When studying Möbius transformations, I failed to understand the transformations involved in the following steps:
\begin{align}&\overline{(z-a, z_2-a, z_3-a, z_4-a)}\\ =& \left(\overline z-\overline a, \frac {R^2}{z_2-a}, \frac {R^2}{z_3-a},\frac {R^2}{z_4-a}\right)\\ =& \left(\frac{R^2}{\overline z-\overline a}, z_2-a, z_3-a, z_4-a\right)\end{align}
where $R$ is a constant (radius of a circle centre $a$).
With
$$\overline{z_2-a}\mapsto\frac{R^2}{z_2-a}$$
I think that Möbius transformations don't include conjugates, hence my confusion, and then why does the first element $\overline{z-a}$ not also undergo the same transformation, also in the second step?
$z_2$, $z_3$ and $z_4$ lie on the circle $\{z\ |\ |z-a|=R\}$, so $$\overline{z_2-a}=\frac{(\overline{z_2-a})(z_2-a)}{z_2-a}=\frac{R^2}{z_2-a}.$$ The second step is $$x\mapsto R^2/x.$$