If an analytic function is identically zero in a domain, can it be always divided by another analytic function in that domain?

Question

Let's say I have two functions $f$ and $g$, both analytic on the same domain (open and connected set) $D$, and suppose also I was able to prove that $f = 0$ on the entire domain.

Question. Is it legitimate to then claim that $$ \frac{f(z)}{g(z)}\equiv 0 $$ identically on that domain? In particular, $g(z)$ may assume a zero value in that domain, but since $f(z)$ will be zero as well the fraction is well defined in any case?

I suppose this question is not strictly about complex analytic functions, nevertheless it arose in a problem where analytic functions are involved.

Answer 1

Strictly speaking, it is not legitimate because $f/g$ would be undetermined, regardless of the value of $f$, in the zeroes of $g$. Particularly, if you cannot prove that $g$ is not identically zero in the domain, then you cannot define $f/g$ anywhere.

However, if $g\not\equiv0$, and $g$ is analytic/holomorphic, then, the zeros of $g$ are isolated singularities of $f/g$ which can be removed. Theorem 3.1 from Lang's state

Theorem 3.1 If $f(z)$ is bounded in some neighborhood of $z_0$, then one can define $f(z_0)$ in a unique way such that the function is also analytic at $z_0$.

Continuity requires $f/g$ to be defined as zero in those points as well, so in a sense your claim is legitimate.

Also related as mentioned in the comments: if $g$ is not identically zero, then $f/g$ is meromorphic. From Wikipedia,

Every meromorphic function on D can be expressed as the ratio between two holomorphic functions (with the denominator not constant 0) ..."

Meromorphic functions have isolated singularities, and in your example $f/g$ is bounded.