An exercise from Miranda's book

An exercise from Miranda's book

pLet $f(z)=\frac{4z^{2}(z-1)^{2}}{(2z-1)^{2}}$ define a holomorphic map of
degree 4 from $\mathbb{P}^{1}(\mathbb{C})$ to itself. Show that there are
three branch points, and that the three corresponding permutations in
$S_4$ are $\sigma_1=(12)(34)$, $\sigma_2=(13)(24)$ and $\sigma_3=(14)(23)$
up to conjugacy./p pUsing Hurwitz' formula, we have $\sum_{P \in X}
\operatorname{mult}_P F=6$: clearly $0$ is a branch point (and its
preimages are $0$ and $1$), also $\infty$ is a branch point with preimages
$\frac{1}{2}$ and $\infty$. According to the fact that
$\operatorname{mult}_P F=1+\operatorname{ord}_{\Phi_\alpha(q)}
\frac{d}{dz} (h_\alpha F \Phi_\alpha^{-1})$ and
$\Phi_\alpha=h_\alpha=Id_{\mathbb{C}}$ (because I have just studied the
case $\infty$), I have to find the zeroes of the derivative. I obtain the
ramification points $\frac{1 \mp i}{2}$, which images are respectively
$\pm 1$. So I have 4 branch points and not three as required. The cyclic
structure is $(2,1,1,2)$, so which are the corresponding permutations?/p