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I am looking for a real-entire function* $f(z)$ such that for any (finite) complex $z$ we have both properties at once:

Property 1)

$$\lim_{t → +\infty} |f(z+t)| = \infty,\quad \lim_{t → +\infty} f(z-t) = 0,$$ where $t$ is real and

Property 2)

$-1$ is not in the range of $f$: $f(z) \ne -1$.

(* real-entire means a function is entire and maps the reals to a subset of the reals. In other words an entire function where the Maclaurin series has all real coefficients )

Any help would be appreciated.

*** Edit ***

Maybe related or helpful:

Entire function $f(z)$ bounded for $\mathrm{Re}(z)^2 > 1$?

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  • $\begingroup$ posted it at mathoverflow too ; mathoverflow.net/questions/394336/searching-for-fz-ne-1 $\endgroup$ – mick Jun 2 at 11:27
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    $\begingroup$ According to Weierstrass, $f(z)+1=e^{g(z)}$ where $g$ is entire. $f(\Bbb{R}) \subseteq \Bbb{R}$ implies $e^{g(\Bbb{R})} \subseteq \Bbb{R}$, so $g(\Bbb{R}) \subseteq \cup_{k \in \Bbb{Z}} (\Bbb{R}+k \pi i)$. Due to the connectness of $\Bbb{R}$, there exists $k_0$ such that $g(\Bbb{R}) \subseteq \Bbb{R}+k_0 \pi i$. Therefore $g-k_0 \pi i$ is a real-analytic function. $\endgroup$ – Zerox Jun 2 at 12:42
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    $\begingroup$ @Zerox yes that is certainly true. However note that $f(z)$ is independent of your choice of $k_0$ since those are just the log branches from taking a log. $\endgroup$ – mick Jun 2 at 14:49
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    $\begingroup$ It would be better if you could edit in the range interpretation of the third condition. $\endgroup$ – Lutz Lehmann Jun 2 at 15:05
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    $\begingroup$ $f(z) = e^{e^z} - 1$ almost works except that the very first property fails if $\mathrm{Im} (z) = \frac{\pi}{2} + \pi k$. $\endgroup$ – Random Jun 2 at 19:37
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I guess this is easy! Define $$g(z)\equiv f(z) + 1$$ This has no roots, therefore we may write $$g(z) = e^{h(z)}$$ Now, $h(z)$ must be an entire-real function that goes to zeros when $|z|\rightarrow\infty$ and $\arg(z)\rightarrow\{0, \pi\}$. There are many examples for this, as an example take $$h(z) = e^{-z^2}$$ which is equivalent to $$\boxed{f(z) = e^{e^{-z^2}}-1}$$

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  • $\begingroup$ the answer is wrong. And the closing of this question is unjustified ! but thanks for trying $\endgroup$ – mick yesterday
  • $\begingroup$ Sure, you're welcome. But could you please also point out the problem with it? $\endgroup$ – K. Sadri 17 hours ago

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