# Prove that there exist $c \in\mathbb C$ such that $|c| \leq 1$ and $f(z)=ce^z$ for every $z\in \mathbb C$

Do you have an idea how to solve this question:

Let $f(z)$ entire function that confirm: $|f(x+iy)|\leq e^x$ for every $x,y \in \mathbb R$.

I need to prove that 1) there exist $c \in \mathbb C$ such that $|c| \leq 1$ and $f(z)=ce^z$ for every $z\in \mathbb C$.

2) what about $|f(z)| \leq e^{|z|}$ for every $z \in \mathbb C$, is 1) stay correct?

Thank You.

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I meant that if we assume $|f(z)| \leq e^{|z|}$, is there exist c such that the conditions in the problem will exist. – bond Dec 5 '11 at 14:38
I see. In retrospect, that's clear. Does my hint for the first part make sense? – Dylan Moreland Dec 5 '11 at 14:47

Let's start with your first question. For a problem like this, you should always try to use Liouville's theorem first. Using the assumption that $|f(z)| \leq e^x$, can you form a bounded entire function involving $f$? Remember that $|e^z| = e^x$.