Conformal mapping of biaxial symmetric shapes into an (unit) circle

Question

I would like to know if there is some kind of coordinate mapping (perhaps conformal mapping) that can map the perimeter of the circles that cross each other (like shown in the picture) into a (unit) circle? figure 1

Answer 1

I am not sure there exists a single transformation of conformal type than can "do the job" for example as the conformal image of a square (because of the four right angles). Nevertheless, when one uses "basic" inversion

$$Z=f(z)=\frac{1}{\overline{z}}\tag{1}$$ on a square ($\overline{z}$ is the complex conjugate of $z$), one gets a kind of gothic "rosace" as represented on Fig. 1,

Fig. 1 : the image of the square (red) is the rosace (blue).

a result which is not so far from what we want. Subsequent transformations (e.g., homographic transforms $g(z)=\frac{az+b}{cz+d}$) can maybe be found that give the desired result.

If we leave aside the quest for a unique transformation, I would like to show here that the inversion idea is still a good one and can bring a satisfactory answer, under the condition to use four of them.

Let us recall that, in the complex plane, the most general inversion is with center $c$ and power $p$ with the following description :

$$Z=c+\frac{p}{\overline{z-c}}\tag{2}$$

(formula (1) is the particular case of (2) when $c=0$ and $p=1$). I will assume that either you already know this transformation or you will study it separately].

The following figure :

Fig. 2.

is the image of the square $A(1,0),B(0,1),C(-1,0),D(0,-1)$ by specific inversions on each of its 4 sides, in such a way in particular that the overall image is connected.

• inversion with center $E$ and power $p=\frac58=0.625$ sends line segment $AB$ onto the green arc of circle,

• inversion with center $F$ and power $p=\frac52=2.5$ sends line segment $BC$ onto the blue arc of circle,

• same kind of operations for the two last ones.

There is a part of arbitrary choice for the centers as long as they are situated on the median lines of the square. Once the center is fixed, the power $p$ is more or less imposed. Moreover centers $H$ and $F$ have to be slightly changed in order that blue and magenta arcs are realy parts of a same circle.

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Edit : a simpler approach :

• in quadrants 2 and 4, we keep the unit circle, whereas

• in quadrants 1 and 3, we transform by inversion the sides of the square $1,i,-1,-i$ ; we take the centers of inversion in $a+ai$ and $-(a+ai)$ resp. for $a \approx 0.29$ and power of inversion such that the different arcs are connected, which is shown possible iff $p=a^2+(a-1)^2$ :

Matlab program :

function Z=inv(z,c,p) % inversion
   Z=c+p/conj(z-c);
end;
%%% main program :
a=0.29;c=a*(1+i);p=a^2+(a-1)^2;P=[];S=[];
for t=-pi:0.05:pi;
   sq=exp(i*t)/(abs(cos(t))+abs(sin(t)));
   S=[S,sq]; % building the Square
   if abs(cos(t-pi/4))<abs(sin(t-pi/4)); % Quadrants 2 or 4
      b=exp(i*t);
   else
      b=inv(sq,sign(t)*c,p);
   end;
   P=[P,b]; % building the Peanut
end;
plot(P,'b','linewidth',5);
plot(S,'r','linewidth',5);
plot([c,-c],'ok')