Let $n \geq 1$. Assume that $\alpha_1, \ldots, \alpha_n \in [0,2\pi)$ are numbers such that \begin{equation} \mathrm{e}^{\mathrm{i} \alpha_1} + \cdots + \mathrm{e}^{\mathrm{i} \alpha_n} = 0. \tag{1} \end{equation} As is well-known, if we set $\alpha_k = \beta + \frac{2 k \pi}{n}$ for some $\alpha$, we obtain a solution. The same of course holds by permuting the indices. Then, writing $n = m_1 + \cdots + m_p$ and if one choses, for each $1 \leq q \leq p$, the $m_q$ equispaced angles including a fixed $\beta_q$, and we set uses those $m_1 + \ldots + m_q = n$ angles for the $\alpha_k$, then \begin{equation} \mathrm{e}^{\mathrm{i} \alpha_1} + \cdots + \mathrm{e}^{\mathrm{i} \alpha_n} = \sum_{q=1}^p \mathrm{e}^{\mathrm{i} \beta_q}\sum_{j=1}^{m_q} \mathrm{e}^{ {2\mathrm{i} \pi j} /{m_q}} =0 , \end{equation} because each $\sum_{j=1}^{m_q} \mathrm{e}^{ {2\mathrm{i} \pi j} /{m_q}} = 0$ as a sum of roots of unity.
Question: Are the described angles (up to permutation) the only solutions to (1)?
Of course, it makes a large set of solutions, but relatively easy to describe and I could not find any other solution.