# Show that if $f(z)=w$ has at most $n$ solutions for each $w \in \mathbb{C}$ then the pole $z_0$ has order at most $n$

Let $$D \subset \mathbb{C}$$ be open and connected, $$z_0 \in D$$ and $$n \in \mathbb{N}$$. Suppose that $$f : D \ \backslash \{z_0\} \to \mathbb{C}$$ is holomorphic in $$D \ \backslash \{z_0\}$$ with the following property:

$$f(z) = w \ \text{has at most} \ n \ \text{solutions for all} \ w \in \mathbb{C}.$$

Show that $$z_0$$ is either a removable singularity or a pole of $$f$$ of order less than or equal to $$n$$.

This is a question that I could not complete in a recent complex analysis example sheet. So I need to prove that $$z_0$$ cannot be an essential singularity or a pole of order greater than $$n$$. I managed to prove that it cannot be essential using the Casorati-Weierstrass Theorem. But despite a lot of effort I couldn't show that $$z_0$$ cannot be a pole of order greater than $$n$$. I'm not sure if this is the best way to approach it, but taking the contrapositive we have the equivalent statement:

If $$g: D \to \mathbb{C}$$ is holomorphic, $$g(z_0) \neq 0$$, $$m>n$$, then there exists $$w \in \mathbb{C}$$ such that the equation $$g(z)=w(z-z_0)^m$$ has more than $$n$$ solutions.

I hope this is correct. But regardless I couldn't find any tools that would help to prove such a statement. Help would be appreciated.

Assume wlog $$g$$ has a pole of order $$m \geq 2$$ at $$0$$, with coefficient $$c \neq 0$$; let $$q(z) = z^{m-1}(g(z)-\frac{c}{z^m})$$ analytic on a disc of radius $$2r$$ near $$0$$, fixing such an $$r$$ and assume $$|q(z)| \leq A, |z| \leq r$$ and pick any large enough $$B$$ s.t. $$Br^m > Ar + |c|$$; then it is obvious that $$Bz^{m}, Bz^{m}-zq(z)-c$$ have the same number of roots, namely $$m$$ on the disc with radius $$r$$ by Rouche and our choice of $$B$$, while since $$c \neq 0$$ none of the roots of $$Bz^{m}-zq(z)-c$$ can be zero and actually they are all distinct for almost all $$B$$ - actually by choosing $$r$$ small enough s.t. $$m|c| > |z^2q'(z)|+(m-1)|zq(z)|, |z| \leq r$$, which we always can do since the RHS of the inequality goes to zero with $$|z|$$, we can insure no double roots for all $$B$$ large enough to satisfy the required inequality as a simple computation shows.

But now $$Bz^{m}-zq(z)-c = 0, z \neq 0$$ means $$g(z)=B$$ since by definition $$g(z) = \frac{c}{z^m}+\frac{q(z)}{z^{m-1}}$$, so we are done showing that $$g$$ having a pole of order $$m \geq 2$$ means $$g$$ takes all large enough values $$m$$ times.

Note that for all analytic or meromorphic functions, double roots of any $$g-B$$ are isolated since they are zeros of the $$g'$$, so as noted above the extra argument above with small $$r$$ etc is not really needed but I included it for completeness