set $\mathfrak{H}$ defined by $$\label{eq:frakH}
\mathfrak{H}=\left\{z\in\mathbb{C}|\quad
\mathrm{Re}\left(\theta_2^N(z)-\theta_3^N(z)\right)=0\right\}$$ is important to the Riemann-Hilbert analysis. Let us now study its properties. First note that it is symmetric across the real axis.

\[le:symm\] Let the branch cut $\mathcal{C}$ between $\lambda_{3}^N$ and $\lambda_4^N$ be a simple contour that is symmetric with respect to the real axis. Then the set $\mathfrak{H}$ in (\[eq:frakH\]) is symmetric with respect to the real axis.

Let us consider the transformation $z\mapsto\overline{z}$ and $\xi(z)\mapsto\overline{\xi}(\overline{z})$ in (\[eq:curveN\]). This gives $$\label{eq:curveNcon}
\begin{split}
\overline{z}a\overline{\xi}(\overline{z})^3&+(A_2\overline{z}+B_2^N)\overline{\xi}(\overline{z})^2+(\overline{z}+B_1^N)\overline{\xi}(\overline{z})+1=0.
\end{split}$$ This means that if $\xi(z)$ is a solution to (\[eq:curveN\]), then so is $\overline{\xi}(\overline{z})$. Since the functions $\overline{\xi}_j^N(\overline{z})$ is analytic away from $\mathcal{C}\cup[\lambda_1^N,\lambda_2^N]$, it must equal either of the functions $\xi_1^N(z)$, $\xi_2^N(z)$ or $\xi_3^N(z)$. By considering the behavior of $\overline{\xi}_j^N(\overline{z})$ near $z=\infty$ using (\[eq:xiinfty\]), we see that $\overline{\xi}_j^N(\overline{z})=\xi_j^N(z)$ for $j=1,2,3$. Let $z_0\in\mathfrak{H}$, then we have $$\label{eq:zoH}
\mathrm{Re}\left(\int_{\lambda^N_3}^{z_0}(\xi_2^N(x)-\xi_3^N(x))dx\right)=0.$$ By taking the complex conjugation of (\[eq:zoH\]) and making use of the fact that $\overline{\xi}_j^N(\overline{z})=\xi_j^N(z)$ for $j=2,3$, we obtain $$\mathrm{Re}\left(\int_{\lambda^N_4}^{\overline{z}_0}(\xi_2^N(x)-\xi_3^N(x))dx\right)=0.$$ Let us now show that $\mathrm{Re}\left(\theta_2^N(\lambda_4^N)-\theta_3^N(\lambda_4^N)\right)=0$. Consider an integration contour $\Gamma$ from $\lambda^N_3$ to $\lambda_4^N$ that is symmetric with respect to $\mathbb{R}$. Then we have $$\begin{split}
\overline{\mathrm{Re}\left(\int_{\lambda_3^N}^{\lambda_4^N}\left(\xi_2^N(x)-\xi_3^N(x)\right)dx\right)}&=
\mathrm{Re}\left(\int_{\lambda_4^N}^{\lambda_3^N}\left(\xi_2^N(x)-\xi_3^N(x)\right)dx\right)\\
&=-\mathrm{Re}\left(\int_{\lambda_3^N}^{\lambda_4^N}\left(\xi_2^N(x)-\xi_3^N(x)\right)dx\right),
\end{split}$$ where we have used $\lambda_3^N=\overline{\lambda}_4^N$ in the above. Hence we have 