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(a) Is there a simple way to characterize the entire functions $f : \mathbb{C} \rightarrow \mathbb{C}$ which have the property that $| \text{Im}(f(z))| < | \text{Im}(z) |$ for all $z \in \mathbb{C} \setminus \mathbb{R}$? Certainly there are the linear functions $f(z) = a_0 + a_1 z$ with $a_0, a_1 \in \mathbb{R}$ and $|a_1| < 1$. Are there others?

(b) Does an analogous result hold with $\text{Im}(z) \rightarrow \text{Im}(z^n)$, $n \geq 1$ an integer? Of course trivial variants with $\text{Im} \rightarrow \text{Re}$ could be considered too.

I have tried to use the techniques that apply when showing that the only entire functions with bounded imaginary part are constants (Liouville's theorem, or the more fancy Picard's little theorem), as in this answer, but I can't find a way to make them work in this case. I am also aware of how to show that when $|f(z)| < C |z|^n$ with $f$ an entire function, for a constant $C$ and integer $n \geq 0$, for all $z \in \mathbb{C}$, then $f$ must be a polynomial of degree at most $n$. The standard argument uses Cauchy's theorem but I also haven't been able to adapt it to this case. Maybe there are other types of functions... Certainly they cannot be polynomials of degree greater than 1 in part (a), though.

Latrace
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  • What do you mean by the expression “$\text{Im}(z) \rightarrow \text{Im}(z^n)$”? Do you mean to replace the $\text{Im}(z)$ in the condition by $\text{Im}(z^n)$? – user1551 Jun 30 '22 at 10:57
  • Yes I mean that. – Latrace Jun 30 '22 at 10:57
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    $| \operatorname{Im}(f(z))| < | \operatorname{Im}(z^n) | \le |z|^n$, and then see https://math.stackexchange.com/q/1013546/42969 to conclude that $f$ is a polynomial of degree at most $n$. – Martin R Jun 30 '22 at 11:05
  • Would you agree that we can close your question as a duplicate of https://math.stackexchange.com/q/1013546/42969 ? – Martin R Jun 30 '22 at 11:39
  • No, I do not understand the proof in that answer. In particular I don't see why one can take $R(z)$ in the representation that is used. – Latrace Jun 30 '22 at 13:08
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    look up Borel caratheodory theorem and see that it trivially implies that if the imaginary part of entire $f$ is bounded by a polynomial (at infinity is enough and the bound needs to be one-sided only, doesn't need absolute values just maximum and a superior bound or minimum and an inferior bound), then the function is a polynomial https://en.wikipedia.org/wiki/Borel%E2%80%93Carath%C3%A9odory_theorem – Conrad Jun 30 '22 at 14:11
  • did you mean $| \text{Im}(f(z))| < | \text{Im}(z) |$ for all $z \in \mathbb C-\mathbb R$? As is you state this holds everywhere in $\mathbb C$ which is impossible for any $z$ on the real line – user8675309 Jun 30 '22 at 16:26
  • Good point. Yes, this should be changed -- I changed it in the question, thank you. – Latrace Jun 30 '22 at 17:08
  • By continuity we would have $|\text{Im}(f(z))| \leq |\text{Im}(z)|$ for all $z \in \mathbb{C}$ though, and we can proceed with the proof, getting first that $f$ is a linear function. After that the bounds on the coefficients are easy to obtain. – Latrace Jun 30 '22 at 17:14
  • for (a) define $h(z):= z -f(z)$ and apply https://math.stackexchange.com/questions/2284674/proving-an-entire-function-that-maps-real-to-real-upper-to-upper-is-a-linear-f to see that $h$ is a degree one polynomial (hence $f$ is as well since $f$ is non-constant) – user8675309 Jul 03 '22 at 18:08

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