modulus of the annulus $A$ is then defined by: $${{\textsf {mod}  \, }}( A ):= \sup_{\rho \in {{\textsf {conf} \,  }}(A)} \,   { l_\rho(\Lambda)^2 \over {{\textsf {area}  }}_\rho (A) } ,$$ where $l_\rho(\Lambda):= \inf_{\lambda \in \Lambda} l_\rho
(\lambda)$.

Statement of the quantified version {#cons}
-----------------------------------

We begin with some notations. Let $0 < \theta <1$ and $\theta_n:= [  \theta  n + { \log
    \tau_* \over  \log d } ] +1$. We introduce this integer in view of applying Briend-Duval theorem with $\eta = d^{-\theta n}$ and $l = \theta_n$ (indeed, $\tau_* / d^{\theta_n} < d^{-\theta
    n}$). Since the degree of $\VV_{\theta_n} = \cup_{i = 1}^{\theta_n} f^i(\CC)$ is at most $\tau_{\theta_n} = (d + \ldots + d^{\theta_n}) \deg \CC$, we have $\tau_{\theta_n} < d^{\theta_n}$ up to a multiplicative constant.\
We let $0 < \theta < \theta' < 1$ and consider $n_0 \geq 1$ satisfying: $$\label{nzero}
\forall n \geq n_0  \ \ \ ,  \ \ \ \theta_n < \theta'n \ \ \textrm{ and }  \ \ \ \tau_{\theta_n} < {d}^{\, \theta' n}.$$ Let us recall that $\VV_{\theta_n}[\delta]$ is the $\delta$-neighbourhood of $\VV_{\theta_n}$ in ${\mathbb {P}}^k$. We fix $\theta'/2 < \zeta < 1$ and define $\DD:= \limsup_{n \geq n_0}
\VV_{\theta_n}[d^{-\zeta n}]$.

\[vold\] 