(\[eq:algeq\]), we see that the Stieltjes transform $m_{\underline{F}}(z)$ satisfies the following equation $$\label{eq:meq}
m_{\underline{F}}(z)=-\left(z-c\int_{\mathbb{R}}\frac{tdH(t)}{1+tm_{\underline{F}}}\right)^{-1}.$$ This equation has an inverse [@BS98], [@BS99] $$\label{eq:zeq}
z(m_{\underline{F}})=-\frac{1}{m_{\underline{F}}}+c\int_{\mathbb{R}}\frac{t}{1+tm_{\underline{F}}}dH(t).$$ The points where $\frac{d z(\xi)}{d\xi}=0$ are of significant interest to us as they are potential end points of the support of $\underline{F}$, due to the following result by Choi and Silverstein.

\[le:CS\]$\cite{CS}$ If $z\notin\mathrm{supp}(\underline{F})$, then $m=m_{\underline{F}}(z)$ satisfies the following.

1.  $m\in\mathbb{R}\setminus\{0\}$;

2.  $-\frac{1}{m}\notin\mathrm{supp}(H)$;

3.  Let $z$ be defined by (\[eq:zeq\]), then $z^{\prime}(m)>0$, where the prime denotes the derivative with respect to $m_{\underline{F}}$ in (\[eq:zeq\]).

Conversely, if $m$ satisfies 1-3, then $z=z(m)\notin\mathrm{supp}(\underline{F})$.

This lemma allows us to identify the complement of $\mathrm{supp}(\underline{F})$ by studying the real points $m$ such that $z^{\prime}(m)>0$.

\[re:baik\] As pointed out in [@Baikspike], Lemma \[le:CS\] applies to any distribution $G$ whose Stieltjes transform $m_{G}(z)$ (\[eq:stie\]) satisfies an equation of the form $$z(m_{G})=-\frac{1}{m_{G}}+c_G\int_{\mathbb{R}}\frac{t}{1+tm_{G}}dH_G(t).$$ for some constant $c_G$ and distribution $H_G(t)$.

Riemann surface and the Stieltjes transform
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We will now restrict ourselves to the case when the matrix $\Sigma_N$ has 2 distinct eigenvalues only. Without lost of generality, we will assume that one of these values is 1 and the other one is $a>0$. Let $0<\beta<1$, we will assume that as $N\rightarrow\infty$, $N_1$ of the 