Let a, b, c, d be complex numbers. We define a Mobius transformation from C (complex numbers) to C as
f(z) = (az+b) / (cz+d).
This family of transformations is also called fractional linear transformations. From the definition it is clear that c and d cannot both be zero. If c is zero we get
f(z) = (a/d)z + (b/d),
just the linear transformations. If c is non-zero then
f(z) = [(a/c)(cz+d) + b - (ad/c)] / (cz+d) = (a/c) + (b – ad/c) / (cz+d).
In both cases we see that if ad = bc the resulting transformation is constant. Therefore we always require ad - bc &ne 0.
If c is non-zero f is not defined for z = - d/c. It is useful to add a new point "at infinity", denoted by &infin, and to define f(-d/c) = &infin , f(&infin) = a/c. If c is zero we define f(&infin) = &infin .
We construct an inverse to a Mobius transformation:
w = (az+b) / (cz+d),
czw+dw = az+b,
z = (dw-b) / (-cw+a) = g(w).
g is again a Mobius transformation (its coefficients satisfy the restriction). If c is zero then f and g are defined in all C and f(&infin) = &infin = g(&infin) completes the claim. If c is non-zero then we check the “infinite” points, f(-d/c) = &infin , g(&infin) = -d/c, f(&infin) = a/c, g(a/c) = &infin .
This is obvious, the constant cancels out. This shows that effectively there are three parameters that define the transformation.
Let f be defined using a, b, c, d, and g be defined by A, B, C, D. We compute the composition
g(f(z)) = (Af(z)+B) / (Cf(z)+D) =
= (A(az+b) / (cz+d)+B) / (C(az+b) / (cz+d)+D) =
= (Aaz+Ab+Bcz+Bd) / (Caz+Cb+Dcz+Dd) =
= ((Aa+Bc)z+(Ab+Bd)) / ((Ca+Dc)z+(Cb+Dd)).
We see that the composition is again a Mobius transformation. Moreover, if we let [f] be the 2x2 matrix with (a, b) in the first row and (c, d) in the second, and similarly for all Mobius transformations, we get
[gf] = [g][f],
where on the left we have composition of functions and on the right matrix multiplication. This defines an action of the group of invertible 2x2 matrices over C on C.
Let f be a Mobius transformation, and suppose f(0)=x, f(1)=y, f(&infin)=z, and that x,y,z are three different points. If z = &infin we have seen that c = 0, and we can normalize so that d = 1. In this case b = x and a = y - b. If z is finite we can normalize so that c = 1, and then a = z. The other two conditions are b/d = x and (z+b) / (1+d) = y. If x = &infin then d = 0 and b = y – z. If y = &infin then d = -1 and b = -x. Otherwise we get b = dx, z+dx = y+dy, d = (z-y) / (y-x).
In all cases we see that x,y,z define a unique Mobius transformation.
As we have seen a Mobius transformation is either a linear transformation (when c is zero), or it can be written as
f(z) = (a/c) + (b-ad/c) / (cz+d).
This can be written as a composition of a linear transformation g(z) = cz+d, followed by an inversion i(z) = 1/z, and then another linear transformation h(z) = (b-ad/c)z+(a/c). Linear transformations are composed of shifts, rotations and scale multiplications, so they transform lines to lines and circles to circles. Let us investigate the inversion i(z).
>> Click here for a geometric proof. <<
First we note that for any constant c
i(z) = ci(cz),
so we can first multiply by a constant when checking the image of the inversion. We will use this to rotate, when |c| = 1.
Let L be any line. Up to rotation we can assume L has the form L = x+it, with x a Real constant and t a Real parameter. The image i(L) consists of the points
1/(x+it) = (x-it)/(x2+t2).
For x = 0 this is just -i/t which is the same line with a different parametrization. Otherwise we look at the distance from 1/2x
|(x-it)/(x2+t2) - 1/2x| = [x/(x2+t2) - 1/2x]2 + t2/(x2+t2)2 =
= x2/(x2+t2)2 + 1/4x2 - 1/(x2+t2) + t2/(x2+t2)2 =
= 1/4x2.
The result is a circle centered at 1/2x with radius 1/2x.
Let C be a circle. Up to rotation we can assume C has the form C = x+reit, with x and r positive Real constants and t a Real parameter. The image i(C) consists of the points
1/(x+reit) = (x+re-it)/(x2+r2+2xrcos(t)).
When x = r, C passes through zero, and i(C) is
(1/2x)(1+e-it)/(1+cos(t)) =
= 1/2x - i(1/2x)sin(t)/(1+cos(t))),
Which is a parametrization of a line. Otherwise we look at the distance from x/(x2-r2)
|(x+re-it)/(x2+r2+2xrcos(t)) - x/(x2-r2)| =
= [(x+rcos(t))/(x2+r2+2xrcos(t)) - x/(x2-r2)]2 +
+ (rsin(t))2/(x2+r2+2xrcos(t))2 =
= (x2+r2cos2(t)+2xrcos(t)+r2sin2(t))/(x2+r2+2xrcos(t))2 +
+ x2/(x2-r2)2 - 2(x2+xrcos(t))/[(x2-r2)(x2+r2+2xrcos(t))] =
= x2/(x2-r2)2 + [1/(x2+r2+2xrcos(t))] [1-(2x2+2xrcos(t))/(x2-r2)] =
= x2/(x2-r2)2 + [1/(x2+r2+2xrcos(t))] [(-x2-r2-2xrcos(t))/(x2-r2)] =
= x2/(x2-r2)2 - 1/(x2-r2) = r2/(x2-r2)2.
The result is a circle centered at x/(x2-r2) with a radius of r/(x2-r2).