Combining Moebius transformations

Question

Moebius transformation in this case $\frac{az+b}{cz+d}$ for complex $z$.

I have several transformations I want to apply to an initial $z$.

For example first transform $f(a,b,z) = z + (a + bi) = \frac{1z + (a + bi)}{0z + 1}$ for given $a,b,z$.

Second transform is $g(a,b,z) = z\cdot(a - bi) = \frac{(a - bi)\cdot z + 0}{0z + 1}$.

Third is $h(z) = \frac{1}{z'} = \frac{0z' + 1}{1z' + 0}$ where z' is complex conjugate.

How to correctly compose $f$, $g$, and $h$ into one transformation? I post this example but I'm after general case for any number of transformations. To my understanding all compositions are of form $\frac{az+b}{cz+d}$.

Answer 1

There is no reason to write $w=a+bi$ in terms of its real and imaginary parts $a,b$. As such, there is good reason not to: introducing superfluous letters clutters up expressions.

Given a matrix $A=[\begin{smallmatrix}a&b\\c&d\end{smallmatrix}]$, we can define $Az:=\frac{az+b}{cz+d}$. If we use invertible matrices, this defines a group action of ${\rm GL}_2(\Bbb C)$ on the Riemann sphere $\widehat{\Bbb C}$. This means $(AB)z=A(Bz)$ (in this notation) where $AB$ is the usual matrix product. (Moreover, one may readily check $\overline{A}\overline{z}=\overline{Az}$.)

You have three transformations:

• $f_w(z)=z+w=\begin{bmatrix}1 & w \\ 0 & 1\end{bmatrix}z$
• $g_w(z)=\overline{w}z=\begin{bmatrix}\overline{w} & 0 \\ 0 & 1\end{bmatrix}z$
• $h(z)=\overline{z}^{-1}=\begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}\overline{z}$

Therefore:

$$(f_w\circ g_w\circ h)(z)=\begin{bmatrix}1 & w \\ 0 & 1\end{bmatrix} \begin{bmatrix}\overline{w} & 0 \\ 0 & 1\end{bmatrix} \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}\overline{z}. $$

I will let you compute the matrix product.