Suppose you are given a holomorphic function on $\{Re(z)>0\}$ and such that $f(z)=\overline{f(\overline{z})}$.
Moreover suppose that for any sequence $\{z_k\}_{k\in\mathbb{N}}$ such that: $\limsup_{k\to\infty}\frac{\lvert z_k\rvert}{Re(z_k)}\leq C$ for some positive $C$ and $Re(z_k)\to 0$. Then we assume that $f$ satisfy for such a sequence and any $n=0,1,\ldots$:
\begin{equation} \lim_{k\to\infty} z_k^n\partial_z^nf(z_k)=0, \end{equation}
where $\partial_z^n$ denotes the nth complex derivative wrt $z$.
Futhermore to avoid trivial counterexamples, we assume that for any $y\in\mathbb{R}$:
\begin{equation} \limsup_{x\to\infty} \lvert f(x+iy)\rvert\leq D_y, \end{equation}
for some $D_y>0$.
Is this function constantly $0$?
EDIT: I alpologize for the editing. Explanation with words: I have an olomorphic function defined on the open right half plane which is bounded on horizontal lines and for which radial limits of $f$ and some manipulation of its derivatives at zero are zero. My first, badly posed question was whether or not such a function should be 0, but one can easily understand that $f(z)=z$ is a counterexaples, with any polynomial for which $p(0)=0$. However, my assumption on horizontal lines exclude this easy cases, yielding the question above.
I believe $f(z) = e^{-1/z}$ is a counterexample.