Sum of complex numbers

Question

Let $N \in \mathbb{N}$. It is well known that $$ \sum_{k=0}^{N-1} e^{2 \pi i \frac{k}{N}} =0. $$ More generally, if $N=pq$ and $P=\{ 0, q,\cdots, (p-1)q \}$, then $$ \sum_{k \in P} e^{2 \pi i \frac{k}{N}} =0. $$

My question :is the converse true?

If $P \subseteq \{0,1,\cdots,N-1\}$ such that
$$ \sum_{k \in P} e^{2 \pi i \frac{k}{N}} =0, $$ can we show that the set $P$ has the form $P=\{ 0, q,\cdots, (p-1)q \}$ where $N=pq$?

---

The above question is not true.

If $P \subseteq \{0,1,\cdots,N-1\}$ such that
$$ \sum_{k \in P} e^{2 \pi i \frac{k}{N}} =0, $$ and any proper subset of $P$ doesn’t satisfy the equality, can we show that the set $P$ has the form $P=\{ 0, q,\cdots, (p-1)q \}$ where $N=pq$?

Answer 1

No, the converse is not true, because you can take unions of such sets (if $\sum_{k\in P}=0$ and $\sum_{k\in P'}=0$, then $\sum_{k\in P \cup P'}=0$.

For example, if $P=\{ 0, q,\cdots, (p-1)q \} \cup \{ 0, p,\cdots, (q-1)p \}$ (with, say $p=3,q=5$) then $P$ is not of the form you gave.

Answer 2

$$S=\sum_{k\in P} e^{2i\pi k/n}=\sum_{j=0}^{p-1} e^{2i\pi j q/n}=\frac{e^{2i\pi p q/(pq)}-1}{e^{2 i \pi q/(pq)}-1}=\frac{e^{2i\pi}-1}{e^{2 i \pi/p}-1}=0.$$ We have trabsformed the index $k$ to $j$ as $k=jq$ where $j$ goes from $0$ to $p-1$. Then we used the sum of finite GP, where $n=pq$.