Finding zeros of a Fourier series I have a short Fourier series on a homework problem,
$$
\begin{align*}
S(v_1, v_2, v_3) & = 
      1 + \exp[-i \pi (v_2 + v_3)]+ \exp[-i \pi (v_1 + v_3)] + \exp[-i \pi (v_1 + v_2)] \\
      & \hspace{3em} + \exp[-i \pi (v_1 + v_2 + v_3) / 2] + \exp[-i \pi (v_2 + v_3) / 2] \\
      & \hspace{3em} + \exp[-i \pi (v_1 + v_3) / 2] + \exp[-i \pi (v_1 + v_2) / 2],
\end{align*}
$$
where $$v_1, v_2, v_3 \in \mathbb{Z}.$$
I'm tasked with finding the zeros of $S$, but I am not sure how to go about breaking up each $v_i$ into cases. I could begin by splitting each $v_i$ into the cases $v_i = 4k_i, 4k_i + 1, 4k_i + 2, 4k_i+3$, but this would mean working out several pages by hand (and I expect the problem to be shorter than that).
How can I go about reducing the complexity of this problem?
 A: Let's show how you can reduce the complexity of your problem. First, notice that your sum is invariant under any permutation: $$S(a,b,c)=S(a,c,b)=S(b,a,c)=S(b,c,a)=S(c,a,b)=S(c,b,a)$$
Now let's use $$v_1+v_2=x\\v_1+v_3=y\\v_2+v_3=z$$
Also note that these are all integers. So
$$S(v_1,v_2,v_3)\to S'(x,y,z)=1+(-1)^x+(-1)^y+(-1)^z+(-i)^x+(-i)^y+(-i)^z+(-i)^{(x+y+z)/2}$$
Two things to notice here. First $S'$ is also invariant under permutations. Second, $x+y+z=2(u_1+u_2+u_3)$ is an even number. That means either all are even or one is even and two odds.
Case 1 all even. Then the first 4 terms are all $+1$, and we want the next 4 terms to be $-1$. That means all $x,y,z$ are of the form $2(2k+1)$. The last term however is the $i$ to some odd power, so the sum is never $0$.
Case 2 $x=2k, y=2l+1, z=2m+1$. In this case$$S'(x,y,z)=1+1-1-1+(-1)^k+(-1)^l(-i)+(-1)^m(-i)+(-i)^{k+l+m+1}$$
Notice that you need $l$ and $m$ to be one even and one odd, otherwise you can't cancel all the terms with $i$. This also yields $k$ is even, so the last term has to be $-1$. Since the order does not matter, $x=4p, y=4q+1, z=4r+3$.
$$v_1+v_2=4p\\v_1+v_3=4q+1\\v_2+v_3=4r+3$$
Subtracting the first from the second, and adding the third yields:
$$2v_3=4(r+p-q)+4$$
Therefore $v_3=2\ (\mod 4)$, $v_2=1\ (\mod 4)$, $v_1=3\ (\mod 4)$. Since the order does not matter, the zeros of $S$ are the form: $$(4\alpha+1,4\beta+2,4\gamma+3)$$
with $\alpha,\beta,\gamma\in\mathbb Z$, and any permutation.
