I've come up with a solution, but it's not so insightful, so I was wondering if anyone has a cleaner not-so-brute-force solution:
Equivalently (since the case where all of them ar zero is trivial), if $\alpha,\beta,\gamma$ are complex numbers in the unitary circle such that $\alpha + \beta + \gamma = 1$ then one of them is equal to one. Then one could write $\alpha = a_\alpha + b_\alpha$ and so on, and solve the system of equations for the real part, the imaginary part and the modulus.
Is there a neater form of solving it? Like using the triangle inequality or something like so?
Best Answer
Replacing $c$ by $-c$, we can rewrite the equation as $a + b = c + d$. Assume that all four of these numbers lie on the unit circle, and (by rotation) assume that $a+b$ is a nonnegative real number. If this is positive, then you can check that it determines the unordered pair $\{a, b\}$ uniquely, and thus either $d = a$ or $d = b$. On the other hand, if $a+b=c+d=0$, then $d = -c$.