and
$|t|<M$},$$ which implies that $F(t^{-1}:t^{-1}\varphi(t):1)=f(t^{-1},t^{-1}\varphi(t))=0$, for $t\in {\Bbb C}$ and $0<|t|<M$. We set $t^{-1}=z$, and we obtain that $$f(z,r(z))=0,\quad \mbox{$z\in {\Bbb C}$\, and
$|z|>M^{-1}$,\qquad where}$$ $$r(z)=z\varphi(z^{-1})=mz+a_1z^{1-N_1/N}+a_2z^{1-N_2/N}+a_3z^{1-N_3/N} + \cdots,\quad a_i\not=0,\, \forall i\in {\Bbb N}$$ $N,N_i\in {\Bbb N},\,\,i=1,\ldots$, and $0<N_1<N_2<\cdots$.

Since $\nu(\varphi)=N$, we get that there are $N$ different series in its conjugacy class. Let $\varphi_1,\ldots,\varphi_N$ be these series, and $$\label{Eq-conjugates}r_i(z)=z\varphi_i(z^{-1})=mz+a_1c_i^{N_1}z^{1-N_1/N}+a_2c_i^{N_2}z^{1-N_2/N}+a_3c_i^{N_3}z^{1-N_3/N}
+ \cdots$$ where $c_1,\ldots,c_N$ are the $N$ complex roots of $x^N=1$. Now we are ready to introduce the notion of infinity branch.

\[D-infinitybranch\] The set $\displaystyle B=\bigcup_{i=1}^NL_i$ where $$L_i=\{(z,r_i(z))\in {\Bbb
C}^2: \,z\in {\Bbb C},\,|z|>M_i\}$$ is called an [*infinity branch*]{} of the affine plane curve $\cal C$. The subsets $L_1,\ldots,L_N$ are called the [*leaves*]{} of the infinity branch $B$.

1.  We observe that an infinity branch is uniquely determined from one leaf, up to conjugation. That is, if $\displaystyle B=\bigcup_{i=1}^NL_i$, where $L_i=\{(z,r_i(z))\in
    {\Bbb C}^2: \,z\in {\Bbb C},\,|z|>M_i\}$, and $$r_i(z)=z\varphi_i(z^{-1})=mz+a_1z^{1-N_1/N}+a_2z^{1-N_2/N}+a_3z^{1-N_3/N} + \cdots$$ then $r_j=r_i,\,j=1,\ldots,N$, up to conjugation; i.e. $$r_j(z)=z\varphi_j(z^{-1})=mz+a_1c_j^{N_1}z^{1-N_1/N}+a_2c_j^{N_2}z^{1-N_2/N}+a_3c_j^{N_3}z^{1-N_3/N}
    + \cdots$$ where $c_j^N=1,\,\,j=1,\ldots,N$, and $N,N_i\in\mathbb{N}$.

2.  Let $M:=\max\{M_1,\ldots,M_N\}$. In the following, we consider $L_i=\{(z,r_i(z))\in {\Bbb
    C}^2: \,z\in {\Bbb C},\,|z|>M\}$.

Let $\varphi(z)=m+a_1z^{N_1/N}+a_2z^{N_2/N}+a_3z^{N_3/N} + \cdots$ be a series expansion for a solution of $g(y,z)=0$. We consider $\psi(t):=\varphi(t^N),$ and we observe that $(1:\psi(t):t^N)$ is a 