Using complex numbers to prove Napoleon's Theorem 
Let $ABC$ be a triangle and erect equilateral triangles on sides $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ outside of $ABC$ with centers $O_A$, $O_B$, $O_C$. Prove that $\bigtriangleup O_AO_BO_C$ is equilateral and that its center coincides with the centroid of triangle $ABC$

I have already seen this answer Proving Napoleon's Theorem with complex numbers
but my doubt is different ,
Now,
In this answer https://artofproblemsolving.com/community/c618937h1650553_proposition_634_napoleons_theorem ($5$th post)
they wrote -
$O_AC$ is a $\frac\pi6$ rotation of $BC$ followed by a dilation with ratio $\frac1{\sqrt3}$ at $C,$ so we have
$\begin{align*}
\frac{o_A-c}{b-c}&=\frac1{\sqrt3}\cdot\frac{\sqrt3+i}{2}\end{align*}$ but i am not able to understand this ,can anybody explain this step please ?
Note-I have solved this problem using simple angle chasing ,but i want to understand properly that how they got co-ordinates of $O_A$

thankyou
 A: Since $O_A$ is the center of an equilateral triangle with $BC$ as one of its side, then $\angle O_ABC=\frac{\pi}{6}$. Furthermore, $\triangle O_ABC$ is isosceles with $\angle O_ABC=\angle O_ACB=\frac{\pi}{6}$.
Hope you can imply the rest from these
A: First of all, for any equilateral triangle $XYZ$ with side $x$ and centroid $G$, then
$$XG= \frac{2}{3} \cdot \frac{\sqrt{3}}{2}x=\frac{x}{\sqrt3} $$
Now locate the triangle $ABC$ with its centroid $O$ on the pole of the complex plane.
Let $z_{1}, z_{2},z_{3}$ be the complex numbers representing respectively the points $A, B $ and $C$, then $$z_{1}+z_{2}+z_{3}=0\cdots (*)$$.
Also denoting the three centroids $G_i$‘s of the equilateral triangles so-formed by the complex numbers $g_1, g_2$ and $g_3$, then $g_i$’s can be found by rotating the sides of the triangle $ABC$ by $\frac{\pi}{6}$ in clockwise direction as below:
$$
\begin{array}{l}
g_{1}=z_{2}+\frac{1}{\sqrt{3}}\left(z_{3}-z_{2}\right) e^{\frac{\pi}{6} i} \cdots(1) \\
g_{2}=z_{3}+\frac{1}{\sqrt{3}}\left(z_{3}-z_{1}\right) e^{\frac{\pi}{6} i} \cdots(2) \\
g_{3}=z_{1}+\frac{1}{\sqrt{3}}\left(z_{2}-z_{1}\right) e^{\frac{\pi}{6} i} \cdots(3)
\end{array}
$$
Adding them together yields
$$
\begin{aligned}
g_{1}+g_{2}+g_{3} &=z_{1}+z_{2}+z_{3}+\frac{1}{\sqrt{3}} e^{\frac{\pi}{6} i}\left(z_{3}-z_{2}+z_{3}-z_{1}+z_{2}-z_{1}\right) \\
&=0+0 \\
&=0
\end{aligned}
$$
Then centroid of the triangle of centroids $=\dfrac{g_{1}+g_{2}+g_{3}}{3}=0$ which coincide with the centroid of the original triangle $ABC$.
