Real and imaginary parts of the Möbius transformation

Question

Given that the Möbius transformation is:

$f(z) = \dfrac{az+b}{cz+d} ,\, (a d-b c) \neq 0$ and with $a,b,c$ and $d$ complex numbers written $a= a_1 + a_2i$ etc.

I think I must be missing something because when separating the Möbius transformation in to its real and imaginary parts I got this:

$\dfrac{(a_1c_1+a_2c_2)z^2 + (b_1c_1+b_2c_2)z +(a_1d_1+a_2d_2)z +(b_1d_1+b_2d_2)}{|c|^2z^2 + |d|^2}$ for the real part... and

$\dfrac{(a_1c_1-a_2c_2)z^2 + (b_1c_1-b_2c_2)z +(a_1d_1-a_2d_2)z +(b_1d_1-b_2d_2)}{|c|^2z^2 + |d|^2}i$ as the imaginary.

It really looks awful, is there a better way to write this?

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Update. Duh. Dumb mistake, reworking them I get this, the point is it's still ugly.

Real: $\dfrac{(a_1x-a_2y+b_1)(c_1x-c_2y+d_1) -(a_2x+a_1y+b_2)(c_2x+c_1y+d_2)}{(c_1x-c_2y+d_1)^2 - (c_2x+c_1y+d_2)^2}$

$\dfrac{R(ac)x^2 - 2(a_1c_2 + a_2c_1)xy + (R(ad) + R(bc))x - (R(ab) + R(bc))y - R(ac)y^2 + R(bd)}{(c_1x-c_2y+d_1)^2 - (c_2x+c_1y+d_2)^2}$

Imag: $\dfrac{(a_2x-a_1y+b_2)(c_1x-c_2y+d_1) -(a_1x+a_2y+b_1)(c_2x+c_1y+d_2)}{(c_1x-c_2y+d_1)^2 - (c_2x+c_1y+d_2)^2}$

Answer 1

Hmm, after trying on my own, what I'm getting is (where $z=x+iy$, as usual):

$$\Re\left(\frac{az+b}{cz+d}\right)=\frac{\mu\,\Re(a)+\nu\,\Im(a)+\xi\,\Re(b)+\eta\,\Im(b)}{\Delta}$$

$$\Im\left(\frac{az+b}{cz+d}\right)=\frac{\mu\,\Im(a)-\nu\,\Re(a)+\xi\,\Im(b)-\eta\,\Re(b)}{\Delta}$$

where

$\mu=|z|^2\Re(c)+\alpha$

$\nu=|z|^2\Im(c)+\beta$

$\xi=x\Re(c)-y\Im(c)+\Re(d)$

$\eta=x\Im(c)+y\Re(c)+\Im(d)$

$\Delta=|z|^2 |c|^2+|d|^2+2\alpha\,\Re(c)+2\beta\,\Im(c)$

and

$\alpha=x\Re(d)+y\Im(d)$

$\beta=x\Im(d)-y\Re(d)$

The usual admonition in programming to "isolate common subexpressions" is a useful way to deal with complexity.

Answer 2

This is not the correct splitting: mind that $z$ is complex too, so you have to write $z=x+iy$, if you want to represent $f(z)=u(x,y)+iv(x,y)$. It will get a lot worse...

Answer 3

Edit: Sorry - this only holds for $a,b,c,d \in \mathbb{R}$!

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Thanks to some help in a question I asked:

\begin{align*} \textrm{Re} (f(z)) &=\textrm{Re} \left(\frac{az+b}{cz+d}\right) \\ &=\textrm{Re} \left( \frac{(az+b)(c \overline{z}+d)}{(cz+d)(c \overline{z}+d)}\right) \\ &=\frac{1}{|cz+d|^2} \textrm{Re} \left( acz\overline{z}+bc\overline{z}+adz+bd\right) \\ &= \frac{\left(ac|z|^2+bd\right)+(ad+bc)\textrm{Re}(z)}{|cz+d|^2} \end{align*} and \begin{align*} \textrm{Im} (f(z)) &=\textrm{Im} \left(\frac{az+b}{cz+d}\right) \\ &=\textrm{Im} \left( \frac{(az+b)(c \overline{z}+d)}{(cz+d)(c \overline{z}+d)}\right) \\ &=\frac{1}{|cz+d|^2} \textrm{Im} \left( acz\overline{z}+bc\overline{z}+adz+bd\right) \\ &= \frac{(ad-bc)}{|cz+d|^2} \textrm{Im}(z) \end{align*}

Answer 4

So this is what I've got.

Given any complex number $z=x+iy$, the general Möbius transformation

\begin{equation} M(z)=\frac{az+b}{cz+d} \end{equation}

where $a, b, c, d$ are complex numbers such that $a_1+ia_2, b_1+ib_2$ etc., can be alternatively expressed as follows:

\begin{equation} M(z)=\frac{\alpha\gamma + \beta\delta}{\gamma^2+\delta^2} + i\frac{\beta\gamma - \alpha\delta}{\gamma^2+\delta^2} \end{equation}

where $\alpha, \beta, \gamma, \delta$ are real numbers such that

$$ \begin{align*} \alpha &= a_1x - a_2y + b_1 \\ \beta &= a_1y + a_2x + b_2 \\ \gamma &= c_1x - c_2y + d_1 \\ \delta &= c_1y + c_2x + d_2 \\ \end{align*} $$

I got this simply by expressing every complex number in the general transformation formula in Cartesian form, multiplying out, rearranging and multiplying everything by the complex conjugate of the denominator over itself.

It is exactly what you had, just expressed in a neater form: for this reason, it could be useful in programming. In fact, with it I managed to teach a basic online graphing utility like Desmos calculator to display Möbius transformations—you can view the graph here.