Let #z_j = r_j e^(i theta_j)#.
The first condition says that #|z_j| = 1# i.e. #r_j = 1 forall j#. Therefore, these are points on the unit circle.
The second condition seems difficult to deal with so let's skip it.
The third condition is very useful. First: a point on the unit circle squared is another point on the unit circle. Second: the only way that three points on the unit circle can add up to 0 is if they are separated equiangularly, i.e. each is #(2pi)/3# apart from each other. This doesn't decide what the initial angle is, therefore we know that
#z_j^2 = e^(i ((2pi)/3 j + phi_0)) #
Therefore, (with #theta = 1/2 phi_0 #):
#z_j = pm e^(itheta)e^(ipi/3 j) #
Since we only care about magnitude, we don't care about "overall phase" which means we can ignore #theta# and at least one #pm_1# (which is an overall phase as well).
By the second condition:
#z_1 + z_2 + z_3 ne 0 #
#e^(itheta) (e^(ipi/3) pm e^(i2pi/3) pm e^(ipi)) ne 0#
#(1/2 + sqrt(3)/2 i) pm_2 (-1/2 + sqrt(3)/2) pm_3 (-1) ne 0 #
#[(1/2) pm_2 (-1/2) pm_3 (-1)] + i [sqrt3/2 pm_2 sqrt3/2] ne 0#
The only way this could become 0 is if
#1 = - pm_2 = - pm_3 #
(which we want to avoid).
Therefore,
#z_1 = e^(ipi/3) #, #z_2 = e^(i(2pi)/3)#, and #z_3 = - e^(ipi) = 1#
Again, there could be some random phases, but since those are constant between all three, they do not matter.
Now, we can prove the initial statement:
#|z_1^n + z_2^n + z_3^n | = l#
#|e^(ipi n / 3) + e^(i2pi n / 3) + 1 | = l #
#|e^(ipi n / 3) + e^(ipi n) e^(-ipi n / 3) + 1 | = l #
Let's consider even #n = 2m#:
#|e^(i(2pi)/ 3 m) + e^(-i (2pi) / 3m) + 1 | = l #
#|2cos(( 2pi) / 3 m) + 1| = l #
Since #cos((2pi)/3m) in {1, -1/2} #, so
#l = |2cos(( 2pi) / 3 m) + 1| in {3, 0} #
Now let's consider odd #n#:
#|2i sin(pi n / 3) + 1 | = l #
#l = sqrt(1 + 4sin^2(pi/3 n)) #
#sin(pi/3n) in {0, pm sqrt(3)/2} implies sin^2((pin)/3) in {0, 3/4}#
Therefore,
#l in {sqrt(1 + 0), sqrt(1 + 3)} = {1, 2} #
Therefore, we know that for any #n ge 2#, #l in {0, 1, 2, 3}#.
