Roots of Unity : Restricted Affine Combinations and Vandermonde Determinants

Question

Fix an integer (not necessarily prime) $p\geq2$. I have two questions:

Question 1) Let $\mathcal{M}_{p}$ be the set of all $p$-tuples $\mathbf{v}=\left(v_{0},v_{1},\ldots,v_{p-1}\right)$ of complex numbers in the closed unit disk (i.e., $\left|v_{k}\right|\leq1$ for all $k$) such that $\sum_{k=0}^{p-1}v_{k}=1$. Define $\Phi:\mathcal{M}_{p}\rightarrow\mathbb{C}$ by: $$\Phi\left(\mathbf{v}\right)\overset{\textrm{def}}{=}\sum_{k=0}^{p-1}v_{k}e^{\frac{2k\pi i}{p}}$$

I'm wondering what the image of $\Phi$ looks like in $\mathbb{C}$. In particular, I'm interested in knowing what $\Phi^{-1}\left(\left\{ 0,1\right\} \right)$ (the pre-image of $\left\{ 0,1\right\}$ under $\Phi$) looks like. What restrictions/estimates (if any) can be put on the $v_{k}$s in this case?

Question 2) I'm investigating the determinant of the vandermonde matrix of the $p$th roots of unity: $$\det\left(\begin{array}{cccccc} 1 & 1 & 1 & 1 & \cdots & 1\\ 1 & \xi_{p} & \xi_{p}^{2} & \xi_{p}^{3} & \cdots & \xi_{p}^{p-1}\\ 1 & \xi_{p}^{2} & \xi_{p}^{4} & \xi_{p}^{6} & \cdots & \xi_{p}^{2\left(p-1\right)}\\ 1 & \xi_{p}^{3} & \xi_{p}^{6} & \xi_{p}^{9} & \cdots & \xi_{p}^{3\left(p-1\right)}\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 & \xi_{p}^{p-1} & \xi_{p}^{2\left(p-1\right)} & \xi_{p}^{3\left(p-1\right)} & \cdots & \xi_{p}^{\left(p-1\right)^{2}} \end{array}\right)$$ where $\xi_{p}=e^{2\pi i/p}$ ; i.e., the quantity: $\prod_{0\leq j<k\leq p-1}\left(\xi_{p}^{k}-\xi_{p}^{j}\right)$

In terms of $p$, the values are $1$,$-\sqrt{2^{2}}$, $-i\sqrt{3^{3}}$, $-i\sqrt{4^{4}}$, $-\sqrt{5^{5}}$, $\sqrt{6^{6}}$, and so on. Clearly, the terms always have magnitude $p^{p/2}$. However, I can't seem to find the pattern in their (complex) arguments. I would love to have a closed formula for these numbers, if such a formula exists. A proof (or reference thereof) would also be nice.

Answer 1

It is just a question of computing the argument of each $\zeta_p^k-\zeta_p^j$, where $0\le j<k\le p-1$. Now $$\zeta_p^k-\zeta_p^j=\exp(2\pi ik/p)-\exp(2\pi ij/p) =\exp(\pi i(j+k)/p)\left[\exp(\pi i(k-j)/p)-\exp(-\pi i(k-j)/p))\right] =2i\exp(\pi i(j+k)/p)\sin\frac{\pi(k-j)}{p}.$$ This sine is positive, so $$\arg(\zeta_p^k-\zeta_p^j)=\frac{\pi(j+k)}p+\frac\pi2$$ (modulo multiples of $2\pi$). So the argument of the Vandermonde determinant is the sum of all these, which is $$\frac{\pi}p\sum_{j=0}^{p-1}(p-1)j+\frac\pi2\frac{p(p-1)}2$$ if I'm not mistaken, which can be simplified with more patience than I have right now.