# Fractals/moebius

Möbius transformation is an example of plane transformation

# definition

A Möbius transformation [1][2][3][4] of extended complex plane ${\displaystyle {\widehat {\mathbb {C} }}=\mathbb {C} \cup \{\infty \}}$  is a rational function f of the form

${\displaystyle f(z)={\frac {az+b}{cz+d}}}$

of one complex variable z.

Here the coefficients a, b, c, d and the result w are complex numbers satisfying

${\displaystyle ad-bc\neq 0}$
${\displaystyle w=f(z)={\frac {az+b}{cz+d}}}$

# Representation or form

• function
• matrix

## Function

${\displaystyle f(z)={\frac {az+b}{cz+d}}}$

${\displaystyle f\ \colon z\to w}$

${\displaystyle w=f(z)}$

## Matrix

In matrix form by using homogeneous coordinates:[5]

${\displaystyle M=\left({\begin{matrix}{\begin{array}{cc}a&b\\c&d\end{array}}\end{matrix}}\right)\,}$

${\displaystyle w=\left({\begin{matrix}{\begin{array}{c}w\\1\end{array}}\end{matrix}}\right)=\left({\begin{matrix}{\begin{array}{cc}a&b\\c&d\end{array}}\end{matrix}}\right)\left({\begin{matrix}{\begin{array}{c}z\\1\end{array}}\end{matrix}}\right)=\left({\begin{matrix}{\begin{array}{c}az+b\\cz+d\end{array}}\end{matrix}}\right)\,}$

Matrix M is a square 2x2 invertible matrix[6]

# Examples

## simple

The following simple transformations are also Möbius transformations:

• ${\displaystyle f(z)=z\quad (a=1,b=0,c=0,d=1)}$  is an identity
• ${\displaystyle f(z)=z+b\quad (a=1,c=0,d=1)}$  is a translation
• ${\displaystyle f(z)=az\quad (b=0,c=0,d=1)}$  is a combination of a homothety and a rotation.
• If ${\displaystyle |a|=1}$  then it is a rotation
• if ${\displaystyle a\in \mathbb {R} }$  then it is a homothety
• ${\displaystyle f(z)=1/z\quad (a=0,b=1,c=1,d=0)}$  inversion and reflection with respect to the real axis)

# How to ...?

## eigenvalue and eigenvector

A number ${\displaystyle \lambda }$  and a non-zero vector ${\displaystyle v}$  satisfying

${\displaystyle Mv=\lambda v}$

are called an eigenvalue and an eigenvector of matrix M, respectively.

For dimensions 2 formulas involving radicals exist that can be used to find the eigenvalues. The eigenvalues can be found by using the quadratic formula:

${\displaystyle \lambda _{\pm }={\frac {{\rm {tr}}(M)\pm {\sqrt {{\rm {tr}}^{2}(M)-4\det(M)}}}{2}}.}$

## diagonalization

a diagonal matrix is a matrix in which the entries outside the main diagonal are all zero. In other words all off-diagonal elements are zero in a diagonal matrix.

the main diagonal of a matrix ${\displaystyle A}$  is the list of entries ${\displaystyle A_{i,j}}$  where ${\displaystyle i=j}$ , here ${\displaystyle a_{11},a_{22}}$

the diagonalization of a matrix M gives a pair of matrices: D, P such that:[8]

• D is diagonal (all elements not on the diagonal are 0)
• ${\displaystyle M=PDP^{-1}}$

For 2x2 matrices there is a simple closed form solution[9]

## Product with a scalar

If A is a matrix and c a scalar, then the matrices ${\displaystyle c\mathbf {A} }$  and ${\displaystyle \mathbf {A} c}$  are obtained by left or right multiplying all entries of A by c.

## trace

The trace ${\displaystyle \operatorname {tr} }$  of a square 2x2 matrix ${\displaystyle \mathbf {A} }$

${\displaystyle \mathbf {A} ={\begin{pmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{pmatrix}}}$

is the sum of its diagonal entries

${\displaystyle \operatorname {tr} (\mathbf {A} )=\sum _{i=1}^{2}a_{ii}=a_{11}+a_{22}}$

So ${\displaystyle \operatorname {tr} (\mathbf {M} )=a+d}$

## determinant

determinant ${\displaystyle \operatorname {det} }$  of matrix ${\displaystyle \mathbf {M} }$

${\displaystyle \operatorname {det} \mathbf {M} =\det {\begin{pmatrix}a&b\\c&d\end{pmatrix}}=ad-bc}$

## inverse

Inverse Möbius transformation[10]

${\displaystyle \mathbf {M} ^{-1}={\begin{pmatrix}a&b\\c&d\end{pmatrix}}^{-1}={\frac {1}{\operatorname {det} \mathbf {M} }}{\begin{pmatrix}d&-b\\-c&a\end{pmatrix}}={\frac {1}{ad-bc}}{\begin{pmatrix}d&-b\\-c&a\end{pmatrix}}}$

${\displaystyle z={\frac {1}{\operatorname {det} \mathbf {M} }}{\begin{pmatrix}d&-b\\-c&a\end{pmatrix}}{\begin{pmatrix}w\\1\end{pmatrix}}}$

${\displaystyle z=f^{-1}(w)}$ .

