If $f, g: \Omega \to \mathbb C$ are holomorphic and $g \not\equiv 0$, is $f/g$ meromorphic?

Question

I know that since $g$ is holomorphic, then the zeros of $g$ are isolated.

But why should $f/g$ then be holomorphic and meromorphic?

I don't understand how to continue because how would we know what happens to $f/g$ at the zeros of $g$?

Answer 1

If $z_0$ is a zero of $g$ of order $m$ and if $f(z_0)\neq0$, then $z_0$ is a pole of order $m$ of $\frac fg$.

And if $z_0$ is a zero of $g$ of order $m$ and zero of order $n$ of $f$, then:

• if $n<m$, $z_0$ is a pole of $\frac fg$ of order $m-n$;
• otherwise, $z_0$ is a removable singularity of $\frac fg$.

So, yes, $\frac fg$ is a meromorphic function.