# Proof Check of a Complex Analysis Result

I have attempted to prove this fact: Let $$f:\mathbb{C}-\mathbb{Z}$$ an injective, holomorphic function. Prove that $$f$$ is a Möbius transformation.

This is my attempt of proof: I observe that the limits $$\lim_{z\to\infty}{f(z)},\qquad \lim_{z\to m}{f(z)}\quad (m\in\mathbb{Z})$$ exist, otherwise, by Casorati-Weierstrass theorem, $$f$$ would not be injective. Hence $$\infty$$ and the integers must be poles or removable singularities. If they were all removable singularities, then, by Liouville theorem, $$f$$ would be constant, and not injective. Hence there is at least a pole. But, since $$f$$ is continuous injective, there is at most one pole, thus there is exactly one pole. We have to cases:

If the pole is at infinity, $$f$$ is a polynomial, and since it is also injective it must be of the form $$f(z)=az+b.$$

If the pole is in one $$m\in\mathbb{Z},$$ we can assume that the pole is at $$m=0$$ (if not we can translate and come back to this case). We can observe, always because $$f$$ is injective, that the pole is of order $$1.$$ We have, in the annulus $$0 the following Laurent expansion of $$f(z)$$ at $$0:$$ $$f(z)=\sum_{k=-1}^{0}{c_kz^k}=\frac{c_{-1}}{z}+c_0=\frac{c_0z+c_{-1}}{z}.$$

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Is this proof correct, or are there some wrong steps? In this case, why are they wrong, and how can this result be proved?

• Apriori, infinity is not an isolated singularity as the integers accumulate there; however, applying your argument for the integers (which are isolated for sure), you remove all but at most one and then the proof is correct. – Conrad Dec 17 '18 at 2:22