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Question: Let $f = g \circ h$ where $f,g$ are both non-constant holomorphic functions on $\mathbb{C}$ and $h$ is continuous on $\mathbb{C}$. Prove or disprove that $h$ is holomorphic on $\mathbb{C}$.

My attempt: I believe this statement is true. Assume for a contradiction that $h$ is not holomorphic in $\mathbb{C}$ then it does not satisfy the Cauchy-Riemann equations for some $z_0 \in \mathbb{C}$. However, I am not sure how to proceed as I cannot assume the differentiability of $\Re f(z)$ and $\Im f(z)$. Any hints are appreciated.

L-JS
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    Compare https://math.stackexchange.com/a/206718/42969 – Martin R Aug 06 '23 at 10:47
  • By assumption, $f$ is holomorphic. So you already know that $\Re f(z)$ and $\Im f(z)$ are differentiable. But a better approach is to use the Inverse function theorem on $g$ locally to rewrite $h = g^{-1}\circ f$ on neighborhoods where $g' \ne 0$. But that leads to a question... – Paul Sinclair Aug 07 '23 at 19:35

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