If $f,g$ are entire functions such that $f(g(z))=0, \forall z, $ then $g$ is constant or $f(z) =0, \forall z \ ?$ Let $f,g$ be entire functions such that $f(g(z))=0, \forall z.$ Could anyone advise me on how to prove/disprove: either $g(z)$ is constant or $f(z) =0, \forall z \ ?$
Hints will suffice, thank you.
 A: If $g$ is not constant, then $U=g(\Bbb C)$ is an open subset, and $f |_U=0$, we get $f=0$. 
A: Suppose $f(z)$ isn't identically zero. Then $f(z_0) \neq 0$ for some $z_0$. The range of $g$ cannot contain $z_0$, because otherwise $f \circ g$. But $f$ is continuous at $z_0$, so it's nonzero is a small open set around $z_0$. $g$ is an entire function, so either it's constant or its image is dense in $\mathbb{C}$; the latter isn't possible (its closure cannot contain $z_0$), so it's constant.
A: Suppose $g$ is not constant, $f$ is not identically zero, and $f \circ g$ is identically zero. Since $f$ is entire and not identically zero, its zero set is discrete (i.e. has no limit points). Since $g$ is not constant, we conclude that $g$ maps $\mathbb{C}$ onto a discrete set which has more than one point. Why is this impossible?
A: Another way to make it: Assume that $g$ is not constant in particular $g'\neq 0$: 
As $f\circ g=0$, then by derivation we get $g'\times (f'\circ g)=0$, but the ring of entires functions ( $\Bbb C$ connected) is an integral domain, it follow that $f'\circ g=0$, another derivation we get $g'\times (f''\circ g)=0$, hence $f''\circ g=0$, by same argument we can show that for all $n\in \Bbb N$; $f^{(n)}\circ g=0$.
Now let $a\in \Bbb C$ and $b=g(a)$, since $f$ is analytic we have :
$$\forall z\in \Bbb C\ \ \ f(z)=\sum_{n=0}^{+\infty}\frac{f^{n}(b)}{n!}(z-b)^n=0$$
because $f^{(n)}(b)=f^{(n)}(g(a))=0$
