# A question about analytic functions.

I'm having trouble with the following question(s) related to analytic functions. We haven't learned about Picard's Theorems in class (which seems to be related to this question), and I'm not sure how to figure it out without using it. The question is as follows:

Why is it true that an entire function $f(z)$ which does not send any $z \in \mathbb{C}$ to half of the plane must be constant?

Would it also be true that an entire function which does not send any $z \in \mathbb{C}$ to an arc must be constant as well?

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You can solve this problem using conformal mapping from the upper-half plane to the unit disk.

Suppose $f$ is a non constant function avoiding $F = \{z \in \mathbb{C} | Im(z) \leq 0\}$. Put $\varphi := \mathbb{H} \longrightarrow \Delta$ be a conformal mapping from the upper half plane to the disk.

$\varphi \circ f$ is an entire function which is bounded so, by Liouville's theorem, it is constant. Since $\varphi$ is a bijection $f$ is constant.

Such a $\varphi$ is given taking the inverse of the function $z \longmapsto \frac{z-i}{z+i}$ restricted to $\Delta$

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For the first question, up to a rotation and a translation you can assume that $f(z) \in \{ \Re z \leq 0 \}$ (other wise replace $f(z)$ by $f(z)e^{i\theta} + z_0$. Hence $\exp f(z)$ is also an entire function, which satisfies that $|\exp f(z)| \in [0,1]$, and by Liouville's theorem must be constant.

The second question has a positive answer as well by Little Picard, but off the top of my head I don't have a good way to prove it without using that theorem.

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For the second part: depending on the arc you can see it without little Picard. The reason is the same as for the half plane: The complement of an arc can be mapped to the unit disc. As an example, suppose that $f$ is entire and maps into $\mathbb{C} \setminus [-1,1]$. The function

$$\sigma: z \mapsto z - \sqrt{z^2-1}$$

is well defined (single valued) on $\mathbb{C} \setminus [-1,1]$ and if you take the positive branch of the square root then $\sigma$ is a 1-to-1 map to the punctured disc $\{ z \mid 0<|z|< 1\}$. Therefore $\sigma \circ f$ is bounded and hence constant.

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Why is it true that an entire function $f(z)$ which does not send any $z \in \mathbb{C}$ to half of the plane must be constant?

Do you know the theorem that a bounded harmonic function must be constant? Can you convert a function mapping to a half-plane into a bounded function?

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