V_{i+1}(z) \quad V_i(z) \big].$$ Then the relation gives us the matrix equation $$W_{i+1}(z)=W_i(z) U_i(z),$$ where $$\begin{aligned}
\label{eq:ui}
    U_i(z):=\begin{bmatrix}
a_i & 1 & 0 \\
z^{-1}b_i & 0 & 1 \\
z^{-1}c_i & 0 & 0
\end{bmatrix}.\end{aligned}$$ We stop explicitly recording the dependence on $z$ as it is notationally cumbersome. Inductively, we have that $$\begin{aligned}
\label{eq11}
W_i = W_0U_0U_1 \dots U_{i-1}.\end{aligned}$$ In particular, $$W_{n}=W_0U$$ where $U:= U_0U_{1}\dots U_{n-1}$. At the same time, we have that $V_{i+ n}=M_zV_i$, where $M_z$ is a matrix representing the monodromy of the polygon defined by the vectors $V_i$. This means that $W_{n}=M_zW_0$. Relating these two expressions for $W_{n}$ we get $$W_0U=M_zW_0  \quad \iff \quad M_z=W_0UW_0^{-1}.$$ Notice that because $V_0,V_1,V_2$ are fixed we have that $W_0=[V_0 \quad V_1 \quad V_2]$ is constant while $z$ varies. This means that all the dependence of $M_z$ on $z$ is contained in the expression for $U$. This gives $$\begin{aligned}
   \frac{dM_z}{dz} &=\frac{d}{dz} \left( W_0 U_0 \dots U_{n-1}  W_0^{-1} \right) 
   \\ &=  \sum_{i=0}^{n-1} W_0 U_0 \dots U_{i-1}\frac{dU_i}{dz}U_{i+1} \dots U_{n-1}   W_0^{-1} 
       = \sum_{i=0}^{n-1} W_i \frac{dU_i}{dz}U_{i+1} \dots U_{n-1}   W_0^{-1},\end{aligned}$$ where the last equality uses that $W_{i}=P_0U_0\dots U_{i-1}$. Further, observe that $$\begin{aligned}
U_{i+1}\dots U_{n-1}&=(U_0\dots U_{i})^{-1}(U_0\dots U_{n-1}) =(W_0^{-1}W_{i+1})^{-1}(W_0^{-1}W_{n})=W_{i+1}^{-1}W_{n}.\end{aligned}$$ Also using that $W_n W_0^{-1}=M_z$, we get $$\begin{aligned}
   \frac{dM_z}{dz} 