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The function $\varphi$ defined by $$ \varphi(z) = \sqrt{i} \, F\bigl(i \sinh^{-1}(\sqrt{i} \, z) \,|\, -1 \bigr). $$ where $F(z | m)$ is the incomplete elliptic integral of the first kind, maps the unit disk to the square with vertices $C \times (\pm 1 \pm i)$ where $C = \varphi(1) \approx 0.92703733865...$, and it is conformal.

I'm trying to use this function to map a checkerboard to a circle. I'm using this pseudo-code:

function colormap(z)
    x = 3 * real(z)
    y = 3 * imag(z)
    out = modulo(x, 2) < 1 ?
        ( modulo(y, 2) < 1 ? "black" : "white" ) :
        ( modulo(y, 2) < 1 ? "white" : "black" )
    return out
end
Z = {0, 0.001, 0.002, ..., 1} x {0, 0.001, 0.002, ..., 1}
for z in Z
    if modulus(z) > 1
        plot( "gray" ) at z
    else
        plot( colormap( phi(z) / 0.92703733865 ) ) at z

Here is the result I get:

enter image description here

I don't understand why the transformed squares at the corners of the checkerboard have only three sides and not four. Could you help me to understand? Am I doing something wrong?

Stéphane Laurent
  • 1,974
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    Each corner square has degenerated into a triangle because three of its vertices lie on the circumference of the circle. The fourth vertex is simply the midpoint of the "base" of the triangle. – Blue Jul 13 '22 at 11:32
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    Both external sides of corner square are transformed to arcs at the boundary of circle with exterior, and as they have common point in the corner then after transformation these arcs will have common point. You just see two adjacent arcs as one arc in your picture. – Ivan Kaznacheyeu Jul 13 '22 at 12:18

0 Answers0