# Prove that if a meromorphic function is constant in an open region then it is constant in all its domain

I am trying to prove that if a meromorphic function is constant on an open subset $$U$$ then it is constant in all its domain $$\Omega$$. My idea is:

Let $$F$$ be meromorphic in $$\Omega$$. Then we have $$F=\frac{f}{g}$$ with $$f$$ and $$g$$ analytic. Since $$F$$ is holomorphic in $$U$$ we have $$f-gK=0$$. Now since $$f-gK=0$$ is holomorphic in $$U$$ by the identity theorem we have $$f-gK=0$$ in $$\Omega$$ and so $$F=\frac{f}{g}=K.$$

Is my proof correct, or I am missing something?

My main concern here is to know if every meromorphic function is of the form $$F=\frac{f}{g}$$ with $$f$$ and $$g$$ analytic.

• You need to assume $\Omega$ is connected. – Robert Israel Jan 12 '20 at 4:03
• @RobertIsrael $\Omega$ is a domain, and usually a domain is defined as a connected open subset of the complex plane. – Paul Frost Jan 12 '20 at 11:29

A meromorphic $$F$$ function on a domain $$\Omega$$ is a function that is holomorphic on all of $$\Omega$$ except for a set $$A$$ of isolated singularities which are poles of the function. The set $$B = \Omega \setminus A$$ is open and connected and contains $$U$$, $$F \mid_B$$ is holomorphic, hence the identity theorem shows that $$F \mid_B$$ must be constant. This shows additionally that all possibly existing singularities are removable, thus $$A$$ must be empty.