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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
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@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
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1 Answer 1

up vote 1 down vote accepted

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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