Suppose $f$ is entire, i.e, $\;f: \Bbb C \to \Bbb C$ is analytic.
Let $\Bbb H:= \{ z: Im(z)>0\}$ be the upper half plane. Suppose that $$\lim_{\substack{z \to \infty \\ z \in \Bbb H}} f(z)=0$$
Then is it true that $\;f \equiv 0$?
I can prove it in the case that $f$ sends reals to reals, because in that case $f(\bar{z}) = \overline{f(z)}$, and it is easy to proceed knowing this, together with the fact that bounded entire functions are constant.
For the general case, I tried all kinds of counterexamples involving exponentials. The closest thing I can find is $f(z)=e^{iz}$. In this case $f(z) \to 0$ as $z \to \infty$ along all directions with $\arg(z) \in (0, \pi)$. But $f(2\pi+\frac{i}{n}) \to 1$ even though $2\pi+\frac{i}{n} \to \infty$ in $\Bbb H$.
I'm wondering, more generally, if $A \subset [0, 2\pi)$ is a set of Lebesgue measure greater than or equal to $\pi$ such that $f(z) \to 0$ within $\{ z: \arg(z) \in A\}$, then $f \equiv 0$? Maybe there's an obvious counterexample...
