values of $\xi$ at $\frac{d z}{d\xi}=0$ correspond to the roots of the quartic equation $$\label{eq:quart}
\begin{split}
a^2(1-c)\xi^4
&+2(a^2(1-c\beta)+a(1-c(1-\beta))\xi^3\\
&+(1-c(1-\beta)+a^2(1-c\beta)+4a)\xi^2 +2(1+a)\xi+1=0
\end{split}$$ Let $\Delta$ be the discriminant of this quartic polynomial, then when $\Delta<0$, the equation (\[eq:quart\]) has 2 distinct real roots $\gamma_1<\gamma_2$ and 2 complex roots $\gamma_3$ and $\gamma_4=\overline{\gamma}_3$. One can check that the coefficients of (\[eq:quart\]) are all positive and hence $\gamma_1<\gamma_2<0$.

Let $\lambda_k$ be the corresponding points in the $z$-plane $$\label{eq:lambda}
\begin{split}
\lambda_k&=-\frac{1}{\gamma_k}+c\frac{1-\beta}{1+\gamma_k}+c\frac{a\beta}{1+a\gamma_k},\quad
k=1,\ldots,4.
\end{split}$$ Then we have the following

\[le:lambdadis\] The points $\lambda_k$, $k=1,\ldots,4$ are all distinct.

The derivative $\frac{d z(\xi)}{d\xi}$ has simple zeros at the points $(\lambda_{k},\gamma_{k})$, $k=1,\ldots,4$, on the Riemann surface defined by (\[eq:curve2\]). Therefore, as a function of $z$, 2 branches of the function $\xi(z)$ behaves as $$\xi(z)=\gamma_{k}\pm
C_{k}(z-\lambda_{k})^{\frac{1}{2}}+O(z-\lambda_{k}),\quad
z\rightarrow\lambda_{k},$$ near the points $\lambda_{k}$. Let $i\neq j$, then Since $\gamma_i\neq\gamma_j$, if $\lambda_i=\lambda_j$, there will be 4 distinct solutions $\xi(z)$ to the equation (\[eq:curve2\]) in a neighborhood of the point $\lambda_{i}=\lambda_j$, which is not possible. Therefore $\lambda_{i}$ and $\lambda_j$ are distinct.

In particular, the points $\lambda_{3}$ and $\lambda_4$ are complex.

\[le:lambdapm\] The points $\lambda_{3}$ and $\lambda_4$ are not real.

Since $\gamma_3=\overline{\gamma}_4$, we see that $\lambda_3=\overline{\lambda}_4$. If $\lambda_{3}$ and $\lambda_4$ are real, then we will have 