mathbb Z}\sum_{k \in \mathbb Z} \psi_j \psi_{j-k} z^{k} \end{align}
where in the pre-last step I choose $k\equiv j-k$ as new index, which is not a problem since the sum is over all $k\in\mathbb{Z}$.
This index-shifting is also used in integrals: for example $$\int_{-\infty}^{\infty}f(x)\,dx=\int_{-\infty}^{\infty}f(x+k)\,dx$$ for any $k$