Real and imaginary parts of the Möbius transformation 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?

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

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

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*}
