# What conformal map transforms two disjoint disks into two parallel lines?

Given two disjoint disks in the complex plane, can the region outside of them be conformally mapped to a region bounded by two parallel lines? I think the answer is yes because the region is simply connected (I do not know further details about the topic).

At first, I thought on using Möbius transformation, but that would work only if they had a tangent point (see this question).

I am thinking of trying to construct a Schwarz-Christoffel mapping that would do it, but I do not know how to achieve that.

The problem I am trying to solve is to find the electric potential distribution of two parallel, non-uniformly charged conductors. I think that finding a solution through conformal mapping would be interesting.

• The complement of two disjoint disks is not simply connected. You can however map that region (using a Möbius) to an annulus (i.e., the region between two concentric circles. – Hagen von Eitzen Oct 29 '19 at 13:28

## 1 Answer

As per a comment of Hagen von Eitzen: the complement of two disjoint disks is not simply connected. But that region can be mapped to an annulus.

Assume two circles, $$C_1$$ and $$C_2$$ with radius $$r_1$$ and $$r_2$$, respectively, such that $$r_1 > r_2$$. Assume yet that $$C_1$$ is centered at the origin and $$C_2$$ centered over the Real axis at a point $$h$$. Lets transform $$C_1$$ and $$C_2$$ such that they form an annulus centered at the origin.

## The algorithm is as follows:

Find parameters $$a,b,c,d$$ of the Möbius transformation (can be by explicit determinant formula) that makes the bigger circle into a line over the Imaginary axis ($$\mathbf{z} \to \mathbf{w}$$).

$$\mathbf{z} = [i\,r_1,\, -i\,r_1,\, r_1] \\ \mathbf{w} = [-i,\, i,\, \infty] \\$$

This transformation is

$$f(z) = \frac{a\,z + b}{c\,z + d}$$

Move both the line and new circle by $$-p$$ such that the circle of radius $$r_w$$ is centered at the origin. Variables $$p$$ and $$r_w$$ satisfy:

$$|w_1|^2 - 2\,\Re(w_1 \,\bar p) + |p|^2 - r_w^2 = 0 \\ |w_2|^2 - 2\,\Re(w_2 \,\bar p) + |p|^2 - r_w^2 = 0 \\$$

Where $$w_1$$ and $$w_2$$ are points over the circle.

This transformation is $$g(z) = z - p$$

Now find $$x_1$$ and $$x_2$$ (zeros of the following equation) that transforms the line into a circle around the origin while also preserving the center of the other circle around the origin.

$$x^2 + 2\,p\,x + r_w^2 = 0$$

This transformation is $$h(z) = \frac{z - x1}{z - x2}$$

Now compose all these transformations:

$$T(z) = h \circ g \circ f (z)$$

$$T(z) = \frac{b + a\,z - (p + x_1) (d + c\,z)}{b + a\,z - (p + x_2) (d + c\,z)}$$

### A Julia code

The following is a Julia code I created that implements the above algorithm.

using LinearAlgebra
using NLsolve
using Polynomials
using Plots

function mobius(z, w)
a = det([z*w  w  1;
z*w  w  1;
z*w  w  1])
b = det([z*w  z  w;
z*w  z  w;
z*w  z  w])
c = det([z  w  1;
z  w  1;
z  w  1])
d = det([z*w  z  1;
z*w  z  1;
z*w  z  1])
return a, b, c, d
end

function find_trans(r1, r2, h, n=200)
infty = 1e10  # "infinity"
θ = range(0, 2π, length=n)
z1 = @. r1*(cos(θ) + 1im*sin(θ))
z2 = @. r2*(cos(θ) + 1im*sin(θ)) + h
# transforms the bigger circle into a line on the imaginary axis
a, b, c, d = mobius([-1im*r1, 1im*r1, r1], [1im, -1im, infty])
f(z) = (a*z + b)/(c*z + d)
# center transformed circle on the origin
w1 = f(z2)
w2 = f(z2[cld(n,2)])
function f!(F, x)
F = abs(w1)^2 - 2*real(w1*conj(x)) + abs(x)^2 - x^2
F = abs(w2)^2 - 2*real(w2*conj(x)) + abs(x)^2 - x^2
end
# sol.zero: center point of the transformed circle and its radius
sol1 = nlsolve(f!, [0.0, 2r1])
p = sol1.zero
r = sol1.zero
# transform the line into a unit circle centered at the origin,
# while also preserving the other at the origin
x1, x2 = roots(Poly([r^2, 2p, 1]))
# Möbius transformation that sends x1 -> 0 and x2 -> oo,
# where x1 and x2 are symmetric points to both the line and circle
T(z) = (b + a*z - (p + x1)*(d + c*z))/(b + a*z - (p + x2)*(d + c*z))
return a, b, c, d, p, r, x1, x2, T
end

begin
n = 500
θ = range(0, 2π, length=n)
r1 = 2
r2 = 1
h = 2r1
z1 = @. r1*(cos(θ) + 1im*sin(θ))
z2 = @. r2*(cos(θ) + 1im*sin(θ)) + h
a, b, c, d, p, r, x1, x2, T = find_trans(r1, r2, h)
w1 = T.(z1)
w2 = T.(z2)
plot(aspect_ratio=1)#, xlim=(-l,l), ylim=(-l,l))
plot!(z1, color=:blue, label="z1")
scatter!(z1[1:1], color=:blue, label="")
plot!(z2, color=:red, label="z2")
scatter!(z2[1:1], color=:red, label="")
plot!(w1, color=:blue, label="w1", linestyle=:dash)
scatter!(w1[1:1], color=:blue, label="")
plot!(w2, color=:red, label="w2", linestyle=:dash)
scatter!(w2[1:1], color=:red, label="")
end


The output of the above code is the following figure. Now the potential problem can be solved in the annulus. See, for example, Conformal Mapping and Bipolar Coordinate for Eccentric Laplace Problems.

## Edit

For the sake of completness, the inverse transformation is $$T^{-1}(z) = \frac{b - d\,(p + x_1) - b\,z + d\,(p + x_2)\,z}{a\,(z - 1) + c\,(p + x_1 - (p + x_2)\,z)}$$