my question is the next.
I've studying meromorphic continuations on the Riemann sphere.
If i have $ g_{n}(z)= e^{\frac{ln(z)}{q}}$ a meromorphic branch of $ z^{\frac{1}{q}}, \ \ q \in \mathbb{N}$ defined on $\mathbb{C}\setminus \{z \in \mathbb{R} : z \leq 0 \}$
I know is not possible a meromorphic continuation at $z=0$ because in any neigbourhood of $z=0$, there is a multivalued function. But we can extend $g_{n}(z)$ by continuity at $0$.
My question is, if we considered $g_{n}( \frac{1}{z})$, at $z=0$
We have $g_{n}(\frac{1}{z})= e^{\frac{ln(\frac{1}{z})}{q}}= e^{\frac{- ln(z)}{q}}$, for this reason is not possible a meromorphic continuation at $z=0$, because $-ln(z)$ isn't a single-valued function in any neighbourhood of $z=0$ But, what is
$$ \mathop{lim}\limits_{z \longrightarrow 0} g_n(\frac{1}{z}) = ?$$
¿Is $\infty$?
But if it is, then $z=0$ is a pole of the function?
Yes, the limit of $g_n(1/z)$ as $z \to 0$ is complex $\infty$, i.e. $|g_n(1/z)| = |z|^{-1/q} \to +\infty$ as $z \to 0$. This does not mean $z=0$ is a pole, again because your function is not analytic in any deleted neighbourhood of $0$.