Inverse of a rational complex function and order of zeroes If we have a complex function $f(z)$ that can be written as
$$
f(z) = \frac{P(z)}{Q(z)}
$$
and $P(z)$ has a single zero $z = z_0$ of order $n > 1$, I've read in ''Conformal field theory'' by P. Di Francesco et al. (section 5.1.2. Global Conformal Transformations) that there's no inverse for $f$ in this case.
I understand that if you have serveral distinct zeroes for $P(z)$, then $f^{-1}$ does not exist because there's an ambiguity regarding the inverse image of $\infty$. But why is it problematic when there's just one zero of order greater than 1? I've read that it is because "the image of a small neighborhood of $z_0$ is wrapped $n$ times around $0$". However, what's the meaning of this?
 A: If $P(z)$ has a zero of order $n > 1$ at $z=z_0$, then the function can be written as $P(z) = (z-z_0)^nR(z)$, where $R(z)$ is analytic and does not vanish at $z = z_0$. Now, both $P(z)$ and its derivative $P'(z) = n(z-z_0)^{n-1}R(z) + (z-z_0)^nR'(z)$ vanish at $z_0$ since $n - 1 \geq 1$, which implies that $f'(z) = \frac{P'(z)Q(z) - P(z)Q'(z)}{(Q(z))^2}$ vanishes at $z_0$. An analytic function whose derivative vanishes at a point cannot be locally one-to-one at that point. Thus, the function $f$ cannot be invertible.
In regard to the last part of your question, the Open Mapping Theorem says that if $f$ is a non-constant analytic function on an open path-connected subset of $\mathbb{C}$, then $f$ is an open map. The map that you have defined is indeed non-constant and analytic in a path-connected neighbourhood $U$ of $z_0$, so $f(U)$ is open in $\mathbb{C}$. By the Argument Principle, it turns out that for some neighbourhood $V$ of $f(z_0) = 0$ contained in $f(U)$, the number of zeros of $f(z) - w$ in the preimage $f^{-1}(V)$ (note that this is not the functional inverse) is constant across all $w \in V$. That is, if $z_0$ is a zero of $f$ of order $n > 1$, then there are exactly $n$ zeros of $f(z) - w$ in $f^{-1}(V)$ for each $w$ in the neighbourhood $V$ of $f(z_0) = 0$. When you were told that "the image of a small neighbourhood of $z_0$ is wrapped $n$ times around $0$," this is presumably what the intended meaning was.
