# Find All Entire Functions that Takes Values Only in the Upper Half Plane. [duplicate]

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!

## marked as duplicate by Martin R, Community♦Jul 25 at 12:41

Hint: $$z\mapsto \frac 1{f(x)+i}$$ is entire and bounded.