1
$\begingroup$

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 $0<z<\infty,$ the following Laurent expansion of $f(z)$ at $0:$ $$f(z)=\sum_{k=-1}^{0}{c_kz^k}=\frac{c_{-1}}{z}+c_0=\frac{c_0z+c_{-1}}{z}.$$

$$———$$

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

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Conrad Dec 17 '18 at 2:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.