The composition of conformal mappings $F$ and $G$ with scaling functions $f$ and $g$ is conformal. and the scaling function of $G\circ F$ is $g\circ F.f$. 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 $A+\la x,v\ra+B|x|^2$. 3. Show that the reciprocal of the scaling funktion of any composition of mappings of the form $I(x-a_j)$, $j=1,\ldots,m$, is a quadratic polynomial of the form $A+\la x,v\ra+B|x|^2$.
Suppose $F:U\rar V$ and $G:V\rar W$ are conformal, then for all $x\in U$, all $y\in V$ and all $u,v\in\R^n$:
$$
\la DF(x)^*DF(x)u,u\ra=f(x)^2\Vert u\Vert^2
\quad\mbox{and}\quad
\la DG(y)^*DG(y)v,v\ra=g(y)^2\Vert v\Vert^2~.
$$
By the chain rule we have $D(G\circ F)(x)=DG(F(x))DF(x)$. Putting $y\colon=F(x)$ and $v\colon=DG(y)u$ we thus get:
\begin{eqnarray*}
\la D(G\circ F)(x)^*D(G\circ F)(x)u,u\ra
&=&\la DG(y)^*DF(x)^*DF(x)DG(y)u,u\ra\\
&=&\la DF(x)^*DF(x)v,v\ra
=f(x)^2\la v,v\ra\\
&=&f(x)^2\la DG(y)u,DG(y)u\ra
=f(x)^2g(y)^2\Vert u\Vert^2~.
\end{eqnarray*}
Proving that $G\circ F$ is conformal with scaling function $h(x)\colon=f(x)g(F(x))$.
2. The reciprocal is given by
$$
\norm{F(x)-b}^2\Vert x-a\Vert^2
=\norm{(x-a)/\Vert x-a\Vert-b\Vert x-a\Vert}^2
=1-\la x-a,b\ra+\norm b^2\Vert x-a\Vert^2
$$
3. Assume the reciprocal of the scaling function of the composition of $m$ maps $F_{m+1},\ldots,F_2$ is given by $A+\la x,v\ra+B\Vert x\Vert^2$, then the reciprocal of the scaling function of the composition of $m+1$ maps $F_{m+1},\ldots,F_2,F$, $F(x)=I(x-a)$, is given by
\begin{eqnarray*}
(A+\la I(x-a),v\ra+B\norm{I(x-a)}^2)\Vert x-a\Vert^2
&=&A\Vert x-a\Vert^2+\la x-a,v\ra+B\norm{(x-a)/\Vert x-a\Vert}^2|^2\Vert x-a\Vert^2\\
&=&A\Vert x-a\Vert^2+\la x-a,v\ra+B
\end{eqnarray*}