I am working on a question stating that
Find all entire functions that takes values only in the upper half plane.
A similar (but not closely similar) question is the part (b) of this post: Existence of a non-constant entire function. However, part (b) only asks to find one entire function that takes real values onto the imaginary axis.
So here is what I attempted:
Let's denote the entire function that takes values only in the upper half plane to be $f(z)$. Then, we have $$f:\mathbb{C}\longrightarrow\mathbb{H}.$$
Since $f$ could take any complex numbers to the upper half plane, the first thing that came to my mind was to use conformal mapping to conjugate $f$ so that we have a automorphism of the unit disc $\mathbb{D}$, and we know that what $Aut(\mathbb{D})$ has a common form, so that we can solve the formula for $f$.
i.e. we have the conformal mappings $$F:\mathbb{H}\longrightarrow\mathbb{D},\ \text{by}\ z\mapsto F(z):=\dfrac{i-z}{i+z},$$ and $$F^{-1}:\mathbb{D}\longrightarrow\mathbb{H},\ \text{by}\ z\mapsto F^{-1}(z):=i\dfrac{1-z}{1+z}.$$
Thus, we have $$F^{-1}\circ f\circ F:\mathbb{D}\longrightarrow\mathbb{D},$$ so that $$F^{-1}\circ f\circ F=e^{i\theta}\dfrac{\alpha-z}{1-\overline{\alpha}z},\ \text{for some}\ \alpha\in\mathbb{C}\ \text{and}\ |\alpha|<1.$$
We then can solve for $f$: $$F^{-1}\circ f\circ F(z)=F^{-1}\circ f\Big(\dfrac{i-z}{i+z}\Big)=i\dfrac{1-f\Big(\dfrac{i-z}{i+z}\Big)}{1+f\Big(\dfrac{i-z}{i+z}\Big)}=e^{i\theta}\dfrac{\alpha-z}{1-\overline{\alpha}z}.$$
We then can replace $w:=\dfrac{i-z}{i+z}$ to solve for $f(w)$.
However, we then have two problems here:
Firstly, $w$ cannot be continuously defined when $z\rightarrow -i$.
Secondly, the final formula was really a mess.
I think the problem here is that when I did the conjugation, I actually restrict the domain of $f$ from $\mathbb{C}$ to $\mathbb{D}$. However, I do not know if there is another way to do it, I believe there is one though.
I think there must be a much more straightforward method for this question. Thank you!
