\begin{array}{l}
        \Phi_t(z,{\lambda}_0) = Mz^Bz^{A_t({\lambda}_0)}P(z),\\
        B^2=0,\\
        P(0) = {\mathrm{I}}_2,\\
        \left[A_t({\lambda}_0),d_zP(0)\right] + d_zP(0) = C_t(0,{\lambda}_0) - B.
        \end{array}
        \right.$$ Compute the monodromy of $\Phi_t$ at ${\lambda}={\lambda}_0$: $${\mathcal{M}}(\Phi_t)({\lambda}_0) = M\exp(2i\pi B)z^B\exp(2i\pi A_t({\lambda}_0))z^{-B}M^{-1} = -M\exp(2i\pi B)M^{-1}.$$ Since the monodromy problem is solved, this quantity equals $-{\mathrm{I}}_2$. Use $B^2=0$ to show that $B=0$ and thus, $$\left\{
    \begin{array}{l}
    \Phi_t(z,{\lambda}_0) = Mz^{A_t({\lambda}_0)}P(z),\\
    P(0) = {\mathrm{I}}_2,\\
    \left[A_t({\lambda}_0),d_zP(0)\right] + d_zP(0) = C_t(0,{\lambda}_0).
    \end{array}
    \right.$$ Diagonalise $A_t({\lambda}_0)$ with Proposition \[propDiagA\] and write $d_zP(0)=\sum p_jX_{t,j}$ to get for all $j\in[1,4]$ $$p_j\left([D_t({\lambda}_0),E_j] + E_j\right) = t\widehat{c}_j(t,{\lambda}_0)E_j.$$ In particular, using $\mu_t({\lambda}_0)=1/2$, $$t\widehat{c}_3(t,{\lambda}_0) = p_3\left([D_t({\lambda}_0),E_3] + E_3\right)=0.$$

Note that with the help of Equations and or , and one can compute the series expansion of $\widehat{c}_3(0)$: $$\widehat{c}_3(0) = \frac{-{\lambda}^{-1}}{2}c_2(0,0) + {\mathcal{O}}({\lambda}^0) \quad \text{if}\quad r<s,$$ $$\widehat{c}_3(0) = \frac{1}{2} c_3(0,0) + {\mathcal{O}}({\lambda}) \quad \text{if} \quad r>s.$$ Hence, $$\label{eqsc2+rc3=0}
    sc_2(t,0)+rc_3(t,0) {\underset{t\to 0}{\longrightarrow}} 0.$$

The following map will be useful in the next section: $$\label{eqP1}
t\in(-T',T')\longmapsto  P^1(t) := t\widehat{c}_1(t)X_{t,1} + \frac{t\widehat{c}_2(t)}{1+2\mu_t}X_{t,2} + \frac{t\widehat{c}_3(t)}{1-2\mu_t}X_{t,3} + t\widehat{c}_4(t)X_{t,4}.$$ For all $t$, Proposition \[propC3\] ensures that the map $P^1(t,{\lambda})$ holomorphically extends to ${\mathbb{A}}_R$. Taking a smaller value of $R$ if necessary, $P^1(t)\in \Lambda{\mathcal{M}}(2,{\mathbb{C}})_R$ for all $t$. Moreover, $${\mathop \mathrm{tr}}P^1(t) = t\widehat c_1 (t) + t\widehat c_4(t) = {\mathop \mathrm{tr}}C_t(0) = 0.$$ Thus, 