Let $p$ be a prime and suppose $e_1,e_2,\cdots,e_p$ denote some $p$-th roots of unity in $\mathbb{C}$, not necessarily distinct.
If
$$e_1 + e_2 + \cdots + e_p=0,$$
then, can we always conclude that all $e_i$'s are the distinct?
Note that for $p=2$ this is true, and so one may assume $p>2$.
On
Let $p$ be prime and let $\zeta:=\exp\frac{2\pi}{p}$. Then the $p$-th rootsof unity are $1,\zeta,\zeta^2,\dots,\zeta^{p-1}$.
Moreover, $$\sum_{0}^{p-1} a_i \zeta^{i}=0 \mbox{ for some } a_i\in\mathbb{Z} \Rightarrow a_i=c\mbox{ for all }.i$$ This is a consequence of the irreducibility of $X^{p-1}+\dots+X+1$, easily seen by Eisenstein.)
If $p$ is a prime, you can indeed only get $0$ as a sum of $p$ such $p$-th roots of unity by taking the $p$ distinct roots once each.
I don't know how much field theory you have done, but here is an explanation of something much more general: Let $\omega \neq 1$ be a complex number satisfying $\omega^{p} = 1.$ Then the $p$-th roots of unity are $\{\omega^{i}: 0 \leq i \leq p-1 \}.$ Furthermore, the polynomial $x^{p}-1 \in \mathbb{Q}[x]$ factors as $(x-1)( 1 + x + \ldots + x^{p-2}+x^{p-1}),$ and the second factor (which we now call $f(x)$) is irreducible in $\mathbb{Q}[x]$ by Eisenstein's criterion. Note that $\omega$ is a root of $f(x).$.
This implies by general theory that $\{1,\omega,\omega^{2},\ldots,\omega^{p-3}, \omega^{p-2} \}$ is linearly independent over $\mathbb{Q}.$ So every non-zero element which is expressible as a $\mathbb{Q}$-linear combination of these elements has a unique such expression. Since $-\omega^{p-1} = 1 + \omega+ \omega^{2}+ \ldots + \omega^{p-3} + \omega^{p-2}$ is the unique way to express $-\omega^{p-1}$ this way, we see that $0 = q(1 + \omega + \omega^{2} + \ldots + \omega^{p-1})$ is the only way to write $0$ as a $\mathbb{Q}$-linear combination of $\{1,\omega,\omega^{2},\ldots,\omega^{p-1} \},$ where $q$ can be any rational number. Hence the only way to write $0$ as a sum of $p$ such $p$-th roots of unity is the case $q = 1.$