# A problem about $e^{2\pi i \alpha_1}+e^{2\pi i \alpha_2}+\cdots+e^{2\pi i \alpha_N}=0$

Let $\alpha_i\in [0,1),\; i\in \{1,\cdots,N\}$ for some positive integer $N$, such that $$e^{2\pi i \alpha_1}+e^{2\pi i \alpha_2}+\cdots+e^{2\pi i \alpha_N}=0$$ and if for any non-empty proper subset $E\subset \{1,\cdots,N\}$ satisfy $\sum_{k\in E}e^{2\pi i\alpha_{k}}\neq 0$, then $N$ be a prime number, and $\{\alpha_i: i\in \{1,\cdots,N\}\}=\rho+\{\frac{i}{N}:i\in\{0,\cdots,N-1\}\}$ for some $\rho \in [0,1)$.

• Fixed the spelling of "problem" in your title; nice question, +1! Apr 24, 2015 at 6:29
• There's a geometric way to think about your hypotheses: each unit complex number is a unit length vector in the plane. The sum of the vectors being zero is saying that they form the sides of an $N$-sided polygon. Can you translate the hypotheses about proper subsets into this geometric language? Apr 24, 2015 at 6:33
• Of course, I also thought about this from N -sided polygon, but it is not easy to see the answer. What more, if $\alpha_i \in \mathbb{Q}$(rational number), from [H. B. Mann, On linears relations between roots of unity, Mathematika 12 (1965),107–117.], we have the proof. Apr 24, 2015 at 7:26
• This is essentially a duplicate of math.stackexchange.com/questions/480008/…, though the equivalence is not so immediate that I voted to close. Jul 8, 2015 at 16:17
• @ErickWong The big difference is that, in this question, the $\alpha_i$ are simply real, whereas in that question they are rational. According to the OP's comment here, Mann answered the rational case, so perhaps the OP should go leave an answer there. Jul 8, 2015 at 17:00

No, this is false. For $N=5$, there are many other solutions. Heuristic reasoning: There are $5$ angles, and the condition they impose is a codimension two condition. Thus, we expect the solutions to the problem to form a three-dimensional set. However, for each way of dividing the $5$ angles into a pair $D$ and a triple $T$, there is only a $2$ dimensional space of ways to have $\sum_{i \in D} \alpha_i = \sum_{i \in T} \alpha_i = 0$. Also, there is only a one-dimensional space of ways to have a rotation of a pentagon. We expect there to be lots of points in a three-dimensional set which are not in finitely many one- and two-dimensional sets.
To give an explicit example, in the picture below, choose $\pi/3 < x < \pi/2$ and $x \neq 2 \pi/5$ and take $y = \cos^{-1}(1/2+\cos x)$, to get an equilateral pentagon which is not regular and no subset of whose sides add up to zero.
Your conjecture is false, see mercio's answer here for a 6-term counterexample with all $\alpha_i$ rational:
$$\alpha_1 = \frac{3}{30},\quad \alpha_2 = \frac9{30},\quad \alpha_3 = \frac{10}{30},\quad \alpha_4 = \frac{20}{30},\quad \alpha_5 = \frac{21}{30},\quad \alpha_6 = \frac{27}{30}.$$