$4 \beta \eta \delta^2 = \alpha^2$, since $\eta \geq \alpha$ we see that $\delta \geq -\sqrt{\alpha/(4\beta)}$ with equality only when $\lambda=0$.

Since all roots of $f_{\alpha,\beta,\lambda}$ are real, the square root $\sqrt{f_{\alpha,\beta,\lambda}(z)}$ may be defined continuously on ${\mathbb{C}}^-\cup{\mathbb{R}}$ so that $\sqrt{f_{\alpha,\beta,\lambda}(0)}=\alpha$. As noted above $\delta<0$, and continuity of $f_{\alpha,\beta,\lambda}$ implies that we have $$\label{RRR}
 \sqrt{f_{\alpha,\beta,\lambda}(z)} = 2(z-\delta)\sqrt{\beta(\eta-z)},$$ where we take the principal value of the square root in the expression $\sqrt{4\beta(\eta-z)}$. Thus finally we arrive at the following form of the $R$-transform $$\label{R}
\begin{split}
r_{\alpha,\beta,\lambda}(z) 
&= \frac{-\alpha + (\lambda+1)z + 2(z-\delta)\sqrt{\beta(\eta-z)}}{2z(\alpha-z)} 
\end{split}$$ which is analytic in ${\mathbb{C}}^-$ and continuous in $({\mathbb{C}}^- \cup{\mathbb{R}})\setminus\{\alpha\}$ as required.

Next we describe the behaviour of the $R$-transform around the singular point $z=\alpha$.

If $\lambda>0$ then $$\label{alpha}
    \begin{split}
    r_{\alpha,\beta,\lambda}(z) 
    = \frac{\lambda}{\alpha-z} -\frac{1}{2\alpha} \left(1+\lambda+\frac{\sqrt{\beta}(2\eta-3\alpha+\delta)}{\sqrt{\eta-\alpha}}\right) + o(1),\qquad\mbox{as } z\to\alpha. 
    \end{split}$$ If $\lambda<0$ then $$\label{alpha2}
    r_{\alpha,\beta,\lambda}(z) = - \frac{1}{2\alpha}\left(1+\lambda+\frac{\sqrt{\beta}(2\eta-3\alpha+\delta)}{\sqrt{\eta-\alpha}}\right)+ o(1), \qquad \mbox{as } z\to\alpha.$$ In the remaining case $\lambda=0$ one has $$\label{alpha3}
    \begin{split}
    r_{\alpha,\beta,0}(z) 
    &= \frac{-\alpha + z + 2(z-\delta)\sqrt{\beta(\alpha-z)}}{2z(\alpha-z)} = -\frac{1}{2z} + \frac{\sqrt{\beta}(z-\delta)}{z\sqrt{\alpha-z}}. 
    \end{split}$$

By the definition we have $f_{\alpha,\beta,\lambda}(\alpha) = (\lambda\alpha)^2$, substituting this in the expression we obtain that $
\alpha |\lambda| = 2(\alpha-\delta)\sqrt{\beta (\eta-\alpha)}. 
$ Taking the Taylor expansion around $z=\alpha$ for $\lambda \neq0$ we obtain 