Plane Geometry using Complex Numbers I am working (showly) through "Visual Complex Analysis" by Tristian Needham.
I am stuck on exercise 23 chapter 1.
Given a triangle with vertices A, B and C draw three new equilateral triangles on the edges of the original triangle with vertices q, p, and r.
We know from exercise 19 that the centroid of a triangle with vertices A, B, C is given by:  
$1/3 (A + B + C)$
We also know from exercise 19 that the centroid of the triangle p,q,r is the same as the centroid of the triangle A,B,C.
Let x denote the centroid of the triangle A,B,p.
Let y denote the centroid of the triangle B,C,q
Let z denote the centroid of the triangle C,A,r
Show (using complex numbers) that the triangle x,y,z is equilateral
 A: Thanks to deyore and wikipedia (Napolean's Theorem):
Since the triangle ABp is equilateral and x is the centroid we have (rotating clockwise around B) that:
$$ p = w x $$
where $w = \sqrt{3} e^{i\pi/6}$
Similarly, working with the triangle rBC we have (still rotating around B) that:
$C = w y$
Thus $|x - y| = |p - C|/\sqrt{3}$ 
Working now with the triangles pAB and rAC, and rotating this time around A we find that:
$p = w x$
and:
$C = w z$
Thus:
$$|x - z| = |p - C| / \sqrt{3} = |x - y|$$
By the symmetry of the problem we also have:
$$|y - z| = |y - z|$$
It follows that the triangle xyz is equilateral, since the three sides are of equal length.
A: 
$\tag{Fig. 1}$
Let $\omega= e^{\frac{2\pi i}{3}}$ be a cube root of unity, then
$$1+\omega+\omega^2=0\implies \omega=-(1+\omega^2)\tag{1}$$ and $$\tag{2} -\omega^2=e^{\frac{\pi i}{3}}.$$ Since $\triangle acp$ is equilateral, a rotation around $c$ through an angle $\pi/3$ maps $p$ to $a$ and so $$-\omega^2(p-c)=a-c$$ $$\implies a -(1+\omega^2)c+\omega^2 p=0$$ $$\tag{3}\implies a+\omega c +\omega^2p=0.$$ Again, since $\triangle acp$ is equilateral, we can permute the vertices in $(3)$ (paying attention to their counterclockwise order) and the result will still hold. Further, since the reasoning in $(3)$ is reversible,

Theorem: $z_1,z_2,z_3$, with counterclockwise order $[z_1,z_2,z_3,z_1]$, are vertices of an equilateral triangle $\iff z_1+\omega z_2 +\omega^2z_3=0.$

Hence $$G_1+\omega G_2+\omega^2G_3=\frac{1}{3}(a+c+p)+\omega\frac{1}{3}(a+q+b)+\omega^2\frac{1}{3}(b+r+c)$$ $$\tag{4}=\frac{1}{3}\big[(a+\omega q+\omega^2b)+(c+\omega b+\omega^2r)+(p+\omega a+\omega^2c)\big]=0. \qquad \square$$ Alternatively, from $[Ex. 19]$, we can use the fact that the centroids of $\triangle abc$ and $\triangle pqr$ coincide. Place the origin at $G$, then $$G_1+G_2+G_3=0 \tag{5}.$$ We will next show that the centroids are all an equal distance from $G$, as Figure 1 strongly suggests. Since the outer triangles are all similar, $$\frac{p-c}{a-c}=\frac{b-c}{r-c}=Re^{i\theta}=A$$ and further, from above we know $$A=-\frac{1}{\omega^2}=e^{\frac{5\pi i}{3}},$$ so that $$p-b=(a-c)A+c-(r-c)A-c=(a-r)A$$ $$\implies |p+a+c|=|p-b|=|a-r||A|=|a+p+q|.\tag{6}$$ Similarly, $|G_3|=|G_1|$ and so the centroids are all a fixed distance from $G$ and they sum to $0$. These conditions are expressed in Figure 2. Noting that the angles between the centroids are invariant under dilation and rotation, you can use the roots of unity as in  this proof to show $\triangle G_1G_2G_3$ is equilateral. More simply, since the triangle in $[B]$ is equilateral, the angle at $G_3$ must be $\pi/3$. Imagine sliding the arrow $G_3$ up so that it is tail to tail with $G_1$, then the angle between them must now be $\pi -\pi/3=2\pi/3$ radians. Hence the angles between the vectors in $[A]$ are all $2\pi/3$ which implies $$|G_1-G_2|=|G_2-G_3|=|G_1-G_3|.\tag{7} \qquad \square$$

$\tag{Fig. 2}$
