Suppose that $f : \{ z \in \mathbb C : \mathrm{Re}(z) > 0 \} \to \mathbb C$ is a holomorphic function in the right half plane with the property that $$ \lim_{z \to 0} f(z) \in \mathbb C $$ i.e. the limit exists. Is it true that if $(z_n)$ is a convergent sequence with limit in $$ \{ z \in \mathbb C : \mathrm{Re}(z) > 0 \} \cup \{ 0 \} $$ such that $f(z_n)=0$ for all $n$, then $f$ vanishes identically? Clearly this holds if the limit lies only in $\{ z \in \mathbb C : \mathrm{Re}(z) > 0 \} $ in view of the uniqueness theorem.

2$\begingroup$ Your use of the term "limit point" is quite confusing. I assume you mean a convergent sequence, with limit in $\{z: \textrm{Re}\: z>0\} \cup \{ 0 \}$. I've taken the liberty to edit accordingly, but of course feel free to edit further if that wasn't your intention. $\endgroup$– Christian RemlingJun 26, 2022 at 16:52
1 Answer
This is false. We can take any function $g\not\equiv 0$ with $\lim_{z\to 0} g(z)=0$ such as $g(z)=z$ and then $f(z)=B(z)g(z)$, where $B$ is a Blaschke product with zeros $z_n$ (the linked page discusses Blaschke products on the unit circle, but of course we can move things over to the half plane by a Cayley transform).
The $z_n=x_n+iy_n$ must satisfy $\sum \frac{x_n}{x_n^2+y_n^2+1}<\infty$, but are otherwise arbitrary, so we can definitely fix a sequence $z_n\to 0$.
Then $f(z_n)=0$ and also still $\lim_{z\to 0}f(z)=0$ since $B\le 1$, but $f$ is not identically zero.