$p$ is adherent value of $\left(\sum_{i=1}^{p} z_{i}^n\right)_{n\in\mathbb{N}}$ where $z_{i}$ are complex of modulus 1

Question

Here is an exercise I'm trying to solve :

Let $z_{1}, \ldots,z_{p}$ be some complex numbers of modulus 1 and, for $n\in\mathbb{N}$, $u_n = \sum\limits_{i=1}^{p} z_{i}^n$. Show that $p$ is adherent value of $(u_n)$

My attempts : Let $v = (\theta_1, \ldots,\theta_p)$ where $\theta_i$ is the argument of $z_i$. Since $\mathbb{Z}v + 2\pi\mathbb{Z}^p$ is a subgroup of $\mathbb{R}^p$, some properties of subgroups of $\mathbb{R}^p$ might help to conclude?

Answer 1

As per the comments, we can use simultaneous version of the Dirichlet's approximation theorem (svDAT). But before, let's suppose $z_k=e^{i\alpha_k}$ (I will use $k$ as the index to avoid confusions with complex $i$) where $|z_k|=1$. From svDAT applied to $\frac{\alpha_k}{2\pi}$ there are integers $t_{1},\ldots ,t_{p}$ and $n\in \mathbb {Z} ,1\leq n\leq N$ such that $$\left|\frac{\alpha_{k}}{2\pi}-{\frac {t_{k}}{n}}\right|\leq {\frac {1}{nN^{1/p}}} \Rightarrow \left|n\alpha_{k}-2\pi t_{k}\right|\leq {\frac {2\pi}{N^{1/p}}}$$ For large enough $N$ we have ${\frac {2\pi}{N^{1/p}}} < \varepsilon$ or $$n\alpha_k \approx 2\pi t_k \Rightarrow z_k^n=e^{i n\alpha_k} \approx e^{i 2\pi t_k}=1 \tag{1}$$ or $$\sum\limits_{k=1}^{p} z_{k}^n \approx p$$

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To motivate $(1)$, if we have $|\alpha -\beta| < \varepsilon$ then $$\left| e^{i\alpha} -e^{i\beta}\right|= \left| \cos{\alpha}+i\sin{\alpha} -\cos{\beta}-i\sin{\beta}\right|= \left| \cos{\alpha}-\cos{\beta}+i(\sin{\alpha}-\sin{\beta})\right|=\\ \sqrt{(\cos{\alpha}-\cos{\beta})^2+(\sin{\alpha}-\sin{\beta})^2}=\\ \sqrt{\left(-2\sin{\frac{\alpha-\beta}{2}}\sin{\frac{\alpha+\beta}{2}}\right)^2+\left(2\sin{\frac{\alpha-\beta}{2}}\cos{\frac{\alpha+\beta}{2}}\right)^2}=\\ 2\left|\sin{\frac{\alpha-\beta}{2}}\right| < 2\left|\frac{\alpha-\beta}{2}\right| < \varepsilon$$ from here.

Answer 2

Sorry to answer my own question. Here is my solution :

Denote by $\theta_{i}$ the argument of $z_i$ for all $i$. We show that for all $\epsilon>0$, there exists $n\in\mathbb{Z}$ nonzero such that $(n\theta_{1}, \ldots,n\theta_{p})\in[0, \epsilon]^{p}$ (mod $2\pi$) :

Fix $\epsilon>0$. Denote $A_{m,k} = [\frac{2\pi(k-1)}{m}, \frac{2\pi k}{m}]$ for all $m, k\in\mathbb{N}, 1\leq k\leq m$. Then $[0, 1]^p\subset\cup_{(k_1, \ldots,k_p)\in[|1, m|]^p}\prod_{i=1}^{p}A_{m, k_{i}}$.

Let $M\in\mathbb{N}$ such that $M>1/\epsilon$ and $N\in\mathbb{N}$ such that $N>M^p$. By pigeonhole principle, there must exists $l, s\in[|1, N+1|]$ distinct such that $(l\theta_{1}, \ldots,l\theta_{p}), (s\theta_{1}, \ldots,s\theta_{p})\in\prod_{i=1}^{p}A_{m, k_{i}}$ (mod $2\pi$) for some $(k_1, \ldots,k_p)\in[|1, M|]^p$. Then $((l-s)\theta_{1}, \ldots,(l-s)\theta_{p})\in[0, \epsilon]^{p}$ (mod $2\pi$). Done.