# Transforming a Bézier Curve by a Complex Function

I have a method to plot a rectangular region and then plot the lattice transformation by some complex function $f[z]$.

latticeTransform[f_: Function, xrange_: Range, yrange_: Range] :=
(curve = Graphics[BezierCurve[RandomPoint[
Rectangle[{xrange[], yrange[]}, {xrange[],yrange[]}], 6]]];
dom = Show[{ParametricPlot[{x, y}, xrange, yrange,
AspectRatio -> 1], curve}];
image = ParametricPlot[{Re[f[z]], Im[f[z]]} /. z -> x + I*y, xrange,
yrange, PlotRange -> All];
GraphicsGrid[{{dom, image}}, ImageSize -> Large]);


I added the Bézier curve to visualize how a random curve will get transformed in the output, but I'm not entirely sure how to map it onto the image. I thought about storing the random points I generated, transforming them with $f[z]$, and then calling BezierCurve[] on the points again, but I believe this wouldn't be a true mapping as Bézier will recalculate and find a new curve.

Is this correct? If so, how would I draw the curve transformed by $f[z]$?

Edit: My only other thought is to get a list of points {x, y} on the Bézier curve and generate a new list by f[x + I y], but again I'm unsure of how to do this and it has a tradeoff of accuracy vs. computation time.

• A Bézier curve after an arbitrary complex transformation will of course not necessarily be a Bézier curve anymore. – J. M. will be back soon Jul 26 '17 at 12:49
• Counter example : A exact circle can't be represented with a Bezier curve, though it is the transformation of a straight line by a complex function. – andre314 Jul 26 '17 at 12:51
• @J.M. thanks, so I can't just call Bezier curve again. But how would I transform the curve by the function? I'm more interested in mapping the curve than the semantics of a Bezier. – Dando18 Jul 26 '17 at 13:02

You'll need to work a bit harder to demonstrate the complex transformation of a Bézier curve; in particular, you need to express the curve explicitly in terms of the Bernstein basis before transforming. Here's an example:

BlockRandom[SeedRandom["bezier"]; (* for reproducibility *)
With[{f = Function[x, ArcTan[x]], xrange = {-5, 5}, yrange = {-4, 4}},

(* generate random Bézier curve explicitly *)
pts = RescalingTransform[{{0, 1}, {0, 1}}, {xrange, yrange}] @ RandomReal[1, {6, 2}];
n = Length[pts] - 1;
{bf[t_], bg[t_]} = BernsteinBasis[n, Range[0, n], t].pts;

GraphicsRow[{Show[ParametricPlot[{x, y}, Prepend[xrange, x] // Evaluate,
Prepend[yrange, y] // Evaluate,
AspectRatio -> Automatic, Mesh -> True,
PlotStyle -> None],
ParametricPlot[{bf[t], bg[t]}, {t, 0, 1}]],
Show[ParametricPlot[ReIm[f[x + I y]], Prepend[xrange, x] // Evaluate,
Prepend[yrange, y] // Evaluate,
AspectRatio -> Automatic, Mesh -> True,
PlotStyle -> None],
ParametricPlot[ReIm[f[bf[t] + I bg[t]]], {t, 0, 1}]]}]]] Turning the entire business into a suitable Manipulate[] is left as an exercise for the interested reader.

• Thanks! I'll have to read up on the Bernstein basis. Would you happen to have a recommendation on a cleaner way to generate a random curve and transform it? maybe without Béziers? – Dando18 Jul 26 '17 at 13:16
• Your "and transform it" is the source of the complication in this case. Some math to ponder: Bézier curves (and more generally B-spline curves) only remain as such under an affine transformation, and such transformations can be done easily by transforming only the control points instead of the entire curve. – J. M. will be back soon Jul 26 '17 at 13:19