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Suppose $f:U\to \mathbb{C}$, where $U$ is just an open set in complex plane. I know that if $f$ is holomorphic and have no-vanishing derivative then $f$ is conformal. Is there a partial 'converse' to this statement? I.e. if $f$ is conformal, will it be the case that $f$ must have no-vanishing derivative on $U$?

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  • $\begingroup$ You should state your definition of a conformal map. According to the standard definition, it is required to be 1-1. $\endgroup$ Feb 26, 2020 at 1:05

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Assume that $f:U\rightarrow \mathbb{C}$ and suppose that $f$ is real differentiable. If $f$ is conformal, then $Df(x)$ preserves angles for every $x$ in $U$. This implies that $Df(x)$ is an orthogonal matrix multiplied by a positive number. Thus $Df(x)$ is a rotation matrix multiplied by a positive number. Hence $Df(x)$ is a multiplication by some nonzero complex number and hence $f$ is a complex differentiable in $x$ with nonvanishing derivative.

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