Showing $f(z)$ is constant if $g(z)=\sqrt{f(z)}$ is entire. 
Concerning an entire function $f$, one is presented with the information that $g(z)=\sqrt{f(z)}$ also defines an entire function.  Show that $f$ must be a constant function.

I'm not sure how to approach this.  I do know that $\sqrt{z}$ is not analytic along $(-\infty,0]$.  This means that $g(z)$ is not entire along $(-\infty,0]$.
So here is what I have so far.  Since $\sqrt{z}$ is analytic only on $\mathbb{C}\setminus (-\infty,0]$ we know that $f(z) \not\in (-\infty,0]$ since $g(z)$ is analytic.  Knowing this we have $g'(z)=\frac{1}{2\sqrt{f(z)}}f'(z)$.  It follows that $f(\mathbb{C})=\mathbb{C}\setminus (-\infty, 0]$.  Then $\sqrt{f(\mathbb{C})}=\{ z \in \mathbb{C} \,| \,Re(z)>0\}.$  I'm not sure about the other half of the plane.
 A: I'm going to rephrase the question. First, though, let me note the objections to the phrasing as given. Take for example $g(z)=z$ and $f(z)=z^2$. Then $g(z)^2=f(z)$, but both $f$ and $g$ are non-constant entire functions. The equation $g(z)=\sqrt{f(z)}$ has no meaning until one choses a branch of the square-root function.
Here's the rephrasing:

Suppose $f$ is an entire function whose range omits the negative real axis, i.e., $f(\mathbb{C})\subseteq \mathbb{C}\smallsetminus (-\infty,0]$. Show that $f$ is constant. Hint: we can define a single-valued analytic branch of the square-root function on $\mathbb{C}\smallsetminus (-\infty,0]$, and using this branch we can define $g(z)=\sqrt{f(z)}$, which must also be entire.

As you've already figured out, the range of $g(z)$ will lie in the right half-plane $\{z: \Re(z)>0\}$. Now compose $g$ with a fractional linear transformation that maps the right-half plane to the interior of the unit disk, say $h(z)=L(g(z))$. What does this tell us about $h$ -- what property does $h$ have? And given that property, what result then tells us that $h$ must be constant?
Incidentally, the usual proof of the famous little Picard theorem (an entire function whose range omits two values is constant) uses the same proof strategy, but replaces the square root with the "inverse" of the elliptic modular function $\lambda$. Now, $\lambda$ doesn't have a single-valued inverse, but one can define a single-valued branch of $\lambda^{-1}(f(z))$.
