$v_{i+1}, v_{i+2}, v_{i+3}$ are not collinear. In addition to that, since $V_{i+n} = V_i$ we have that the sequences $a_i,b_i,c_i$ are $n$-periodic. Finally, notice that for fixed $a_i, b_i, c_i$ the sequence $V_i$ is uniquely determined by equation and initial condition $V_0, V_1, V_2$. Indeed, given $V_0, V_1, V_2$ and using that $c_i \neq 0$, we can successively find all $V_i$’s from . This gives us a way to deform the polygon $P$: keeping $V_0, V_1, V_2$ unchanged, we deform the coefficients in . Namely, consider the following equation $$\begin{aligned}
\label{eq6}
     V_{i+3} = a_iV_{i+2} +z^{-1}\big( b_i V_{i+1}+c_iV_i \big), \end{aligned}$$ We assume that the vectors $V_0, V_1, V_2$ do not depend on $z$ and coincide with the above-constructed lifts of vertices of $P$. For any $z \neq 0$, equation has a unique solution with such initial condition. For $z = 1$ we recover the initial polygon, while for other values of $z$ we get its deformation. Note that for $i \neq 0,1,2$ the solutions $V_i$ of are actually functions of the parameter $z$, i.e. $V_i = V_i(z)$.

Taking the solution of such that $V_0, V_1, V_2$ are fixed lifts of vertices $v_0, v_1, v_2$ of $P$ and projecting the vectors $V_i 