I've spent quite a bit of time on google trying to find information on whether or not Bezier curves are invariant under conformal mapping (i.e. a conformal mapping of all points on the curve is the same as a curve formed with just the conformal mapped control points for the curve). It feels like this should be true, but I can't seem to find anything that either proves or disproves this.

Does anyone know whether Bezier curves are invariant under conformal mapping?

Edit: based on Hagen's observation of a straight line becoming a circle, it no longer feels like this "should" be true! Although that does raise the question "which conformal transforms are also affine transforms", which I'm also having a bit of trouble googling (although I did find http://www.leptonica.com/affine.html)

  • $\begingroup$ Could you give some examples, I find the claim suspicious, but haven't experimented with it. In particular, if one thinks about the control polygon, most conformal maps wouldn't preserve the straight lines of the control polygon. If you can understand the behavior of the control polygon, perhaps you will have some success. $\endgroup$ Commented Nov 29, 2015 at 15:49
  • $\begingroup$ A Bezier curve cannot lie on a sphere unless it degnerates to a point. $\endgroup$
    – bubba
    Commented Dec 3, 2015 at 13:57

2 Answers 2


No. For example a straight line (a Bezier curve) might be conformally mapped to a circle (a non-Bezier curve).

However, Bezier curves are invariant under affine linear transformations.

  • $\begingroup$ So, then, the question becomes which conformal maps preserve Bezier curves... $\endgroup$ Commented Nov 29, 2015 at 15:50
  • $\begingroup$ (+1) Affine linear transformation are the most general such transformations: If $T$ is not affine, pick $q_{1}$, $q_{2}$ so that the segment from $q_{1}$ to $q_{2}$ is not mapped into a line, and consider the control points $p_{1} = q_{1}$, $p_{2} = p_{3} = p_{4} = q_{2}$. $\endgroup$ Commented Nov 29, 2015 at 16:23

The straight line is usually give as a counter example for this question. However, I think that the straight line is not a proper Bezier curve but a degenerate one. To resolve this, simply break up the line into smaller segments, and you will have circles and all other shapes. Smaller the line segments, better the final result.

  • 1
    $\begingroup$ That's all the counter example needed though; if the degenerate case does not hold, then the original claim doesn't hold. I'm not sure about breaking up the line into segments, since that turns the Bezier curve into a set of, in the limit, points, which makes any claims about them irrelevant to any higher order curve description. $\endgroup$ Commented Dec 2, 2015 at 19:05
  • $\begingroup$ Agreed! (I take back my comment) $\endgroup$ Commented Dec 5, 2015 at 12:51

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