where $\displaystyle D(z)=\left(\frac{z}{\sqrt{z^2-1}}+1\right)^{1/2}, \quad D(\infty)=\sqrt{2}$ and $\displaystyle a(z)= \frac{(z-1)^{1/4}}{(z+1)^{1/4}}.$ Hence $N$ is the solution of the following Riemann-Hilbert problem

1.  $N \in \mathcal{H}(\mathbb{C}\setminus [-1,\,1])$,

2.  $\displaystyle N_+(x)=N_-(x)\left(\begin{array}{c c}
    0 & \displaystyle (1-x^2)^{-1/2} \\
    -(1-x^2)^{1/2} &0 
    \end{array}
    \right)$, $x\in (-1,\,1),$

3.  $N(z)=\mathbb{I}+\mathcal{O}\left(1/z\right)$ as $z \to \infty$,

4.  $\displaystyle N(z)=\mathcal{O}\left(\begin{array}{ c c }
    1 & \left| z\pm 1\right|^{-1/2} \\
    1 & \left| z \pm 1\right|^{-1/2} 
    \end{array}\right)$ as $z \to \mp 1$.

Introduce the matrix function $R(z)=S(z)N^{-1}$. Taking into account that $R$ and $S$ satisfy the same jump conditions across $(-1,\,1)$ we have that $R_+(x)=R_-(x)$. So $R \in \mathcal{H}(\mathbb{C} \setminus (\gamma \cup \{-1,1\}))$. Since $\det N=1$ and from (\[Ndef\]) we have that $$N^{-1}(z)=\mathcal{O}\left(\begin{array}{ c c }
\left| z\pm 1\right|^{-1/2} & \left| z\pm 1\right|^{-1/2} \\
1 & 1 
\end{array}\right) \quad \mbox{as} \quad z \to \mp 1.$$ Thus, when $z \to \mp 1$ $$R(z)=\mathcal{O}\left(\begin{array}{c c}
1 & \left| z\pm 1\right|^{-1/2}  \\
1 & \left| z\pm 1\right|^{-1/2}  \end{array}\right)\mathcal{O}\left(\begin{array}{ c c }
\left| z\pm 1\right|^{-1/2}  & \left| z\pm 1\right|^{-1/2}  \\
1 & 1 
\end{array}\right).$$ This implies $$R(z)= \mathcal{O}\left(\begin{array}{ c c }
\left| z\pm 1\right|^{-1/2} & \left| z\pm 1\right|^{-1/2} \\
\left| z\pm 1\right|^{-1/2} & \left| z \pm 1\right|^{-1/2} 
\end{array}\right),$$ which means that each entry of $R$ has isolated 