= \sum_{i=0}^{n-1} W_i \frac{dU_i}{dz} W_{i+1}^{-1}W_{n}  W_0^{-1} = \left(\sum_{i=0}^{n-1} W_i \frac{dU_i}{dz} W_{i+1}^{-1}\right)M_z.\end{aligned}$$ Further, using that the monodromy satisfies $M_1={ {I}}$ because we started with a closed $n$-gon, we arrive at $$\left.\frac{dM_z}{dz}\right|_{z=1} \!\!\!= \,\,\sum_{i=0}^{n-1} S_i,$$ where $$S_i := \left.\left(W_i \frac{dU_i}{dz} W_{i+1}^{-1}\right)\,\right\vert_{z=1}.$$ Now, we will show that summing these $S_i$ with $i=0,1,\ldots,n-1$ gives up to a scalar matrix. Using , we get $$\begin{aligned}
\left.\frac{dU_i}{dz}\right|_{z=1} \!\!=\,\, 
\begin{bmatrix}
0 & 0 & 0 \\
-b_i & 0 & 0 \\
-c_i & 0 & 0
\end{bmatrix}.\end{aligned}$$ Further, observe that for $z=1$ the matrix $W_i$ sends the standard basis to the lifts $V_{i+2}, V_{i+1}, V_i$ of the vertices of $P$. Therefore $W_{i+1}^{-1}$ takes the vectors $V_{i+3},V_{i+2},V_{i+1}$ to the standard basis, from which we find that the matrix $S_i$ acts on these vectors as $$V_{i+3} \mapsto -b_iV_{i+1} - c_iV_i,
       \quad V_{i+2} \mapsto 0
       \quad V_{i+1} \mapsto 0.$$ Using also , we find that $$S_i(V_i) = \frac{1}{c_i}S_i(V_{i+3}) =  -\frac{b_i}{c_i} V_{i+1} - V_i,$$ which means that $$\begin{aligned}
    S_i(V)=\frac{|V,V_{i+1},V_{i+2}|}{|V_i,V_{i+1},V_{i+2}|}\Big(-V_i-\frac{b_i}{c_i}V_{i+1}\Big) \quad \forall \,\,V \in {\mathbb R}^3, \label{eq13}\end{aligned}$$ where $|A,B,C|$ is the determinant of the matrix with columns $A, B, C$. Further, rewriting as $$-V_i-\frac{b_i}{c_i}V_{i+1} = \frac{a_i}{c_i}V_{i+2} -\frac{1}{c_i}V_{i+3}$$ we get $$\begin{aligned}
    S_i(V)=
    \frac{|V_{i+1},V_{i+2},V|}{|V_{i+1},V_{i+2},V_i|}\Big( \frac{a_i}{c_i}V_{i+2} -\frac{1}{c_i}V_{i+3} \Big) = \frac{|V_{i+1},V_{i+2},V|}{|V_{i+1},V_{i+2},c_i^{-1}V_{i+3}|}\Big( \frac{a_i}{c_i}V_{i+2} -\frac{1}{c_i}V_{i+3} \Big),\end{aligned}$$ where 