Is there any bound of $\Big|\sum_{n=1}^\infty z_n^p\Big|\leq C_{p,N}\Big|\sum_{n=1}^Nz_n\Big|^p$?

Question

Let $\{z_n\}_{n\in\mathbb{N}}\subset S^1\subset\mathbb{C}$ and $p\in \mathbb{R}$. I want to know if there is any $C_{p,N}>0$ such that $$\Big|\sum_{n=1}^N z_n^p\Big|\leq C_{p,N}\Big|\sum_{n=1}^Nz_n\Big|^p.$$ If it exists, can it be bounded in terms of $N$, i.e. $\sup_NC_{p,N}<\infty$?

Answer 1

Since the comments got too long let me summarize them as an answer:

In order to have some bound of the type asked, we need the arguments $z_n$ to be in a sector of width $\delta < \pi$. The counterexample given by the OP where the arguments are $0, \pi$ shows that is the best we can do.

In the case that the arguments are in such a sector, there is a $c_{\delta}>0$ s.t $|\sum_{n=1}^N{z_n}| \ge c_{\delta}\sum_1^N|z_n|=c_{\delta}N$

(if the sector is given by $|\arg z_n| \le \frac{\delta}{2}$ then $\Re z_n =|z_n|\cos \arg z_n \ge |z_n| \cos \frac{\delta}{2}$, so $|\sum z_n| \ge \Re \sum {z_n}=\sum {\Re z_n} \ge c_{\delta} \sum {|z_n|}$, otherwise rotate by a fixed $e^{i\theta}$ to take all $z_n$ there and obviously nothing changes in the estimates above)

But now LHS is at most $N$, while obviously $c_{\delta}^pN^p \le $ RHS $ \le N^p$, so if $p \ge 1$ we get $C_{p,N}=c_{\delta}^{-p}N^{1-p} \le c_{\delta}^{-p}$, so indeed we get an inequality LHS $\le c_{\delta}^{-p}$ RHS and for $p>1$ we can have it even more precise when $N$ is large as $C_{p,N} \to 0$ fast

However if $p<1$, choosing $z_k=1$, LHS is actually $N$, so $C_{p,N}$ cannot be better than $N^{1-p}$ and that goes to infinity, so we do not get something that useful now

Answer 2

No, at least for general $p\in \mathbb{R}$. Consider $p=2$ and $z_1=1,z_2=-1,z_3=1,\cdots$. Then L.H.S be $N$, while R.H.S is either $1$ or $0$ depends on $N$.