Prove or disprove: analytic $N$-th roots are unique up to multiplication by an $N$-th root of unity

Question

Let $\Omega$ be a a convex open subset of $\mathbb{C}$. Suppose that $f : \Omega \to \mathbb{C}$ is analytic such that $f$ has an analytic $N$-th root, where $N \ge 2$. That is, there is some analytic $g : \Omega \to \mathbb{C}$ so that $g^N = f$.

I would like to prove or disprove the following statement: if $g_1$ and $g_2$ are two analytic $N$-th roots of $f$ on $\Omega$, then $g_1 = \lambda g_2$ for $ \lambda \in \mathbb{C}$ with $\lambda^N = 1$

If $f \equiv 0$ on $\Omega$, then this statement definitely holds. Furthermore, if $f$ is nonvanishing $\Omega$, then so are $g_1$ and $g_2$. We can then define analytic $h(z) = \frac{g_1(z)}{g_2(z)}$ on $\Omega$, (with $(\frac{g_1(z)}{g_2(z)})^N \equiv 1$ showing that $h(z)$ is always equal to some $N$-th root of unity). But $h(\Omega)$ is connected, and thus $h(\Omega)$ must be a singleton, because the $N$-th roots of unity are discrete.

My trouble now comes with trying to find a proof when $f \not\equiv 0$ but $f$ does have some isolated zeroes. Are there counterexamples that can be constructed in this situation?

Hints or solutions are greatly appreciated.

Answer 1

On the set $\Omega\setminus f^{-1}(0)$, the map $\lambda\colon z\mapsto \frac{g_2(z)}{g_1(z)}$ is continuous and by discretenss of the roots of unity it is locally constant. If $f$ is not identically zero, then $f^{-1}(0)$ is discrete and therefore $U$ is still connected, hence $\lambda$ constant (and can be extended constant and continuously to all of $\Omega$).