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

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.