have $\gamma<0$.

The following remark was used in [@Szp17 Remark 2.1] without a proof. We give a proof here.

\[rem:F\_lambda\_sym\] We have $f_{\alpha,\beta,\lambda}(z)=f_{\alpha,\beta,-\lambda}(z)$, where $\alpha,\beta>0,\lambda\in\mathbb{R}$.

To see this one has to insert the definition of $\gamma$ into to obtain $$f_{\alpha,\beta,\lambda}(z)=\alpha z \lambda^2+\left(\left(ab\alpha^2-2\alpha\beta\frac{a+b}{\sqrt{ab}}+\frac{\beta^2}{ab}+2\alpha\beta\right)z-4 \beta z^2-\alpha\right) (z-\alpha),$$ where $a=a(\alpha,\beta,\lambda)$ and $b=b(\alpha,\beta,\lambda)$. Thus it suffices to show that the quantity $g(\alpha,\beta,\lambda):=ab\alpha^2-2\alpha\beta\frac{a+b}{\sqrt{ab}}+\frac{\beta^2}{ab}$ does not depend on the sign of $\lambda$. To see this, observe from the system of equations and that $a(\alpha,\beta,-\lambda)=\frac{\beta}{\alpha b(\alpha,\beta,\lambda)}$ and $b(\alpha,\beta,-\lambda)=\frac{\beta}{\alpha a(\alpha,\beta,\lambda)}$. It is then straightforward to check that $
g(\alpha,\beta,-\lambda) = g(\alpha,\beta,\lambda). 
$

The $R$-transform of the measure $\mu(\alpha,\beta,\lambda)$ can be extended to a function (still denoted by $r_{\alpha,\beta,\lambda}$) which is analytic on $\mathbb{C}^{-}$ and continuous on $({\mathbb{C}}^- \cup{\mathbb{R}})\setminus\{\alpha\}$.

A direct calculation shows that using parameters $A,B$ defined by the polynomial $f_{\alpha,\beta,\lambda}$ under the square root factors as $$f_{\alpha,\beta,\lambda}(z) =  \frac{(B-A)^2(B-\lambda A)}{2 A B}\left[z +\frac{2(B+\lambda A)}{B(B-A)}\right]^2  \left[  \frac{2B}{A(B-\lambda A)} - z   \right].$$ Thus we can write $$\label{eq:pol_par}
f_{\alpha,\beta,\lambda}(z) = 4\beta (z-\delta)^2(\eta-z),$$ where $$\begin{aligned}
&\delta = - \frac{2(B+\lambda A)}{B(B-A)}<0, \label{eq5}\\
&\eta = \frac{2B}{A(B-\lambda A)} >0. \label{eq6}\end{aligned}$$ It is straightforward to verify that implies $\eta \geq \alpha$ with equality valid only when $\lambda=0$.\
Calculating $f_{\alpha,\beta,\lambda}(0)$ using first and then we get 