\eta \in \mu_e \}
= P \cap \mathcal{M}(e,r)$. Then $\tau_1,\tau'_1$ lie in the plane $P$, and $\tau'_1$ is homotopic in $P^0$ to

where the half-circle represents the positive detour from $(\zeta^{-1} b_1,b_2)$ to $(\zeta^{-1} b_2,b_1)$. We let now $t_0 = \tau_1$, $t_1 = \tau'_1$, $t_{i+1} = t_i^{-1} t_{i-1}t_i$ for $1 \leq i \leq e-2$. A way to understand paths in $P \simeq \C^2$ is to use the projection $\C^2 \to \mathbbm{P}^1(\C)$ given by $(z_1,z_2) \mapsto
z_2/z_1$. Note for example that two paths $\gamma_1,\gamma_2$ in $P$ with the same endpoints whose images are homotopic in $\mathbbm{P}^1(\C) \setminus
\mu_e$ are homotopic in $P^0$ as soon as, writing $\gamma_i(t)
=(x_i(t),y_i(t))$, the set $x_1([0,1]) \cup x_2([0,1])$ is contained in some simply connected subspace of $\C \setminus \{ 0 \}$. We let $\alpha = b_2/b_1 \gg 1$. Then the positive detour $t_0$ is mapped to a path from $\alpha$ to $\alpha^{-1}$ close to the line segment, with image in the positive half-plane. The line segments of the form $\gamma(t) = (z_1,z_2(t)$ are mapped to line segments, and lines form $\gamma(t) = (z_1(t),z_2)$ are mapped to images of a line under $z \mapsto 1/z$, which is the composite of the complex conjugation with the geometric inversion 