$\overline{L}\subset
\overline{B}$.

\[R-common-inf-point\] From Remark \[R-DistVertical\], statement 1, we get that two convergent infinity branches are associated to the same infinity point.

\[P-convergent-branches\] Two infinity branches $B$ and $\overline{B}$ are convergent if and only if for each leaf $L\subset B$ there exists a leaf $\overline{L}\subset
\overline{B}$ convergent with $L$, and reciprocally.

**Proof:** Let $B$ and $\overline{B}$ be two convergent infinity branches, and let us prove that for any $L_i\subset B$ there exists $\overline{L}_j\subset \overline{B}$ convergent with $L_i$ (using Definition \[D-distance0\], we clearly have the reciprocal). From Definition \[D-distance0\], there exist two leaves $L=\{(z,r(z))\in {\Bbb C}^2:\,z\in {\Bbb
C},\,|z|>M\}\subset B$, and $\overline{L}=\{(z,\overline{r}(z))\in
{\Bbb C}^2:\,z\in {\Bbb C},\,|z|>\overline{M}\}\subset \overline{B}$ convergent. Let $$r(z)=z\varphi(z^{-1})=mz+u_1z^{1-\frac{N_1}{N}}+\cdots+u_kz^{1-\frac{N_k}{N}}+u_{k+1}z^{1-\frac{N_{k+1}}{N}}+\cdots,\, \, u_i\not=0,$$ $$\overline{r}(z)=z\overline{\varphi}(z^{-1})=mz+\overline{u}_1z^{1-\frac{\overline{N}_1}{\overline{N}}}
+\cdots+\overline{u}_kz^{1-\frac{\overline{N}_k}{\overline{N}}}
+\overline{u}_{k+1}z^{1-\frac{\overline{N}_{k+1}}{\overline{N}}}+\cdots,\,
\, \overline{u}_i\not=0,$$ where $\nu(B)=N$, $\nu(\overline{B})=\overline{N}$, $N_k\leq N<N_{k+1}$ and $\overline{N}_k\leq \overline{N}<\overline{N}_{k+1}$.

From Lemma \[L-DistVertical\], we deduce that the terms with non negative exponent in $r$ and $\overline{r}$ must coincide. Thus, $u_l=\overline{u}_l=a_l$, for $l=1,\ldots,k$, and $$r(z)=mz+a_1z^{1-\frac{n_1}{n}}+\cdots+a_kz^{1-\frac{n_k}{n}}+u_{k+1}z^{1-\frac{N_{k+1}}{N}}+\cdots,\, \, a_i,u_i\not=0$$ $$\overline{r}(z)=mz+a_1z^{1-\frac{n_1}{n}}+\cdots+a_kz^{1-\frac{n_k}{n}}+\overline{u}_{k+1}z^{1-\frac{\overline{N}_{k+1}}{\overline{N}}}
+\cdots,\, \, a_i,\overline{u}_i\not=0,$$ where $n, n_i\in {\Bbb
N},\, 0<n_1<\cdots<n_k<n$, $N_{k+1}>N$,$\overline{N}_{k+1}>\overline{N}$. Observe that we have simplified the non negative exponents such that $\gcd(n,n_1,\ldots,n_k)=1$. That is, for $l=1,\ldots,k$, there are $b,\overline{b}\in\mathbb{N}$ such that $N_l=bn_l$, $N=bn$, $\overline{N}_l=\overline{b}n_l$, and $\overline{N}=\overline{b}n$.

Under these conditions, we observe that the different leaves of $B$ and $\overline{B}$ are 