$f$ an analytic function on $\mathbb{C}$ which takes values in $\mathbb{C}\backslash(-\infty,0]$ implies constant

Question

Let $f$ be an analytic function on $\mathbb{C}$ which takes values in $\mathbb{C}\backslash(-\infty,0]$, i.e. takes values in the complement of the nonpositive part of the real axis. Show that $f$ is constant.

My first observation is that there is a branch of the square root defined on $\mathbb{C}\backslash(-\infty,0]$ which takes positive real values along the positive part of the real axis. But I am not sure what do from thereon out. Any help would be appreciated!

Answer 1

So composing the branch of the square root you mentioned in your observation with $f$ gives a function $\sqrt{f}$ on $\mathbb{C}$ which takes values in the right half-plane $\{z \text{ }|\text{ Re}(z) > 0\}$. There is a fractional linear transformation $g$ which maps the right half-plane to the unit disk. Then $g(\sqrt{f(z)})$ is a bounded analytic function on $\mathbb{C}$, so it's constant by Liouville's theorem. Applying the inverse of $g$, we find that $\sqrt{f}$ is constant; the square of a constant function is also constant, so $f$ itself must be constant.

Answer 2

Compose it with the square root which halves the argument on $-\pi <\arg z<\pi$, i.e. with $\text{Log}$ the principal branch of the logarithm,

$$z\mapsto \sqrt{z}=\exp\left({1\over 2}\text{Log } z\right).$$

Then this is well-defined and analytic on the image of $f$, so that $\sqrt{f}$ is entire and misses an open set, namely the left half-plane. Either compose with the Möbius transform ${z-1\over z+1}$ to get a bounded entire function, or if you already know the proof that missing an open set implies constant, quote that.

Answer 3

This is killing a fly with a sledgehammer, but the fact that $f$ is constant follows immediately from either one of Picard's theorems.

Of course, Adam's solution is the "right" one!