Proving Napoleon's Theorem with complex numbers I have gotten stuck at an exercise that deals with proving Napoleon's Theorem via complex numbers. 

Consider the complex plane $\mathbb{C}$ to be identified standard Euclidean plane. 
a) Given two complex numbers $z_1, \ z_2$ find an expression for a third complex number that forms an equilateral triangle with them. 
b) Use complex numbers to show that if equilateral triangles are constructed on the sides of any triangle, then their centers form an equilateral triangle themselves. 

My problem lies with b). I started my attempt as follows:
Let $z_1,z_2,z_3$ be the vertices of a triangle. Let $d=e^{i \frac{\pi}{3}}$. Let $\Delta az_1z_3, \Delta bz_2z_3, \Delta bz_2z_3$ be the equilateral triangles constructed on the sides of $\Delta z_1z_2z_3$.
We may assume that 
$$ a=d(z_3-z_1)+z_1 \\
b=d(z_2-z_3)+z_3 \\
c=d(z_1-z_2)+z_2.$$ 
Let A, B, C denote the centers of $\Delta az_1z_3, \Delta bz_2z_3, \Delta bz_2z_3$. We have 
$$A=\frac{a+z_1+z_3}{3} \\
B=\frac{b+z_2+z_3}{3} \\
C=\frac{c+z_1+z_2}{3} $$ 
I want to show that 
$$ A-C=d(B-C)  \tag{1} $$
$$ B-A=d(C-A)  \tag{2} $$
as this will imply that $\Delta ABC$ is an equilateral triangle. I know that 
$$a-c=d(z_3-2z_1+z_2)+z_1-z_2  \\
  b-c=d(-z_1+2z_2-z_3)+z_3-z_2 \\
     =-d(z_1-2z_2+z_3)+z_3-z_2. $$
It follows 
$$ A-C=\frac{a-c+z_3-z_2}{3}                 \\
=\frac{d(z_3-2z_1+z_2)+z_1-z_2+z_3-z_2}{3}   \\
=\frac{d(z_3-2z_1+z_2)+z_1+z_3-2z_2}{3}      \\
=\frac{d(z_2-2z_1+z_3)+z_1+z_3-2z_2}{3}      \\
d(B-C)=d \frac{b-c+z_3-z_1}{3}               \\
=d \frac{-d(z_1-2z_2+z_3)+z_3-z_2+z_3-z_1}{3}\\
=d \frac{-d(z_1-2z_2+z_3)+2z_3-z_2-z_1}{3}   \\ 
=d \frac{-(d(z_1-2z_2+z_3)+z_1+z_2-2z_3)}{3} $$
But I fail to see how to proceed in order to prove (1). 
 A: Here's a simpler approach. Let's find the centroid of the equilateral triangle erected outward on the side from $z_1$ to $z_2$. We follow an altitude from the midpoint $\frac{z_1+z_2}{2}$ to the centroid,$$\frac{z_1+z_2}{2}+\frac{i}{2\sqrt{3}}(z_2-z_1)=\frac{\sqrt{3}-i}{2\sqrt{3}}z_1+\frac{\sqrt{3}+i}{2\sqrt{3}}z_2=\frac{\zeta^\ast z_1+\zeta z_2}{\sqrt{3}},\,\zeta:=\exp\frac{\pi i}{6}.$$The displacement from this centroid to the analogous one on the side from $z_2$ to $z_3$ is$$\frac{\zeta^\ast z_1+(\zeta-\zeta^\ast)z_2-\zeta z_3}{\sqrt{3}}=\frac{\zeta^{-1}z_1+iz_2+i^2\zeta z_3}{\sqrt{3}}.$$But the coefficients form a geometric progression whose common ratio is a third root of unity, so the next displacement going round all three centroids is of the same length and rotated by $2\pi/3$. And that proves the centroids are the vertices of an equilateral triangle.
A: I would solve this as follows: Given $z_{1,2}$, find $z_3$ so as to form an equilateral triangle. Thus, geometrically
$$
z_3=z_2+|z_2-z_1|e^{i(2\pi/3+\alpha)}
$$
where $\alpha=\arg(z_2-z_1)$. The argument in the exponential makes the angle between $z_1,z_2,z_3$ equal to $\pi/3$.
