# $\sum\limits_{i=1}^n z_i=0\Longleftrightarrow z_1,\ldots ,z_n$ are the vertices of …

Suppose for $n\geq 3$ we have, $z_1,\ldots ,z_n\in\mathbb{C}$ and $|z_1|=|z_2|=\cdots=|z_n|=1$. Now I need to determine a property $P$ such that the following is true :

$$\sum_{i=1}^n z_i=0\Longleftrightarrow z_1,\ldots ,z_n\mbox{ are the vertices of a polygon satisfying } P$$

I have solved the cases for $n=3$ and $n=4$, in the first case, $P$ is equilateral triangle and for the second, $P$ is rectangle. But my methods does not generalize for general case. Can we solve it for general $n$ ? At least I would like to know what happens for $n=5$.

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Two (venial) sins that I hope you will not repeat: 1. using a title that is entirely in $\LaTeX$, and 2. having the $\LaTeX$ in the title in \displaystyle or enclosed in $$. – Ｊ. Ｍ. Aug 4 '12 at 4:49 If you know physics, consider this: how must n people be arranged around a circle if each one is pulling at an object at the center of the circle with a force of 1 \mbox{N}, such that the net force on the object is zero (i.e., their tugging effectively gets cancelled out). – ladaghini Aug 4 '12 at 4:54 @J.M.:Thanks for the suggestions, I will not repeat these. But may I know why is this important – pritam Aug 4 '12 at 5:02 1. Titles that are entirely in \LaTeX are annoying to right-click on for, say, opening in new windows or tabs. 2. Titles with \displaystyle \LaTeX look rather disruptive on the front page. – Ｊ. Ｍ. Aug 4 '12 at 5:21 @lada, that's not even true for n=4, where, as OP knows, rectangles work. It gets worse for larger n. – Gerry Myerson Aug 5 '12 at 12:33 show 1 more comment ## 2 Answers You can get all the solutions for n=5, but it isn't pretty. There must be a sector of size 4\pi/5 containing at least 3 of the points. By rotating and reflecting, if necessary, we may assume z_j=e^{2\pi i\theta_j} with \theta_1=0, 0\le\theta_1\le1/5, \theta_1\le\theta_2\le2/5. Now provided only that$$|z_1+z_2+z_3|\le2$$we can find unique \theta_4\le\theta_5 to make \sum^5z_j=0. Thinking of \theta_1 and \theta_2 as parameters, this gives a 2-parameter family of solutions to$$z_1+z_2+z_3+z_4+z_5=0,\qquad|z_j|=1

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For any $n$ put $z_k = e^{\frac{2\pi i(k-1)}{n}}$ then $|z_k|=1$ and $\sum_{k=1}^nz_k = 0$ and $z_k$ will be the edges of a regular n-gon.

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Of course, but that's only a special case. The question is looking for a characterization. –  Gilles Feb 13 at 7:08