Preimage of $[0,1]$ under $f:\mathbb{C}\rightarrow\mathbb{C}, z\mapsto \frac{-27(1+\frac{1}{x^{3}-3})^{2}}{x^{3}-3}$

Question

I want to find $$ f^{-1}([0,1]) $$ where $$f:\mathbb{C}\rightarrow\mathbb{C}, z\mapsto \frac{-27(1+\frac{1}{x^{3}-3})^{2}}{x^{3}-3}.$$ I have to do this in order to find a dessins d'enfant associated to certain Riemann Surface, but I find very difficult to deal with this map.

Answer 1

I'm not knowledgeable about dessins d'enfant or Riemann surfaces, but I think that this may be helpful. Define $g:\mathbb{C} \to \mathbb{C}$ by $$g(z) = -27z(1+z)^2$$ and also $h:\mathbb{C} \to \mathbb{C}$ by $$h(z) = \frac{1}{z^3-3}$$ Factor $f$ as $f = g \circ h$. Now you have $f^{-1}([0,1]) = h^{-1}(g^{-1}([0,1]))$ but this way is easier to find the preimages.

For each $w$ in $(0,1]$, the equation $w = g(z) = -27z(1+z)^2$ has three different real solutions and for $w=0$ it has a double real solution and a simple, real one. You can check this by considering the graph of the polynomial defining $g$, and analyzing intersections with horizontal lines at heights $w \in [0,1]$. So you have $g^{-1}([0,1]) = [\delta ; 0]$ where $\delta = \min \{x \in \mathbb{R} : g(x)=1\} = -1.1774\dots$

Now $f^{-1}([0,1]) = h^{-1}(g^{-1}([0,1])) = h^{-1}([\delta ; 0])$. By considering $h$ you have $$f^{-1}([0,1]) = \Big\{ \xi \cdot x\ : \xi \in G_3 \, ,\; x \in \Big(-\infty , \sqrt[3]{3 + 1/\delta} \, \Big] \subseteq \mathbb{R} \Big\}$$

Just in case, $$\delta = -\frac{2}{3} - \frac{1}{3}\cos(\pi/9) - \frac{\sqrt{3}}{3}\sin(\pi/9)$$ or $$\delta= -1.17736296207931869013492843370361111595722163805359683056...$$