Identities involving derangements and roots of unity

Question

For a positive integer $n$, a derangement of $\{1,\ldots,n\}$ is a permutation $\tau$ of $\{1,\ldots,n\}$ with $\tau(j)\not=j$ for all $j=1,\ldots,n$. For convenience, we let $D(n)$ denote the set of all derangements of $\{1,\ldots,n\}$.

I have the following conjecture on identities involving both derangements and roots of unity.

Conjecture. Let $n>1$ be an integer and let $\zeta$ be a primitive $n$-th root of unity.

(i) If $n$ is even, then $$\sum_{\tau\in D(n)}\prod_{j=1}^n\frac1{1-\zeta^{j-\tau(j)}}=\frac{((n-1)!!)^2}{2^n}.\tag{1}$$ If $n$ is odd, then $$\sum_{\tau\in D(n-1)}\prod_{j=1}^{n-1}\frac1{1-\zeta^{j-\tau(j)}}=\frac1n\left(\frac{n-1}2!\right)^2.\tag{2}$$

(ii) If $n$ is even, then $$\sum_{\tau\in D(n)}\mathrm{sign}(\tau)\prod_{j=1}^n\frac1{1-\zeta^{j-\tau(j)}}=(-1)^{n/2}\frac{((n-1)!!)^2}{2^n}.\tag{3}$$ If $n$ is odd, then $$\sum_{\tau\in D(n-1)}\mathrm{sign}(\tau)\prod_{j=1}^{n-1}\frac1{1-\zeta^{j-\tau(j)}}=\frac{(-1)^{(n-1)/2}}n\left(\frac{n-1}2!\right)^2.\tag{4}$$

My numerical computation suggests that the conjecture should be true. The first assertion in part (i) of the conjecture appeared in my recent preprint available from arXiv:2108.07723. Part (ii) of the conjecture involves determinants, hence it might be not so difficult.

QUESTION. How to prove the above conjecture? Any ideas?

Your comments are welcome!

Answer 1

My idea is to define a $(n-1)\times(n-1)$ Hermitian matrix with diagonal elements equal to zero and off-diagonal elements $a_{ij}= 1/(1-x_{i-j})$, where $x_k=\zeta^k$. So the left-hand side of (4) is equal to $\det(A)$. Multiply $A$ from right by a $(n-1)\times(n-1)$ diagonal matrix $b_{ii}=(1-x_{is})$ for any fixed $s\in\{-(n-1)/2,\ldots,-1,1,(n-1)/2\}$. We can show the resulting matrix $C_s$ has an eigenvalue $s$ with eigenvector whose $j$-th element is equal to $x_{j((n-1)/2-s)}$. The difficulty is to show such an eigenvalue of $C_s$ is also an eigenvalue of $C_1$ for all s, even though numerical evaluation has suggested it. Prof. Fedor Petrov has a very elegant proof of this. See here.