Entire reflection symmetric function which is near $i$ when $\textrm{Im }z$ is big Is there an analytic entire function $f:\mathbb{C} \rightarrow \mathbb{C}$ such that

*

*$f(z)=\overline{f(\overline{z})},$

*For every $\varepsilon>0$ there is a $\delta >0$ such that if $\textrm{Im }z > \delta,$ then $\|f(z)-i\|<\varepsilon,$

*$f(0)=0?$
(Note that the tangent function satisfies the three properties, but fails to be entire.)
 A: The construction is as follows. Property 1 will follow from the fact that $f$ maps the real line to itself, which is easier to verify. First consider the building block,
$$g(z)=\int_\gamma\frac{e^{e^\zeta}}{\zeta-z}d\zeta.$$
where $\gamma$ consists of two parallel rays $\{ z=\pm(\pi/2+\epsilon)i+t:t\geq 0\}$
and a segment of imaginary axis connecting their endpoints. It is not difficult to
check that integral is convergent, and defines an analytic function outside of the
half-strip bounded by $\gamma$, and such that $g(z)\to 0$, $z\to\infty$ outside
of the half-strip.
This is called sometimes the Mittag-Leffler function, or the generalized Mittag Leffler function. The point is that it is entire: to perform an analytic continuation
to the inside of the half-strip one deforms the contour and applies Cauchy theorem.
This entire function $g_0$ grows like the double exponential inside the strip, and tends
to $0$ outside. For all these properties rigorously checked I can refer to the book of Hayman, Meromorphic functions, Chap. 4.1 but it occurs in many papers (a translation of this function gives you the simplest example of an entire function
which tends to $0$ on every ray from the origin). The more familiar Mittag-Leffler function $E_\alpha$ is obtained by a similar construction
where the double exponential is replaced by $e^{z^\alpha}$ and the half-strip by an appropriate sector.
Now consider $g(z)+g(-z)$ with some real $w$ to be chosen later. It grows in a horizontal strip and tends to 0 outside
of it, as $z\to\infty$. Convergence to $0$ is $O(1/z)$, which could be too slow for our purposes, so
find two complex conjugate roots of this function, say $z_1,z_2$ and divide on
$(z-z_1)(z-z_2)$. (It cannot be that all roots are real, but we don't care: in this hypothetical situation, we just take two real zeros for $z_1$ and $z_2$)
Now we want it to converge to different values in the upper and lower half-planes.
For this we take the integral
$$h_w(z)=\int_0^z \frac{(1+zw)(g_0(\zeta)+g_0(-\zeta))}{(\zeta-z_1)(\zeta-z_2)}d\zeta,$$
with some real $w$.
Since the integral on  vertical ray converges,
this function tends to two constants: say to $ia$ az $z\to\infty$ in the lower half-plane, and to $-ia$ in the upper half-plane, where $a$ is a real number.
(This is easily checked by symmetry of the integrand). So $h(z)/z$
has all the required properties,
if we can assure that $a\neq 0$. Of course $a=0$ is very unlikely to happen,
but to make sure, I introduced the additional parameter $w$.
One can show that $a$ depends on $w$ analytically and is not constant in $w$,
so one can choose $w$ so that $a\neq 0$.
Added. I was somewhat overoptimistic when I said "one can prove that $a(w)\not\equiv 0$". I don't see a simple, elementary way to guarantee this. There are two ways to go around this annoying issue: one way is to use Hayman, On integral functions with distinct asymptotic values,
Proc. Cambridge Philos. Soc. 66 (1969), 301–315. Another way I can explain if needed: it is a simple argument but relies on some deep results.
