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

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 $f(z)=0$.

• You have $|\exp(-if(z))|=\exp(Im(f(z))$, not $\exp(Im(z))$ Jul 4, 2017 at 19:37
• Hint: Put $m=\log(C)$, $h(z)=\exp(-imz)$ and show that $|f(z)|\leq |h(z)|$ for all $z$. Jul 4, 2017 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. Jul 4, 2017 at 20:09
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 $$a\in\mathbb R$$.
• It will be $|\,f(z)|\le C^{\mathrm{Im}\,z}=|C^{-iz}|$ Jul 5, 2017 at 7:34