# countable number of zeros

#### dwsmith

##### Well-known member
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.