Why Does a Meromorphic Function Have Only Countably Many Zeros?

[Dustinsfl]

Let $f$ be a nonzero meromorphic function on the complex plane. Prove that $f$ has at most a countable number of zeros.

Since $f$ is meromorphic on $\mathbb{C}$, $f$ is holomorphic on $\mathbb{C}$ except for some isolated singularities which are poles. Aslo, $f$ being meromorphic we can write $f$ as $g(z)/h(z)$ both holomorphic with $h\neq 0$.

Now does multiplying through help lead to the conclusion?

So we would have $fh = g$. If so, I am not sure with what to do next.

 

[Jhoneine]

Hint: Use the fact that a holomorphic function is locally constant.Let $z_0$ be a zero of $f$. Then $f(z_0) = 0$, so by the definition of a holomorphic function, $f$ is locally constant in a neighborhood of $z_0$. That is, there exists an open set $U$ containing $z_0$ such that $f$ is constant on $U$. Since $f$ is nonzero, this constant must be nonzero, and thus $f$ has no other zeros on $U$. Hence, for each zero of $f$, there corresponds a distinct open set on which $f$ is nonzero. Thus, the set of zeros of $f$ is at most countable.