$f(z)$ and $g\circ f(z)$ holomorphic implies $g(z)$ holomorphic as well Let $\Omega\subset\mathbb{C}$ be a connected open set and let $f,g$ be non-constant functions defined on $\Omega$ and $f(\Omega)$ respectively.
Suppose that both $f$ and $g\circ f\in H(\Omega)$. Prove that $g\in H(f(\Omega))$.

I've made progress on this problem from two directions; one using the limit definition of holomorphicity, and the other via the open mapping theorem. However, I don't have enough faith in the rigor of my techniques and am looking for an airtight approach or suggestion from others.

Inspired by problem $10.14$ in Rudin's Real and Complex Analysis
 A: If $f'(z_0)\neq 0$ for some $z_0$ then $f$ is invertible in a neighbourhood $U$ of $z_0$, its inverse is holomorphic, so $g=(g\circ f)\circ f^{-1}$ is holomorphic on $f(U)$. If $f'(z_0)=0$ then $f'(z)\neq0$ in a neighbourhood of $z_0$ (except for $z_0$ itself) - so $g$ is holomorphic in a neighbourhood of $f(z_0)$, except possibly at $z_0$. As a result, $g$ is holomorphic on $f(\Omega)$ minus a discrete subset. From here it's easy to see that $g$ is continuous, so just apply Riemann's removable singularity theorem to see that $g$ is holomorphic$.
A: I was able to come up with a proof today; here's a sketch:
After applying the Open Mapping Theorem and some manipulations, we can reduce the original statement to showing $$g(z^n)\in H(\{0\})\implies g(z)\in H(\{0\})$$
Now expand the power series for $g(z^n)$ at $0$ and substitute $z\to \lambda z$ where $\lambda^n=1$. By comparing coefficients, we can construct a power series by hand that converges to $g(z)$ and the result follows.
A: Use the complex variant of the chain rule. See, for example, my response to criterions for holomorphic functions.
Edit: I was assuming $g$ was real differentiable. I sit corrected. :) Still a useful technique to have in one's bag of tricks.
