eigenvalues take the value $a$ and $N_0=N-N_1$ of the eigenvalues are 1 and that $\frac{N_1}{N}\rightarrow\beta$. That is, as $N\rightarrow\infty$, the e.d.f $H_N(x)$ converges to the following $$\label{eq:limedf}
dH_N(x)\rightarrow dH(x)=(1-\beta)\delta_1+\beta\delta_{a}.$$ By substituting this back into (\[eq:zeq\]), we see that the Stieltjes transform $\xi(z)=m_{\underline{F}}(z)$ is a solution to the following algebraic equation $$\label{eq:curve1}
z=-\frac{1}{\xi}+c\frac{1-\beta}{1+\xi}+c\frac{a\beta}{1+a\xi}.$$ Rearranging the terms, we see that $\xi=m_{\underline{F}}$ solves the following $$\label{eq:curve2}
\begin{split}
za\xi^3&+(A_2z+B_2)\xi^2+(z+B_1)\xi+1=0,\\
A_2&=(1+a),\quad B_2=a(1-c),\\
B_1&=1-c(1-\beta)+a(1-c\beta).
\end{split}$$ This defines a Riemann surface ${\mathcal{L}}$ as a 3-folded cover of the complex plane.

By solving the cubic equation (\[eq:curve2\]) or by analyzing the asymptotic behavior of the equation as $z\rightarrow\infty$, we see that these solutions have the behavior given by (\[eq:xiinfty\]) as $z\rightarrow\infty$. On the other hand, as $z\rightarrow 0$, the 3 branches of $\xi(z)$ behave as follows $$\label{eq:zeroasym}
\begin{split}
\xi_{\alpha}(z)&=-\frac{1-c}{z}+O(1),\quad z\rightarrow 0,\\
\xi_{\beta}(z)&=R_1+O(z),\quad z\rightarrow 0,\\
\xi_{\gamma}(z)&=R_2+O(z),\quad z\rightarrow 0.
\end{split}$$ where the order of the indices $\alpha$, $\beta$ and $\gamma$ does not necessarily coincide with the ones in (\[eq:xiinfty\]) (i.e. we do not necessarily have $\alpha=1$, $\beta=2$ and $\gamma=3$). The constants $R_1$ and $R_2$ are the two roots of the quadratic equation $$\label{eq:quad}
a(1-c)x^2+(1-c(1-\beta)+a(1-c\beta))x+1=0.$$ The functions $\xi_j(z)$ will not be analytic at the branch points of ${\mathcal{L}}$ and they will 