map $z(\xi)$ given in (\[eq:mN\]).

For the time being, we will choose the branch cut of $\xi_1^N(z)$ to be the interval $[\lambda_1^N,\lambda_2^N]$, and the branch cut between $\lambda_{3}^N$ and $\lambda_4^N$ to be a simple contour $\mathcal{C}$ that is symmetric with respect to the real axis and oriented upwards. Across $\mathcal{C}$, the two branches $\xi_2^N(z)$ and $\xi_3^N(z)$ change into each other. The branch cut $\mathcal{C}$ will be chosen such that it intersects the real axis at exactly one point $x^{\ast}$ that is not equal to $\lambda_1^N$ or $\lambda_2^N$.

We will now define the functions $\theta_j^N(z)$ to be the the integrals of $\xi_j^N(z)$. $$\label{eq:theta}
\begin{split}
\theta_1^N(z)&=\int_{\lambda_l^N}^z\xi_1^N(x)dx,\quad
\theta_2^N(z)=\int_{\lambda_{3}^N}^z\xi_2^N(x)dx,\quad
\theta_3^N(z)=\int_{\lambda_{3}^N}^z\xi_3^N(x)dx,\\
\textrm{if }x^{\ast}&<\lambda_2^N,\quad l=2;\quad \textrm{if
$x^{\ast}>\lambda_2^N,\quad l=1.$}
\end{split}$$ The integration paths of the above integrals are chosen as follows. If $x^{\ast}<\lambda_2^N$, then the integration path will not intersect the set $\mathcal{C}\cup(-\infty,\lambda_2^N)$ and if $x^{\ast}>\lambda_2^N$, then the integration path will not intersect the set $\mathcal{C}\cup(\lambda_1^N,\infty)$.

Then from (\[eq:xiinfty\]), (\[eq:zeroasym\]) and (\[eq:xizero\]), we see that the integrals (\[eq:theta\]) have the following behavior at $z=\infty$ and $z=0$. $$\label{eq:asymtheta}
\begin{split}
\theta_1^N(z)&=-\log z+l_1^N+O\left(z^{-1}\right),\quad
z\rightarrow\infty,\\
\theta_1^N(z)&=-(1-c_N)\log z+O\left(1\right),\quad
z\rightarrow 0,\\
\theta_2^N(z)&=-z+c_N(1-\beta_N)\log
z+l_2^N+O\left(z^{-1}\right),\quad
z\rightarrow\infty,\\
\theta_2^N(z)&=O\left(1\right),\quad
z\rightarrow 0,\\
\theta_3^N(z)&=-\frac{z}{a}+c_N\beta_N\log
z+l_3^N+O\left(z^{-1}\right),\quad
z\rightarrow\infty,\\
\theta_3^N(z)&=O\left(1\right),\quad z\rightarrow 0.
\end{split}$$ for some constants $l_1^N$, $l_2^N$ and $l_3^N$.

Then the 