TITLE: coefficients in the expansion of multivariable expression
QUESTION [8 upvotes]: Consider the expansion of the following $N$ variable expression
$$ D_N(z_1,\ldots,z_N)=\prod_{1\leq j<k\leq N}\left(1-\frac{z_j}{z_k}\right)\left( 1-\frac{z_k}{z_j} \right) $$
For example, in the case of $N=3$ we have
$$ 
\begin{align*}
D_3(z_1,z_2,z_3) & = \left(1-\frac{z_1}{z_2}\right)\left( 1-\frac{z_2}{z_1} \right)  \left(1-\frac{z_2}{z_3}\right)\left( 1-\frac{z_3}{z_2} \right) \left(1-\frac{z_3}{z_1}\right)\left( 1-\frac{z_1}{z_3} \right) = \\
& = 6 -2\frac{z_1}{z_2}-2\frac{z_2}{z_1} -2\frac{z_2}{z_3}-2\frac{z_3}{z_2} -2\frac{z_3}{z_1}-2\frac{z_1}{z_3} + \\
& +2\frac{z_1z_2}{z_3^2} +2\frac{z_3^2}{z_1z_2} +2\frac{z_2z_3}{z_1^2} +2\frac{z_1^2}{z_2z_3} +2\frac{z_3z_1}{z_2^2} +2\frac{z_2^2}{z_3z_1} - \\
& -\frac{z_1^2}{z_2^2} -\frac{z_2^2}{z_1^2} -\frac{z_2^2}{z_3^2} -\frac{z_3^2}{z_2^2} -\frac{z_3^2}{z_1^2} -\frac{z_1^2}{z_3^2}
\end{align*}
$$
It is known that the constant term of $D_N$ is $N!$ (this can be proved using the formula for the Dyson integral). I am interested in finding out if there are any patterns that describe the coefficients of the other (non-constant) terms in the expansion of $D_N$ ?
As an analogy for what I'm hoping to obtain, in the case of the well-kown multinomial expansion
$$ (z_1+z_2+\ldots+z_M)^N $$
the coefficient of any term $z_1^{k_1}z_2^{k_2}\ldots z_M^{k_M}$ can be simply expressed as
$$ \frac{N!}{k_1! k_2!\ldots k_M!} $$

REPLY [3 votes]: You may play with this method and find formulae for many coefficients. To be more concrete, below I prove using it that the coefficient of $(z_p/z_q)^c$ equals $-(N-c) (N-2)!$ and the coefficient of $z_1^c(z_2\dots z_{c+1})^{-1} $ equals $(-1)^c c! (N-c)!$ for all $c=1,2,\dots,N-1$.
Assume without loss of generality that $c_1\geqslant c_2\geqslant \ldots \geqslant c_N$, $\sum c_i=0$, and we look for $M:=[\prod z_i^{c_i}]\prod_{i<j} (1-z_i/z_j)(1-z_j/z_i)$. We have $$M=\left[z_1^{N-1+c_1}z_2^{N-1+c_2}\ldots z_N^{N-1+c_N}\right]F,\quad F=\prod_{i<j} (z_j-z_i)(z_i-z_j-\lambda_{ij}),$$
for any constants $\lambda_{ij}$.
Denote $A_i=[0,1,\dots,N-1+c_i]$. Look how can it happen that $z_i\in A_i$ and $F\ne 0$. At first, all $z_i's$ must be distinct. Choose all $\lambda_{ij}=1$. Consider the set $Z=\{z_1,\dots,z_N\}\subset A_1$. It consists of several segments of consecutive integer points, and in each segment the indices of corresponding $z_i$'s must increase (else some bracket $z_i-z_j-1$, with $i<j$, does vanish). 
If all $c_i$'s are equal to 0, this already implies that $z_i=i-1$, so the RHS of CN formula has unique summand and we get $N!$ formula for the constant term. If $c_1=c>0$, $c_{N}=c_{N-1}=\dots =c_{N-c+1}=-1$, the same choice of $\lambda$'s gives you unique non-zero value, that yields the aforementioned answer $(-1)^c c! (N-c)! $. 
Now consider, say, $c_1=c>0$, $c_2=\dots=c_{N-1}=0$, $c_N=-c$, as I said in the beginning. (This includes, in particular, the case $c_1=2,c_{N}=-2$ mentioned in the answer of Gjergji Zaimi.)
It makes sense to change $\lambda$'s: take $\lambda_{iN}=1$ for $i=1,\dots,N-1$; $\lambda_{1j}=c$ for $j=2,3,\dots,N-1$, other $\lambda$'s are equal to 0. We have $z_{N}\leqslant N-1-c$. It implies that $z_{N}+1\notin Z$. Therefore $Z=\{0,1,\dots,N-1\}\cup z_1\setminus (z_N+1)$ and $z_1\in \{N,N+1,\dots,N+c-1\}$. But $z_1-c\notin \{z_2,z_3,\dots,z_{N-1}\}$. It may happen only if $z_1-c=z_N$ or $z_1-c=z_N+1$. But $z_1-c\geqslant N-c\geqslant z_N+1$, thus we must have $z_N=N-1-c$, $z_1=N$, other $z_i$'s form a permutation of $\{0,1,\dots,N-1\}\setminus \{N-c\}$. All permutations give the same contribution to the CN formula, and after careful but straightforward cancellation we get the above answer.