$\lambda_3=\lambda_4$. This contradicts Lemma \[le:lambdadis\] and hence the points $\lambda_{3}$ and $\lambda_4$ are not real.

We will label the points $\gamma_3$ and $\gamma_4$ such that $\mathrm{Im}(\lambda_3)>0$.

Note that, from the behavior of $z(\xi)$ in (\[eq:curve1\]), we see that near the points $-1$ and $-\frac{1}{a}$, the function $z(\xi)$ behaves as $$\label{eq:zsing}
\begin{split}
z(\xi)&=\frac{c(1-\beta)}{1+\xi}+O(1),\quad \xi\rightarrow -1,\\
z(\xi)&=\frac{ca\beta}{1+a\xi}+O(1),\quad \xi\rightarrow
-\frac{1}{a}.
\end{split}$$ and hence $z^{\prime}(\xi)$ is negative near these points. From this and (\[eq:zprime\]), we see that $z^{\prime}(\xi)>0$ on the intervals $(-\infty,\gamma_1)$, $(\gamma_2,0)$ and $(0,\infty)$ and none of the points $-1$ or $-\frac{1}{a}$ belongs to these intervals. On $(\gamma_1,\gamma_2)$, $z^{\prime}(\xi)$ is negative.

The images of these intervals under the map (\[eq:curve1\]) then give us the complement of $\mathrm{supp}(\underline{F})$ in the $z$-plane. Let us study these images

\[le:image\] The intervals $(-\infty,\gamma_1)$, $(\gamma_2,0)$ and $(0,\infty)$ are mapped by $z(\xi)$ to $(0,\lambda_1)$, $(\lambda_2,\infty)$ and $(-\infty,0)$ respectively.

Since none of the points $-1$, $-\frac{1}{a}$ and $0$ belongs to these intervals both $z(\xi)$ and $z^{\prime}(\xi)$ are continuous on these intervals. Moreover, $z(\xi)$ is strictly increasing on these intervals. Therefore the images of these intervals are given by $$\begin{split}
z\left((-\infty,\gamma_1)\right)&=(z(-\infty),z(\gamma_1))=(0,\lambda_1)\\
z\left((\gamma_2,0)\right)&=(z(\gamma_2),z(0^-))=(\lambda_2,\infty)\\
z\left((0,\infty)\right)&=(z(0^+),z(\infty))=(-\infty,0).
\end{split}$$ where the $\pm$ superscripts in the above indicates that the function is evaluated at $\pm\epsilon$ for $\epsilon\rightarrow 