# Entire Functions and Lacunary Values.

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1. Aug 9, 2015

### WWGD

#Hi All,
Let $f: \mathbb C \rightarrow \mathbb C$ be entire, i.e., analytic in the whole Complex plane. By one of Picard's theorems, $f$ must be onto , except possibly for one value, called the lacunary value.
Question: say $0$ is the lacunary value of $f$. Must $f$ be of the form $e^{g(z)}$ , with $g(z)$ analytic?.
Clearly if $g(z)$ has no lacunary values, then $e^{g(z)}$ will have only $0$ as its lacunary value,
and if $g(z)$ has only $w$ as its lacunary value, then this value will be assumed in $w+i2\pi n$, so $0$ will still be the lacunary value of $e^{g(z)}$. Maybe we can consider composing functions with known lacunary values, but I don't see offhand how, since I don't know the lacunary values of general entire functions.
I considered using Weirstrass factorization thm, but it seems overkill and has led nowhere.
Thanks.

Last edited: Aug 9, 2015
2. Aug 9, 2015

### micromass

Theorem: Let $D$ be any elementary domain (for example $D = \mathbb{C}$), let $f:D\rightarrow \mathbb{C}$ be an analytic function such that $f(z) \neq 0$ for each $z\in D$, then there exist an analytic function $h$ such that $f(z) = e^{h(z)}$ for all $z\in D$.

Proof: Let $F$ be a primitive of $f^\prime / f$. Then put $G(z) = \frac{e^{F(z)}}{f(z)}$. One checks easily that $G^\prime (z) = 0$ for all $z\in D$. Therefore, $e^{F(z)} = Cf(z)$ for some nonzero constant $C$. Since $e^z$ is surjective, we can write $C = e^c$. Then $h(z) = F(z) - c$ does the trick.

3. Aug 9, 2015

### WWGD

Thanks, that shows the side I am aware of, but does it show that all entire functions that do not hit zero are of this form? Sorry, I know I am being lazy and using composition of entire functions would give an answer, but I wondered if someone knew the answer offhand.

4. Aug 9, 2015

### Staff: Mentor

They all can be expressed in that way. That's what the theorem says.

5. Aug 9, 2015

6. Aug 9, 2015

### WWGD

Sorry for not being clear: I am looking for a common way of generating/describing all such functions, I would say with an iff rule. I know that every function _can_ be expressed that way. What I meant to say is: does that form exhaust all possible functions with lacunary value 1?. I am looking for something along the lines of $f: \mathbb C \rightarrow \mathbb C$ is an automorphism (global diffeomorphism) iff $f(z)= az+b$. Yes, you are right that any function $f(z)$ with $f(z)$ entire and never $0$ does have a logarithm $g(z)$ in a simply-connected region so that $e^{g(z)}=f(z)$ , but what I should have asked is whether this exhausts all possible such functions. Hope I was clearer this time

7. Aug 9, 2015

### micromass

Sorry, I really don't get it.

8. Aug 9, 2015

### WWGD

Let me think it through and I will try to be more clear. Thanks.

9. Aug 10, 2015

### Staff: Mentor

"For every entire nonzero non-constant function f there exists g with [property]" is as exhaustive as it can get.
The other direction is also true:
For every non-constant entire g there is a non-constant nonzero entire f with f(z)=eg(z).