The action of $\mathbb{Z}_{p}$ on $\mathbb{R}^{2}$

Consider the action of $\mathbb{Z}_{p}$ on $\mathbb{R}^{2}$ generated by a rotation of angle $2 \pi / p$ around the origin. The quotient is a cone with cone angle $2 \pi / p$. For $p=3$, we get the following cone :

I didn't understand how they find this results ! Let $u\in \mathbb R^2,$ and $p=3$,

\begin{align*}[u]&=\{v \in \mathbb R^2 : u \sim v\} \\&=\{v \in \mathbb R^2 /: R(O,2\pi/3)(u)=v \} \\&= \{ v=\begin{pmatrix} -1/2u_1-\sqrt 3/2 u_2 \\ \sqrt 3/2u_1-1/2u_2 \end{pmatrix} \in \mathbb R^2 \}\end{align*} I get stuck here and I am trying to get a formal proof so any help is highlty appreciated !

The idea is the same like that of Steven Creech.

The homemorphism map is the following:

Let $C:=\{(x,y,z) : z=-\sqrt{x^2+y^2}\}$ be a cone on $\mathbb{R}^3$. Then

$\phi: \mathbb{R}^2/ \mathbb{Z}_p\to C$

is defined by

$\phi([(\rho (cos(2\pi\theta), sin(2\pi\theta))]):= \rho ( cos(2\pi \theta p) , sin(2\pi\theta p) , -1) $

In this case is better to define the map using polar coordinates, but you can define it also using the Cartesian coordinates.