## interpolation

How to smootly interpolate between möbius transformations?[11][12]

If you have two Möbius transformations represented as:

${\displaystyle f(z)={\frac {az+b}{cz+d}}}$

${\displaystyle g(z)={\frac {pz+q}{rz+s}}}$

where coefficients are complex numbers

${\displaystyle a,b,c,d,p,q,r,s,z\in \mathbb {C} }$

Is it possible to derive a third function ${\displaystyle h(z,t)}$ , where ${\displaystyle t\in \mathbb {R} }$  and ${\displaystyle 0\leq t\leq 1}$ , which "smoothly" interpolates between the transformations represented by ${\displaystyle f(z)}$  and ${\displaystyle g(z)}$ ?

The solution:

${\displaystyle h(z,t)=fe^{t\log(f^{-1}g)}=f\operatorname {exp} (t\log(f^{-1}g))}$

## Specifying a transformation by three points

Given a set of three distinct points z1, z2, z3 on the one Riemann sphere ( let's call it z-sphere) and a second set of distinct points w1, w2, w3 on the second sphere ( w-sphere) , there exists precisely one Möbius transformation f(z) with :

• ${\displaystyle f(z_{i})=w_{i}}$
• ${\displaystyle f^{-1}(w)=z_{i}}$

for i=1,2,3

### Mapping to 0, 1, infinity

The Möbius transformation with an explicit formula :[13]

${\displaystyle f(z)={\frac {(z-z_{1})(z_{2}-z_{3})}{(z-z_{3})(z_{2}-z_{1})}}}$

maps :

• z1 to w1= 0
• z2 to w2= 1
• z3 to w3= ∞

#### the unit circle to the real axis - first method

Let's choose 3 z points on a circle :

• z1= -1
• z2= i
• z3= 1

then the Möbius transformation will be :

${\displaystyle f(z)={\frac {(z+1)(i-1)}{(z-1)(i+1)}}}$

Knowing that :[14]

${\displaystyle i+1=-i(i-1)}$

one can simplify this to :

${\displaystyle f(z)=i{\frac {z+1}{z-1}}}$

In Maxima CAS one can do it :

(%i1) rectform((z+1)*(%i-1)/((z-1)*(%i+1)));
(%o1) (%i*(z+1))/(z−1)


where coefficients of the general form are :

${\displaystyle a=i}$
${\displaystyle b=i}$
${\displaystyle c=1}$
${\displaystyle d=-1}$

so inverse function can be computed using general form :

${\displaystyle f^{-1}(w)={\frac {dw-b}{-cw+a}}={\frac {-i-w}{i-w}}}$

Lets check it using Maxima CAS :

(%i3) fi(w):=(-%i-w)/(%i-w);
(%o3) fi(w):=−%i−w/%i−w
(%i4) fi(0);
(%o4) −1
(%i5) fi(1);
(%o5) −%i−1/%i−1
(%i6) rectform(%);
(%o6) %i


Find how to compute it without symbolic computation program (CAS)  :

(%i3) fi(w):=(-%i-w)/(%i-w);
(%o3) fi(w):=−%i−w/%i−w
(%i8) z:x+y*%i;
(%o8) %i*y+x
(%i9) z1:fi(w);
(%o9) (−%i*y−x−%i)/(−%i*y−x+%i)
(%i10) realpart(z1);
(%o10) ((−y−1)*(1−y))/((1−y)^2+x^2)+x^2/((1−y)^2+x^2)
(%i11) imagpart(z1);
(%o11) (x*(1−y))/((1−y)^2+x^2)−(x*(−y−1))/((1−y)^2+x^2)
(%i13) ratsimp(realpart(z1));
(%o13) (y^2+x^2−1)/(y^2−2*y+x^2+1)
(%i14) ratsimp(imagpart(z1));
(%o14) (2*x)/(y^2−2*y+x^2+1)


So using notation :

${\displaystyle z=x+yi=f^{-1}(w)}$

one gets :

${\displaystyle x=\operatorname {Re} (z)=\operatorname {Re} (f^{-1}(w))={\frac {y^{2}+x^{2}-1}{y^{2}-2y+x^{2}+1}}}$

${\displaystyle y=\operatorname {Im} (z)=\operatorname {Im} (f^{-1}(w))={\frac {2x}{y^{2}-2y+x^{2}+1}}}$

It can be used for unrolling the Mandelbrot set components [15]

#### the unit circle to the real axis - second method

Function :

${\displaystyle f(z)=i{\frac {z-1}{z+1}}}$

sends the unit circle to the real axis :

• z=1 to w=0
• z=i to w=1
• z=-1 to ${\displaystyle w=\infty }$

### Mapping to the imaginary axis

Function ${\displaystyle f(z)={\frac {z-1}{z+1}}}$  sends the unit circle to the imaginary axis.[16]