By scaling, it is sufficient to consider systems of non-zero complex numbers $z_1,z_2,\ldots,z_n$ that satisfy $|z_{k+1}-z_k|^2\ge1$ for all $k$, and to determine \[ \lambda(n) = \min \sum_{k=1}^n |z_k|^2 \] over all such systems. For even $n$ of the form $n=2k$, the answer clearly is $\lambda(n)=k/2$.

I claim that the answer for even $ n $ is $ \lambda(n) = \frac{n}{4}. $ To see that this works, write $ z_k = a_k + b_ki $ for some $ a_k, b_k \in \mathbb{R} $ for all $ k \in \{1, 2, \dots, n\} $ (for the remainder of the proof, indices will be taken mod $ n $). Then $ \sum_{k = 1}^{n}|z_k|^2 = \sum_{k = 1}^{n}a_k^2 + \sum_{k = 1}^{n}b_k^2 \ge \frac{1}{2}\left(\sum_{k = 1}^{n}a_k^2 + \sum_{k = 1}^{n}b_k^2 - \sum_{k = 1}^{n}a_ka_{k + 1} - \sum_{k = 1}^{n}b_kb_{k + 1}\right) = \frac{1}{4}\sum_{k = 1}^{n}|z_k - z_{k + 1}|^2 \ge \frac{n}{4}\min\limits_{1\le k\le n}\{|z_{k+1}-z_k|^2\}. $ To show that this is optimal, we can let $ a_k = - a_{k + 1} $ and $ b_k = -b_{k + 1} $ for all $ k \in \{1, 2, \dots, n\} $ which gives equality. I also conjecture that $ \lambda(2k + 1) = \frac{2k + 1}{4\sin^2\left(\frac{k\pi}{2k + 1}\right)} $

And here is hopefully a proof of the odd case, discovered by mathocean97 and myself. I claim that $ \lambda(2n + 1) = \frac{2n + 1}{4\sin^2\left(\frac{n\pi}{2n + 1}\right)} $ for all $ n \in \mathbb{N}. $ The equality case is when you let the $ z_k $ be vertices of regular "star" $ n $-gon centered at the origin where every "spoke" of the star has radian measure $ \frac{\pi}{2n + 1}. $ Consider a configuration where the $ |z_{k + 1} - z_k| $ are fixed and we want to minimize $ \sum_{k = 1}^{2n + 1}|z_k|^2. $ It is clear that we can translate the configuration so that its center of mass is at the origin. Now, consider a configuration of $ z_k $ where the $ |z_k| $ are fixed and we want to maximize $ \min\limits_{1\le k\le 2n + 1}\{|z_{k+1}-z_k|^2\}. $ Assume for the sake of contradiction that $ 1 \le m < 2n + 1 $ of the differences $ |z_{k + 1} - z_k|^2 $ attain this maximum minimum. Then there must be a sequence $ z_k, z_{k + 1}, z_{k + 2} $ such that $ |z_{k + 1} - z_k|^2 $ attains the maximum minimum and $ |z_{k + 2} - z_{k + 1}|^2 $ does not. But then we can smooth $ z_{k + 1} $ by rotating it around the origin so that $ |z_{k + 1} - z_k| = |z_{k + 2} - z_{k + 1}| $ which means that now $ m - 1 $ of these $ |z_{k + 1} - z_k| $ attain the maximum minimum. We keep doing this until $ 0 $ of the $ |z_{k + 1} - z_k| $ attain the maximum minimum so we can increase the size of the maximum minimum, contradicting its maximality. Therefore we must have that $ |z_{k + 1} - z_k| $ is the same for all $ k \in \{1, 2, \dots, 2n + 1\}. $ Denote its value by $ x. $ Now, let $ \theta_k $ be the angle formed by the vectors associated with $ z_k $ and $ z_{k + 1}. $ It's clear that $ \min\limits_{1\le k\le 2n + 1}\{|z_{k+1}-z_k|^2\} $ will be maximized when each of these angles is obtuse and where $ \sum_{k = 1}^{2n + 1}\theta_k = 2{\pi}n. $ By the Law of Cosines we have that $ x^2 = |z_k|^2 + |z_{k + 1}|^2 - 2|z_k||z_{k + 1}|\cos{\theta_k} \le \left(|z_k|^2 + |z_{k + 1}|^2\right)(1 - \cos{\theta_k}) $ which implies that $ \sum_{k = 1}^{2n + 1}|z_k|^2 \ge \frac{x^2}{2}\sum_{k = 1}^{2n + 1}\frac{1}{1 - \cos{\theta_k}} \ge \frac{x^2}{2}\left(\frac{2n + 1}{1 - \cos{\left(\frac{2{\pi}n}{2n + 1}\right)}}\right) = \left(\frac{2n + 1}{4\sin^2\left(\frac{n\pi}{2n + 1}\right)}\right)x^2 $ which implies the desired result. Note: For the middle inequality, I used Jensen since if we let $ f(x) = \frac{1}{1 - \cos{x}} $ then $ f''(x) = \frac{1}{4}(\cos{x} + 2)\csc^4{\left(\frac{x}{2}\right)} \ge 0 $ so $ f $ is convex everywhere and for the final equality I used the identity $ \cos{2x} = 1 - 2\sin^2{x}. $

I think the following solution, which is not very substantial, is still worth mentioning. This is for the odd case, and it is far less involved than the above solution. From http://ssmr.ro/gazeta/gmb/2009/4/articol.pdf we observe that Wolstenholme-Lenard inequality, which is: For reals $x_1, x_2, \cdots x_n$ and positive angles $\theta_1, \theta_2, \cdots \theta_n$, if $\theta_1 + \theta_2 \cdots + \theta_n = \pi$, it is true that \[ \cos{\frac{\pi}{n}} \sum_{i = 1}^{n} x_i^2 \ge \sum_{i = 1}^{n} x_ix_{i+1}\cos{\theta_i} \] Let $\theta_1 = \pi - \angle{A_1OA_2}$, $\theta_2 = \pi - \angle{A_2OA_3}$, and so on. Choose the orientation such that these are positive angles. It is clear that $\theta_1 + \theta_2 + \cdots \theta_n = n\pi - 2k\pi$, where $k$ is the number of loops and $n$ is odd, so the sum is at least $\pi$. Now, by the definition of $\lambda$, assume that \begin{align*} \sum_{k=1}^n |z_k|^2 < \lambda |z_{2} - z_1|^2 &= |z_2|^2 + |z_1|^2 + 2|z_1||z_2|\cos{\theta_1}\lambda \\ \sum_{k=1}^n |z_k|^2 < \lambda |z_{3} - z_2|^2 &= |z_3|^2 + |z_2|^2 + 2|z_2||z_3|\cos{\theta_2}\lambda \\ \cdots \\ \sum_{k=1}^n |z_k|^2 < \lambda |z_{n} - z_1|^2 &= |z_n|^2 + |z_1|^2 + 2|z_1||z_n|\cos{\theta_n}\lambda \end{align*} The finish is fairly clear now -- just sum up both sides, yielding the inequality \[ (n - 2\lambda) \sum_{k=1}^n |z_k|^2 < \lambda\sum_{i = 1}^{n}2 |z_i||z_{i+1}|\cos{\theta_i} \] \[ \frac{n - 2\lambda}{2\lambda} \sum_{k=1}^n |z_k|^2 < \sum_{i = 1}^{n} |z_i||z_{i+1}|\cos{\theta_i} \] If the sum of the $\theta$'s is larger than $\pi$, uniformly scale and/or subtract $2\pi$ until it exactly equals $\pi$. This only increases the RHS, so its fine. Now, by Wolstenholme-Lenard, the above is a contradiction if $\frac{n-2\lambda}{2\lambda} > \cos{\frac{\pi}{n}}$, so we obtain that $\frac{n}{2\lambda} > 1 + \cos{\frac{\pi}{n}}$, and $\lambda \ge \frac{n}{2(1 + \cos{\frac{\pi}{n}})}$. An example is easy to come by and was mentioned above: take the spokes in an $n$-gon.

