is given by (\[eq:D3\]) and $r(z)$ is given by $$\begin{split}
r(z)&=\frac{1}{27}\Bigg(-\frac{2B_2^3}{a^3}z^{-3}+\left(\frac{9B_1B_2}{a^2}-\frac{6A_2B_2^2}{a^3}\right)z^{-2}+\left(\frac{9B_2}{a^2}+\frac{9B_1A_2}{a^2}-\frac{27}{a}-\frac{6A_2^2B_2}{a^3}\right)z^{-1}\\
&+\left(\frac{9A_2}{a^2}-\frac{2A_2^3}{a^3}\right)\Bigg).
\end{split}$$ The cube root in (\[eq:density\]) is chosen such that $\sqrt[3]{A}\in\mathbb{R}$ for $A\in\mathbb{R}$ and the square root is chosen such that $\sqrt{A}>0$ for $A>0$.

The proof of this theorem is the same as Theorem 6 in [@Mo].

Let us now show that the function $\xi_1(z)$ is in fact the Stieltjes transform $m_{\underline{F}}(z)$. From the asymptotic behavior of the $\xi_j(z)$ (\[eq:xiinfty\]) and the fact that the Stieltjes transform $m_{\underline{F}}(z)$ is the unique solution of (\[eq:curve2\]) that vanishes as $z\rightarrow\infty$, we see that $$\label{eq:stiexi}
m_{\underline{F}}(z)=\xi_1(z).$$ For $c<1$, it was shown in [@CS] that $F$ has a continuous density and hence the Stieltjes transform $m_F(z)$ does not have any poles. Therefore, by (\[eq:stielim\]), we see that $m_{\underline{F}}(z)$, and hence $\xi_1(z)$, has the following singularity at $z=0$. $$\label{eq:xizero}
\xi_1(z)=-\frac{1-c}{z}+O(1),\quad z\rightarrow 0.$$ Since $m_{\underline{F}}(z)$ is the Stieltjes transform of a measure supported on the real axis, it is analytic away from the real axis and hence by (\[eq:stiexi\]), $\xi_1(z)$ is analytic at the points $\lambda_3$ and $\lambda_4$ with a branch cut on $[\lambda_1,\lambda_2]$. Since $\lambda_3$, $\lambda_4$ are not branch points of the function $\xi_1(z)$, they must be branch points of the functions $\xi_2(z)$ 