# Is a linear conformal mapping same as a similarity transformation?

1. For a mapping between two Euclidean spaces, is it a linear conformal mapping if and only if it is a similarity transformation?

My answer is yes, because the Jacobian matrix of a conformal transformation is everywhere a scalar times a rotation matrix.

Note that both allow reflection, i.e. change of orientation.

2. Is it correct that a conformal mapping may not be an affine nor projective transformation, because it may not be linear?

Thanks and regards!

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Could you please explain why the conformal map is a scalar times a rotation matrix? I can show it for the opposite direction but I'm stuck in the regular direction of the proof. – Amontillado Feb 20 at 16:32

1. Yes, for elementary reasons. Let $f$ be a linear conformal map and apply this to any triangle $ABC$. Then $f(AB),f(BC),f(CA)$ will be lines by linearity, and by conformality $f(ABC)$ will have the angles of $ABC$ so they will be similar therefore $f$ is a similarity mapping.