I have stumbled across the following conjecture. For $m\in\Bbb Z_{\ge2}$ let $\zeta_m=e^{2i\pi/m}$. Thus we have $\zeta_m^m=1$ and $\sum_{l=0}^{m-1}\zeta_m^l=0$.
Conjecture: for all integers $j$ with $0\le j\le m-1$, we have $$p_{m,j}(z):=\sum_{k=0}^{m-1}\frac{\zeta_m^{-kj}}{1-\zeta_m^kz}=\frac{mz^j}{1-z^m}.\tag1$$
Thoughts: I have shown that $(1)$ holds for all $m$ when $j=0$, but no other general cases.
The conjecture is easy to show true for $m=2$, so that is left to the reader as a fun exercise. The case $m=3$ is a bit more messy, so I present it here. The messiness grows quadratically (the case-by-case proof for the $m=n$ involves $n^2-n$ terms at least) with $m$, and I have thus far not had the patience to check any $m\ge4$.
Proof: ($m=3$)
Since $1-z^m=\prod_{k=0}^{m-1}(1-\zeta_m^kz)$, we have that $$ (1-z^3)p_{3,1}(z)=(1-\zeta_3z)(1-\zeta_3^2z)+\zeta_3^2(1-z)(1-\zeta_3^2z)+\zeta_3(1-z)(1-\zeta_3z). $$ This is $$\begin{align} (1-z^3)p_{3,1}(z)=&1-(\zeta_3+\zeta_3^2)z+\zeta_3^3z^2\\ &+\zeta_3^2(1-(1+\zeta_3^2)z+\zeta_3^2z^2)\\ &+\zeta_3(1-(1+\zeta_3)z+\zeta_3z^2)\\ =&1+\zeta_3+\zeta_3^2\\ &-3(\zeta_3+\zeta_3^2)z\\ &+(1+\zeta_3+\zeta_3^2)z^2\\ =&3z. \end{align}$$ Next up, we have $$(1-z^3)p_{3,2}(z)=(1-\zeta_3z)(1-\zeta_3^2z)+\zeta_3(1-z)(1-\zeta_3z)+\zeta_3^2(1-z)(1-\zeta_3^2z).$$ This is $$\begin{align} (1-z^3)p_{3,2}(z)=&1-(\zeta_3+\zeta_3^2)z+z^2\\ &+\zeta_3(1-(1+\zeta_3^2)z+\zeta_3^2z^2)\\ &+\zeta_3^2(1-(1+\zeta_3)z+\zeta_3z^2)\\ =&1+\zeta_3+\zeta_3^2\\ &+(1+\zeta_3+\zeta_3^2)z\\ &+(1+1+1)z^2\\ =&3z^2. \end{align}$$ And we are done. $\square$
Progress on Proof: Suppose that $A\subset \Bbb C\setminus\{0\}$ is a finite set of size $n$. Then we have $$\prod_{a\in A}(1-az)=\sum_{k=0}^{n}(-z)^k\sum_{{U\subset A}\atop{\#U=k}}\prod_{p\in U}p.$$ Taking $A=S_n=\{\zeta_n^k:k=0,1,...,n-1\}$ we have that $\prod_{a\in A}(1-az)=1-z^n$, so that $$\sum_{{U\subset S_n}\atop{\#U=k}}\prod_{p\in U}p=0,\qquad \text{for }1\le k\le n-1.$$ Perhaps this will be useful when we note that $$(1-z^m)p_{m,j}(z)=\sum_{k=0}^{m-1}\sum_{v=0}^{m-1}(-z)^v\zeta_{m}^{-kj}\sum_{{U\subset S_m\setminus \{\zeta_m^k\}}\atop{\#U=v}}\prod_{p\in U}p.$$ Specifically, it could be used to show that $$(-1)^v\sum_{k=0}^{m-1}\zeta_m^{-kj}\sum_{{U\subset S_m\setminus \{\zeta_m^k\}}\atop{\#U=v}}\prod_{p\in U}p=m\delta_{v,j},\tag2$$ with $\delta_{a,b}$ being the Kronecker delta. From this point I am stuck.
Can anyone help me prove $(2)$? or $(1)$? Thank you very much!
$(1)$ is simply an application of PFD. Since $1-z^m=\prod_{k=0}^{m-1}(1-\zeta_m^k z)$, we have (for fixed $m$ and $j$) $$\frac{mz^j}{1-z^m}=\sum_{k=0}^{m-1}\frac{\alpha_k}{1-\zeta_m^k z},\quad\alpha_k=\lim_{z\to\zeta_m^{-k}}\frac{mz^j}{1-z^m}(1-\zeta_m^k z),$$ giving $(1)$ after evaluation of the limit.