Coursera: Analysis of a Complex Kind. 4.3 Mobius Transformations I

Jul 27, 2016

Slide 2 Definition: of form f(z) \frac{az + b}{cz+d}

a, b, c , d \in \textrm{C where ad-bc} \neq 0

Properties

As z \rightarrow \; \infty f(z) \rightarrow \; \frac{a}{c} leading to f(z) \rightarrow \; \infty if c = 0

Define f(\infty) = \frac{a}{c} \textrm{ if } c \neq 0 and f(\infty) = \infty \textrm{ if }c = 0

Also, by looking at denominator, f(- \frac{d}{c}) = \infty

We say f is a mapping from the extended complex plane: \widehat{C} = C \cup \{ \infty \} to the extended plane \widehat{C}

Slide 3 Properties of Mobius Transformation

Differentiate: Use quotient rule

f ‘ (z) = \frac{ad-bc}{{(cz + d)}^2} so ad-dc neq 0 guarantees f not constant.

Mobius is one-one in \widehat{C}

Show: Rearrange w = \frac{az + b}{cz+d} to z = \frac{wd - b}{-wc+a}

Therefore Mobius are (the only) conformal mapping from \widehat{C} \textrm{ to } \widehat{C}

Special case

c = 0, d = 1, f(z) = az + b (a \neq 0), called affine transformations. Since \infty \textrm{ maps to } \infty then by excluding this point from domain C maps to C. Therefore affine transformations are (only) transforms maps C to C.

f(az) is a rotation and dilation

f(z + b) is a translation

Slide 5 Inversion f(z) = 1/z

Image a circle under inversion?

z = re^{i \theta}, \; 1/z = (\frac{1}{r})e^{-i \theta}. The image is equation of circle centre origin, radius 1/r.

Since \theta \textrm{ maps to }  - \theta and r \textrm{ maps to } \frac{1}{r}  f  interchanges the inside and outside of the unit circle  and circle centre origin, radius r goes to circle centre origin radius 1/r.

Other circles?

Slide 6. Inversion on general circle *Important example*

K={z: |z- 3| = 1}, circle centre 3, radius 1.

Image under inversion?

w \in f(K) <=> 1/w \in K, so put this point 1/w in definition of K

| 1/w – 3 | = 1

multiply both sides by |w|

| 1 - 3w| = |w|

square both sides ***

| 1 - 3w|^2 = |w|^2 (*)

Since z^2 = z*\overline{z}, (*) becomes ***

( 1 - 3w)( 1 - 3\overline{w}) = |w|^2

 1 - 3w + 3\overline{w} + 9|w|^2 = |w|^2

8|w|^2 - 3w + 3\overline{w} = -1 

( w-\frac{3}{8})( \overline{w}-\frac{3}{8}) = \frac{9}{64} - \frac{1}{8}

| w-\frac{3}{8}| = \frac{1}{8}

which is a circle centre 3/8, radius 1/8.

So a circle has been inverted to a circle.

Slide 8 Reciprocal of circle whose circumference passes through origin

K={z: |z – 1| = 1}, unit circle centre (1,0), cuts through origin

<< diagram >>

w \in f(K) <=> 1/w \in K <=> |1/w – 1| = 1

|1 – w| = |w|

 |1 - w|^2 = |w| ^2 (*)

Since {|z|}^2 = z \overline{z}

(*)  (1 - w)(1- \overline{w})= |w| ^2

 1 -  \overline{w}-w + |w| ^2= |w| ^2

 1 =  \overline{w}+w

Now, imaginary parts of complements cancel, and real parts of complements are same so double

Re(w) = 1/2 (there is no constraint or condition on imaginary part , so can take any value)

Plot is vertical line through Re(w) = 1/2

<< to continue >>

Slide 12

IMPORTANT RESULT

The particular Mobius transformation f(z) = \frac{z - z_1}{z - z_3} . \frac{z_2 - z_3}{z_2 - z_1}

maps z_1, z_2, z_3 \textrm{ to } 0, 1, \infty

i.e. f(z_1) \textrm{ to } 0

f(z_2) \textrm{ to }  1

f(z_3) \textrm{ to } \infty

 

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