Let $N\geq3$. Let $w_{N k} = \exp (i 2 \pi k/N)$, $k = 0 \cdots N-1$, be the powers of the N-th root of 1, i.e. the solutions of $w_{N k}^N = 1$. Consider an array of coefficients $(a_j)$, $j = 0 \cdots N-1$, where the values are $a_j = \pm 1$, i.e. there is only a binary choice.
Define the function $$f((a_j),w_{N k}) = \sum_{j=0}^{N-1} a_j \;w_{N k}^j $$
For which choices of $(a_j)$ and for which $k$ exists a solution $f((a_j),w_{N k}) = 0$ ?
Note, for the motivation of this question: The function $f$ computes the eigenvalues of a circulant matrix with row vector $(a_j)$ from its associated polynomial. If there are $N_k$ zeroes of $f$ with different $k$, then the rank of the circulant matrix is $N - N_k$. In other words, if you have another method which directly gives the rank of a circulant matrix with binary ($\pm1$) row vector $(a_j)$, I'm happy as well.
I can see two obvious symmetries:
a) sign inversion: if there exists a solution for some $(a_j)$ and some $k$ then there is also a solution for $(-a_j)$ and $k$.
b) cyclic shift: if there exists a solution for some $(a_j)$ and some $k$ then, by multiplying with $w_{N k}^q$, there is also a solution for that $k$ and for $(a_{((j+q)\mod N)})$ with $q=1\cdots N-1$, i.e. for all cyclic shifts of $(a_j)$.
Also, clearly $k=0$ will appear only if $N$ is even.
I performed computer simulations which suggest:
a) If $N$ is prime, then the only solution to $f((a_j),w_{N k}) = 0$ is to set all $a_j=1$. This holds for $k = 1\cdots N-1$, and also, by the first symmetry above, setting all $a_j=-1$ holds.
Can one prove that this is the only solution for $N$ prime?
b) If $N$ is not prime, things get considerably more involved. Let us index the array $(a_j)$ by a unique decimal number $A = \sum_{j=0}^{N-1} 2^{j-1} (1+a_j)$. So the array with all $a_j=-1$ has $A=0$, and the array with all $a_j=1$ has $A=2^N -1$. If the $a_j$ were $(0,1)$, then $(a_j)$ were the binary representation of $A$. Switching all signs of the $a_j$ results in $B = \sum_{j=0}^{N-1} 2^{j-1} (1-a_j) = - \sum_{j=0}^{N-1} 2^{j-1} (-2 + 1+a_j) = 2^N - 1- A$, so by the first symmetry above, only $A = 0 \cdots 2^{N-1}$ needs to be considered. Further, cyclic shifts in $(a_j)$ - the second symmetry above - produce, from $A$, further solutions with the same $k$s for $(2^q A \mod (2^N-1))$.
Can you give a general result which $A$ and $k$ lead to a solution of $f(A,w_{N k}) = 0$?
Here are some results where solutions occur:
$N=4$ (contains $k=0$):
A=0 at k = 1 2 3
A=3 at k = 0 2 3
A=5 at k = 0 1 3
A=6 at k = 0 2 3
and, for completeness,
A=9 at k = 0 2 3
A=10 at k = 0 1 3
A=12 at k = 0 2 3
A=15 at k = 1 2 3
$N=9$ (does not contain $k=0$):
A=0 at k = 1 2 3 4 5 6 7 8
A=7 at k = 3 6
A=14 at k = 3 6
A=21 at k = 3 6
A=28 at k = 3 6
A=35 at k = 3 6
A=42 at k = 3 6
A=49 at k = 3 6
A=56 at k = 3 6
A=63 at k = 3 6
A=70 at k = 3 6
A=73 at k = 1 2 4 5 7 8
A=84 at k = 3 6
A=98 at k = 3 6
A=112 at k = 3 6
A=119 at k = 3 6
A=126 at k = 3 6
A=133 at k = 3 6
A=140 at k = 3 6
A=146 at k = 1 2 4 5 7 8
A=161 at k = 3 6
A=168 at k = 3 6
A=175 at k = 3 6
A=189 at k = 3 6
A=196 at k = 3 6
A=219 at k = 1 2 4 5 7 8
A=224 at k = 3 6
A=231 at k = 3 6
A=238 at k = 3 6
A=245 at k = 3 6
A=252 at k = 3 6
and symmetric continuation until A=511.