## 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$(z_3) \textrm{ to } \infty$