Linear Fractional Transformations

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Contents

Definition and basic facts

Let M be a Möbius transformation. Then

  • M can be expressed as the composition of a finite number of rotations, translations, magnifications and inversions
  • M maps the extended complex-plane  \{ C \cup \infty \} to itself
  • M maps the class of circles and lines to circles and lines
  • M is conformal at every point besides its pole

Poles and Fixed Points

A pole (regular singularity) of a function is a point z0 where \lim_{z \rightarrow z_{0}} f(z) = \infty.
A fixed point of a function f(z) is a point z0 such that f(z0) = z0 That is the point gets mapped to the same spot in the UV-plane.

1. Find the poles of M = \frac{az+b}{cz+d}. How many poles does it have? How many fixed points does it have? (These answers should depend on a, b, c and d.) Show your work (computations).


2. A. If a line or circle passes through the pole of M then it must be mapped to what shape?
B. If a line or circle does NOT pass thru the pole of M it must get mapped to what kind of shape?
C. Where does M map the point at infinity?

Inverses of linear fractional transformations

Since M is a one-to-one mapping on the extended complex plane, it has an inverse.
If you solve w = M then M^{-1}=\frac{dz-b}{-cz+a} where ad-bc \neq 0
Note that the inverse is also an LFT. In general, if S and T are two LFTs, then S(T(z)) is also an LFT.

3. A. Let M=\frac{z}{2z-8}. Find the inverse of f.
B. Find the image of the interior of the circle | z − 2 | = 2 under the LFT f.
C. Sketch the image and pre-image of C under w=f(z).

Some special cases

4. Show that every Möbius transformation of the form M(z) =\frac{pz+q}{\bar{q}z+\bar{p}} where | p | > | q | can be rewritten in the form M(z) = e^{i \theta}\frac{z-a}{\bar{a}z-1}  where | a | < 1.
5. Show that Möbius transformation of the form M(z) = e^{i \theta}\frac{z-a}{\bar{a}z-1} where | a | < 1 map the unit circle and unit disk onto itself. Is this also true for the inverse of M?
6. Show that Möbius transformation of the form M(z) =\frac{az+2b}{\bar{b}z+\bar{a}} where | a | 2 − 2 | b | 2 = 1 map the circle | z | 2 = 2 to itself. Is this also true for the inverse of M? What happens to the circle if we apply two different Möbius transformations of this type?

Group structure

7. We can associate a matrix to a Möbius transformation. In problem 6 we would have the matrix \mathbf{M} = \begin{pmatrix}
a & 2b \\
\bar{b} & \bar{a} \end{pmatrix}. We would like to show that the collection of matrices \{ \mathbf{M} = \begin{pmatrix}
a & 2b \\
\bar{b} & \bar{a}\end{pmatrix}  | |a|^2-2|b|^2=1 \} forms a group. We would need to show that the collection is closed under multiplication, has an identity element, has an inverse and is associative.
a. Closed under multiplication: Show that given two arbitrary elements, the product belongs to the collection. I.e. show that \begin{pmatrix}
a_1 & 2b_1 \\
\bar{b}_1 & \bar{a}_1 \end{pmatrix} * \begin{pmatrix}
a_2 & 2b_2 \\
\bar{b}_2 & \bar{a}_2 \end{pmatrix} is a matrix that belongs to the collection.
b. Identity element: Show that \mathbf{I} = \begin{pmatrix}
1 & 0 \\
0 & 1 \end{pmatrix} belongs to our collection.
c. Inverses: Given \mathbf{M} = \begin{pmatrix}
a & 2b \\
\bar{b} & \bar{a} \end{pmatrix}, show that the inverse also belongs to the collection.
d. Associativity: Show that  (\mathbf{M_1} \mathbf{M_2})\mathbf{M_3} = \mathbf{M_1} (\mathbf{M_2} \mathbf{M_3})
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