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Are conformal mappings biholomorphic?

$f$ is considered a conformal mapping when $f:D \rightarrow G$ is holomorphic and injective and holds $f(D) = G$. This implies that $f$ is also surjective on $G$ right?

We know that when $f:D \rightarrow G$ is holomorphic then $f^{-1} : G \rightarrow D$ is also holomorphic on $G$. So doesnt this implie that conformal mappings are biholomorphic?

This would mean that the conformal mappings are exactly the biholomorphic functions?

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That is correct.

Be aware of that some authors have a different definition for conformal functions that is not equivalent to the one you use. Namely, sometimes a conformal function is defined to be a function with everywhere nonzero derivative.

With your definition the argument is correct.

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