Exam

The composition of conformal mappings

F

and

G

with scaling functions

f

and

g

is conformal. and the scaling function of

G \circ F

g \circ F . f

. 2. Show that the reciprocal of the scaling funktion of the composition of

F (x) = I (x - a)

and

G (x) = I (x - b)

is a quadratic polynomial of the form

A + ⟨ x, v ⟩ + B | x |^{2}

. 3. Show that the reciprocal of the scaling funktion of any composition of mappings of the form

I (x - a_{j})

j = 1, \dots, m

, is a quadratic polynomial of the form

A + ⟨ x, v ⟩ + B | x |^{2}

Suppose

F : U \to V

and

G : V \to W

are conformal, then for all

x \in U

, all

y \in V

and all

u, v \in R^{n}

⟨ D F (x)^{*} D F (x) u, u ⟩ = f (x)^{2} ‖ u ‖^{2} and ⟨ D G (y)^{*} D G (y) v, v ⟩ = g (y)^{2} ‖ v ‖^{2} .

By the chain rule we have

D (G \circ F) (x) = D G (F (x)) D F (x)

. Putting

y : = F (x)

and

v : = D G (y) u

we thus get:

\begin{array}{rcl} ⟨ D (G \circ F) (x)^{*} D (G \circ F) (x) u, u ⟩ & = & ⟨ D G (y)^{*} D F (x)^{*} D F (x) D G (y) u, u ⟩ \\ = & ⟨ D F (x)^{*} D F (x) v, v ⟩ = f (x)^{2} ⟨ v, v ⟩ \\ = & f (x)^{2} ⟨ D G (y) u, D G (y) u ⟩ = f (x)^{2} g (y)^{2} ‖ u ‖^{2} . \end{array}

Proving that

G \circ F

is conformal with scaling function

h (x) : = f (x) g (F (x))

.
2. The reciprocal is given by

{‖ F (x) - b ‖}^{2} ‖ x - a ‖^{2} = {‖ (x - a) / ‖ x - a ‖ - b ‖ x - a ‖ ‖}^{2} = 1 - ⟨ x - a, b ⟩ + {‖ b ‖}^{2} ‖ x - a ‖^{2}

3. Assume the reciprocal of the scaling function of the composition of

m

maps

F_{m + 1}, \dots, F_{2}

is given by

A + ⟨ x, v ⟩ + B ‖ x ‖^{2}

, then the reciprocal of the scaling function of the composition of

m + 1

maps

F_{m + 1}, \dots, F_{2}, F

F (x) = I (x - a)

, is given by

\begin{array}{rcl} (A + ⟨ I (x - a), v ⟩ + B {‖ I (x - a) ‖}^{2}) ‖ x - a ‖^{2} & = & A ‖ x - a ‖^{2} + ⟨ x - a, v ⟩ + B {‖ (x - a) / ‖ x - a ‖ ‖}^{2} |^{2} ‖ x - a ‖^{2} \\ = & A ‖ x - a ‖^{2} + ⟨ x - a, v ⟩ + B \end{array}