Find all the entire functions that satisfy : $$|f(z)| \le C^{Im(z)}$$ for a positive $C$

My solution:

I said that if $f(z)$ is entire, then also $e^{-if}$ is entire, and also: $h(z)=\frac{f}{e^{-if}}$ is entire. ($|e^{-if}|>0$) then: $$|h(z)|=\frac{|f|}{|e^{-if}|}=\frac{|f|}{e^{Im(z)}}<c$$

so according to Liouville h(z) is bounded and entire. so its constant. if I take the derivative of $h(z)$:


and from here we get that $f'(z)=0$ and $f(z)=0$.

any comments?

  • 3
    $\begingroup$ You have $|\exp(-if(z))|=\exp(Im(f(z))$, not $\exp(Im(z))$ $\endgroup$
    – Kelenner
    Jul 4, 2017 at 19:37
  • 3
    $\begingroup$ Hint: Put $m=\log(C)$, $h(z)=\exp(-imz)$ and show that $|f(z)|\leq |h(z)|$ for all $z$. $\endgroup$
    – Kelenner
    Jul 4, 2017 at 19:47
  • $\begingroup$ @Kelenner thanks. I proved what u wrote. the rest I guess is like I did at the end right? $\frac{f(z)}{h(z)}$ is entire and bounded by 1. $\endgroup$ Jul 4, 2017 at 20:09
  • $\begingroup$ Right, you can now apply Liouville's Theorem. $\endgroup$
    – Kelenner
    Jul 4, 2017 at 20:10

1 Answer 1


Let $C=e^a$,where $a\in\mathbb R$. We have $$ |\,f(z)|\le C^{\mathrm{Im}\,z}=|e^{iaz}| $$ and hence $$ |\,e^{iaz}f(z)|\le 1. $$ Thus, by virtue of Liouville's Theorem, $e^{iaz}f(z)$ is constant.

Therefore, $f(z)=ce^{-iaz}$, for some $|c|\le 1$, and $a\in\mathbb R$.

  • 1
    $\begingroup$ It will be $|\,f(z)|\le C^{\mathrm{Im}\,z}=|C^{-iz}|$ $\endgroup$
    – Empty
    Jul 5, 2017 at 7:34

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