find all the entire functions that satisfy $|f(z)| \le C^{Im(z)}$

Question

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 $h(z)$:

$$h'(z)=\frac{f'e^{-if}+ie^{-if}f}{(e^{-if})^2}=0$$

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

any comments?

Hint: Put $m=\log(C)$, $h(z)=\exp(-imz)$ and show that $|f(z)|\leq |h(z)|$ for all $z$. — Kelenner, Jul 04 '17 at 19:47
@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. — Firas Abd El Gani, Jul 04 '17 at 20:09

Yiorgos S. Smyrlis · Answer 1 · 2022-10-03T15:33:10.547

2

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 $a\in\mathbb R$.

edited Oct 03 '22 at 15:33

answered Jul 05 '17 at 06:35

Yiorgos S. Smyrlis

83,933

1

It will be $|,f(z)|\le C^{\mathrm{Im},z}=|C^{-iz}|$ – Empty Jul 05 '17 at 07:34

find all the entire functions that satisfy $|f(z)| \le C^{Im(z)}$

1 Answers1