Complex no. (Not too hard)

Set
z2=z1(cos120+isin120)=z1(-1/2+√3/2i)
z3=z2(cos120+isin120)=z2(-1/2+√3/2i)
z3=z1(cos240+isin240)=z1(-1/2-√3/2i)
So
z1.z2=z1^2(-1/2+√3/2i)
z2.z3=z1^2(-1/2+√3/2i)(-1/2-√3/2i)=z1^2(1/4+3/4)=z1^2
z3.z1=z1^2(-1/2-√3/2i)
z1.z2+z2.z3+z3.z1
=0
On the other hand
z2^2=z1^2(-1/2+√3/2i)^2
z3^2=z1^2(-1/2-√3/2i)^2
z1^2 + z2^2 + z3^2
=z1^2+z1^2(-1/2+√3/2i)^2+z1^2(-1/2-√3/2i)^2
=z1^2(1+1/4-√3/4i-3/4+1/4+√3/4i-3/4)
=0
So, z1^2 + z2^2 + z3^2 = z1.z2+z2.z3+z3.z1

To Ivan,
The proposition holds for arbitrary equilateral triangles in a complex plane indeed.
However, the proof given by myisland8132 presumed that the triangle is symmetric about the origin.
For the points 0, 1, w=cis(pi/6),
1^2 + w^2 = 1+w^2 and 1*w = w, which are equal by w^3+1=0.

can think in this way also:
if they are an equilateral triangle over the origin, then
they can be represented by cube root of unity 1, w, w^2

Then the equation above is just:
1+w^2 + w^4 = 1*w + w*w^2 + w^2*1
both sides equal 1+w+w^2 ( = 0)

2008-09-16 03:22:38 補充：
Cube root of unity means w^3 = 1.
If the triangle is rotated, or scaled, there is only an extra factor, e.g. becomes c, cw, cw^2, the calculation is the same and it will not affect anything.

2008-09-16 03:23:14 補充：
Finally this is not true for general equilateral triangle, only true for those symmetric with respect to the origin. So next time please write the question more carefully.
(for example, take 0, 1, rt(3)/2 + i/2)