singularities at $z=-1$ and $z=1$ with $R(z)=\mathcal{O}|z\pm 1|^{-1/2}$ as $z \to \mp 1$, and they are removable. So $R$ satisfies the following Riemann-Hilbert conditions:

1.  $R \in \mathcal{H}(\mathbb{C} \setminus \gamma)$,

2.  $\displaystyle R_+(\zeta)=R_-(\zeta) \left(\begin{array}{c c}
    1 & 0 \\
    e^{-2 n \mathcal{A}(\zeta)} & 1 
    \end{array}
    \right),$ when $\zeta \in \gamma.$

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

From (\[defwp\]) $2\mathcal{A}\in \mathcal{H}(D\setminus[-\infty,\,1])$ and $\mathcal{R}e(2\mathcal{A}_{\pm}(x))=0$, $x \in [-1,\,1]$. Using the fact $\displaystyle 2\mathcal{A}_{\pm}^{\prime}(x)=\pm 2 i \overline{\nu}^{\prime}(x)=\mp 2 \pi i\frac{\psi(x)}{\sqrt{1-x^2}}$, $x \in (-1,\,1)$ and the Cauchy-Riemann conditions we have that $\displaystyle \frac{\partial \mathcal{R}e(2\mathcal{A}_\pm)}{\partial y}(x)>0$, $x \in [-1,\,1]$. Since $\mathcal{R}e(2\mathcal{A})$ is a harmonic function on $D\setminus [-1,\,1]$ we have $\mathcal{R}e(2\mathcal{A}(z))>0,$ $z\in D\setminus [-1,\, 1]$. So given an arbitrary compact set $K \subset D\setminus [-1,\,1]$ there exists a constant $c(K)>0$ and an $N\in \mathbb{N}$ large enough such that for every $n\geq N$ the function $\mathcal{R}e(2\mathcal{A}(z))(z))>c(K)$, $z \in K$ and $n\geq N$. Note also that $\phi_n \to 0$ as $n \to \infty$. So according [@AV] we arrive at $R(z)=\mathbb{I}+\mathcal{O}(e^{-c n})$ uniformly as $n \to \infty$ for each compact set $K \subset \mathbb{C} \setminus [-1,\,1]$. Take $z \in \mbox{Int}(\gamma)$. Going back now from $R$ to $Y$, and considering just the first column, we have that: $$e^{n g(z, \overline{\nu})} 