We denote $\tau = \widehat{(-1)\,1\,\mathrm{i}}$, $\tau_0 = \widehat{0\,(-1)\,(-\mathrm{i})}$, $\tau_1 = \widehat{0\,1\,\mathrm{i}}$. We denote by $\phi_0$ and $\phi_1$ the maps from $\mathbf{C}$ to $\mathbf{C}$ defined by: $\phi_0(z) = \frac{1+\mathrm{i}}{2}\bar{z} + \frac{-1+\mathrm{i}}{2}$ and $\phi_1(z) = \frac{1-\mathrm{i}}{2}\bar{z} + \frac{1+\mathrm{i}}{2}$. The map $f\in\mathcal{E}$ satisfies $Tf = f$ and $f([0,1])\subset\tau$.
Conversely, let $z\in\tau$.
a) Show that we can define two sequences $(z_n)_{n\geq 0}$ and $(r_n)_{n\geq 1}$ in the following way:
- $z_0 = z$ and, if $n\geq 1$:
- if $z_{n-1}\in\tau_0$ then $r_n = 0$ and $z_n = (\phi_0)^{-1}(z_{n-1})$
- otherwise $r_n = 1$ and $z_n = (\phi_1)^{-1}(z_{n-1})$.
Prove that, for every integer $n\in\mathbb{N}$, $z_n$ belongs to $\tau$.
b) Prove that $f\left(\sum_{n=1}^{\infty}\frac{r_n}{2^n}\right) = z$ (one may express $z$ in terms of $z_n$ and the $\phi_{r_i}$).
c) Write a function that takes as argument a complex number $z$ (which we will assume is in $\tau$) and a real number $\epsilon$ and which returns an approximate value to within $\epsilon$ of a preimage of $z$.