obtained by conjugation on $r(z)$ and $\overline{r}(z)$. That is (see equation (\[Eq-conjugates\])), $$r_i(z)=mz+u_1c_i^{N_1}z^{1-\frac{N_1}{N}}+\cdots+u_kc_i^{N_k}z^{1-\frac{N_k}{N}}+u_{k+1}c_i^{N_{k+1}}z^{1-\frac{N_{k+1}}{N}}+\cdots$$ $$\overline{r}_j(z)=mz+\overline{u}_1d_j^{\overline{N}_1}z^{1-\frac{\overline{N}_1}{\overline{N}}}+\cdots+
\overline{u}_kd_j^{\overline{N}_k}z^{1-\frac{\overline{N}_k}{\overline{N}}}
+\overline{u}_{k+1}d_j^{\overline{N}_{k+1}}z^{1-\frac{\overline{N}_{k+1}}{\overline{N}}}+\cdots,$$ where $c_1,\ldots,c_{N}$ are the $N$ complex roots of $x^{N}=1$, and $d_1,\ldots,d_{\overline{N}}$ are the $\overline{N}$ complex roots of $x^{\overline{N}}=1$.

We simplify the exponents and, using that $u_l=\overline{u}_l=a_l,\,
l=1,\ldots,k$, we get that: $$r_i(z)=mz+a_1c_i^{N_1}z^{1-\frac{n_1}{n}}+\cdots+a_kc_i^{N_k}z^{1-\frac{n_k}{n}}+u_{k+1}c_i^{N_{k+1}}z^{1-\frac{N_{k+1}}{N}}+\cdots$$ $$\overline{r}_j(z)=mz+a_1d_j^{\overline{N}_1}z^{1-\frac{n_1}{n}}+\cdots+a_kd_j^{\overline{N}_k}z^{1-\frac{n_k}{n}}
+\overline{u}_{k+1}d_j^{\overline{N}_{k+1}}z^{1-\frac{\overline{N}_{k+1}}{\overline{N}}}+\cdots.$$

Hence, we only have to show that for each $i\in\{1,\ldots,N\}$ there exists $j\in\{1,\ldots,\overline{N}\}$ such that $c_i^{N_l}=d_j^{\overline{N}_l}$ for every $l=1,\ldots,k$. Indeed: since $c_i,\,i=1,\ldots,N$ are the $N$ complex roots of $x^{N}=1$, we have that $c_i=e^{\frac{2(i-1)\pi I}{N}}$, where $I$ is the imaginary unit. Taking into account that $N=bn$, we deduce that $c_i^b=e^{\frac{2(i-1)\pi I}{n}}$,$i=1,\ldots,N$, and $c_i^b=c_{i+(m-1)n}^b$ for each $i=1,\ldots,n$, and $m=1,\ldots,b$. That is, $(c_i^{b})^n=1$,$i=1,\ldots,n$. Reasoning similarly, we have that $d_j^{\overline{b}}=e^{\frac{2(j-1)\pi
I}{n}}$,$j=1,\ldots,\overline{N}$, and $d_j^{\overline{b}}=d_{j+(m-1)n}^{\overline{b}}$ for each $j=1,\ldots,n$, and $m=1,\ldots,\overline{b}$. That is, $(d_j^{\overline{b}})^n=1$,$j=1,\ldots,n$. Therefore, $c_i^b=d_{i+(m-1)n}^{\overline{b}}$,$m=1,\ldots,\overline{b}$, and using that $N_l=bn_l$ and $\overline{N}_l=\overline{b}n_l$,$l=1,\ldots,k$, it follows that $c_i^{N_l}=d_{j}^{\overline{N}_l}$,$j=i+(m-1)n,\,m=1,\ldots,\overline{b}$. $\Box$

Two convergent infinity branches may have different ramification indexes i.e., they may have different number of leaves. However, the value $n\in\mathbb{N}$ obtained by simplifying the non negative exponents, is the same in both branches. We refer to it the [*degree of the infinity branch*]{}. Observe that the proof of Proposition \[P-convergent-branches\] implies that two convergent infinity branches have the same degree.\