simplyconnected43 wrote: I think the following solution, which is not very substantial, is still worth mentioning. This is for the odd case, and it is far less involved than the above solution. From http://ssmr.ro/gazeta/gmb/2009/4/articol.pdf we observe that Wolstenholme-Lenard inequality, which is: For reals $x_1, x_2, \cdots x_n$ and positive angles $\theta_1, \theta_2, \cdots \theta_n$, if $\theta_1 + \theta_2 \cdots + \theta_n = \pi$, it is true that \[ \cos{\frac{\pi}{n}} \sum_{i = 1}^{n} x_i^2 \ge \sum_{i = 1}^{n} x_ix_{i+1}\cos{\theta_i} \] Let $\theta_1 = \pi - \angle{A_1OA_2}$, $\theta_2 = \pi - \angle{A_2OA_3}$, and so on. Choose the orientation such that these are positive angles. It is clear that $\theta_1 + \theta_2 + \cdots \theta_n = n\pi - 2k\pi$, where $k$ is the number of loops and $n$ is odd, so the sum is at least $\pi$. Now, by the definition of $\lambda$, assume that \begin{align*} \sum_{k=1}^n |z_k|^2 < \lambda |z_{2} - z_1|^2 &= |z_2|^2 + |z_1|^2 + 2|z_1||z_2|\cos{\theta_1}\lambda \\ \sum_{k=1}^n |z_k|^2 < \lambda |z_{3} - z_2|^2 &= |z_3|^2 + |z_2|^2 + 2|z_2||z_3|\cos{\theta_2}\lambda \\ \cdots \\ \sum_{k=1}^n |z_k|^2 < \lambda |z_{n} - z_1|^2 &= |z_n|^2 + |z_1|^2 + 2|z_1||z_n|\cos{\theta_n}\lambda \end{align*} The finish is fairly clear now -- just sum up both sides, yielding the inequality \[ (n - 2\lambda) \sum_{k=1}^n |z_k|^2 < \lambda\sum_{i = 1}^{n}2 |z_i||z_{i+1}|\cos{\theta_i} \] \[ \frac{n - 2\lambda}{2\lambda} \sum_{k=1}^n |z_k|^2 < \sum_{i = 1}^{n} |z_i||z_{i+1}|\cos{\theta_i} \] If the sum of the $\theta$'s is larger than $\pi$, uniformly scale and/or subtract $2\pi$ until it exactly equals $\pi$. This only increases the RHS, so its fine. Now, by Wolstenholme-Lenard, the above is a contradiction if $\frac{n-2\lambda}{2\lambda} > \cos{\frac{\pi}{n}}$, so we obtain that $\frac{n}{2\lambda} > 1 + \cos{\frac{\pi}{n}}$, and $\lambda \ge \frac{n}{2(1 + \cos{\frac{\pi}{n}})}$. An example is easy to come by and was mentioned above: take the spokes in an $n$-gon. simplyconnected43 wrote: I think the following solution, which is not very substantial, is still worth mentioning. This is for the odd case, and it is far less involved than the above solution. From http://ssmr.ro/gazeta/gmb/2009/4/articol.pdf we observe that Wolstenholme-Lenard inequality, which is: For reals $x_1, x_2, \cdots x_n$ and positive angles $\theta_1, \theta_2, \cdots \theta_n$, if $\theta_1 + \theta_2 \cdots + \theta_n = \pi$, it is true that \[ \cos{\frac{\pi}{n}} \sum_{i = 1}^{n} x_i^2 \ge \sum_{i = 1}^{n} x_ix_{i+1}\cos{\theta_i} \] Let $\theta_1 = \pi - \angle{A_1OA_2}$, $\theta_2 = \pi - \angle{A_2OA_3}$, and so on. Choose the orientation such that these are positive angles. It is clear that $\theta_1 + \theta_2 + \cdots \theta_n = n\pi - 2k\pi$, where $k$ is the number of loops and $n$ is odd, so the sum is at least $\pi$. Now, by the definition of $\lambda$, assume that \begin{align*} \sum_{k=1}^n |z_k|^2 < \lambda |z_{2} - z_1|^2 &= |z_2|^2 + |z_1|^2 + 2|z_1||z_2|\cos{\theta_1}\lambda \\ \sum_{k=1}^n |z_k|^2 < \lambda |z_{3} - z_2|^2 &= |z_3|^2 + |z_2|^2 + 2|z_2||z_3|\cos{\theta_2}\lambda \\ \cdots \\ \sum_{k=1}^n |z_k|^2 < \lambda |z_{n} - z_1|^2 &= |z_n|^2 + |z_1|^2 + 2|z_1||z_n|\cos{\theta_n}\lambda \end{align*} The finish is fairly clear now -- just sum up both sides, yielding the inequality \[ (n - 2\lambda) \sum_{k=1}^n |z_k|^2 < \lambda\sum_{i = 1}^{n}2 |z_i||z_{i+1}|\cos{\theta_i} \] \[ \frac{n - 2\lambda}{2\lambda} \sum_{k=1}^n |z_k|^2 < \sum_{i = 1}^{n} |z_i||z_{i+1}|\cos{\theta_i} \] If the sum of the $\theta$'s is larger than $\pi$, uniformly scale and/or subtract $2\pi$ until it exactly equals $\pi$. This only increases the RHS, so its fine. Now, by Wolstenholme-Lenard, the above is a contradiction if $\frac{n-2\lambda}{2\lambda} > \cos{\frac{\pi}{n}}$, so we obtain that $\frac{n}{2\lambda} > 1 + \cos{\frac{\pi}{n}}$, and $\lambda \ge \frac{n}{2(1 + \cos{\frac{\pi}{n}})}$. An example is easy to come by and was mentioned above: take the spokes in an $n$-gon. thank you,and what is the name of the book which the following article belongs?