How to find orthogonal vector sets Assume a vector expressed as
    $$v=\left[e^{j\theta_1},e^{j\theta_2},e^{j\theta_3}\right]^T$$
$\theta_k, k=1..3$ can change. I want to find a set of vectors  $\{v_1,v_2,...v_N\}$, so that $v_i^Hv_j=0$, if $i\neq j$. $H$ represent  conjugate transpose.  
The questions are: 


*

*How to find the vectors?

*What is the maximum number of $N$? 3 for the vector defined above?

*If I change the length of vector $v$ from 3 to $L$, is there a relation between $L$ and $N$?
 A: Clearly, as Theo pointed out, $N \leq 3$. However, as you're requiring that vectors be of the form $v = [z_1, z_2, z_3]$ with $|z_1| = |z_2| = |z_3| = 1$, it would theoretically be possible that $N < 3$.
That is not the case. Take $v_1 = [1,1,1], v_2 = [1,e^{2i\pi/3},e^{4i\pi/3}], v_3 = [1, e^{4i\pi/3},e^{2i\pi/3}]$. Therefore $N = 3$.
Now if you take the above solution and multiply any vector by a complex number of norm 1, you have a solution. Also, swapping the $i$-th and $j$-th coordinates ($1 \leq i < j \leq 3$) in every vector of the solution gives a solution as well.
Hope this clears things up.
EDIT: Let us now generalize for $n$ coordinates. In fact, an analogous solution can be found in a straightforward manner:
Let $\omega = e^{2\pi i/n}$. Notice that $1 + \omega + \omega^2 + \ldots + \omega^{n-1} = 0 = 1 + \omega^2 + \omega^4 + \ldots + \omega^{2(n-1)} = \ldots = 1 + \omega^{n-1} + \ldots + \omega^{(n-1)^2}$. In other words, $$\sum_{j=0}^{n-1}\omega^{kj} = 0 \text{ for } k = 1,2,\ldots,n-1$$
Using this, if we take $v_k = [1,\omega^k,\omega^{2k},\ldots,\omega^{(n-1)k}]$ for $0 \leq k < n$, the $n$ vectors $v_0,v_1,\ldots,v_{n-1}$ are a solution to the problem.
