I am trying to make sense of the following:
A mapping of the form $S(z)=\frac{az+b}{cz+d}$ is called a linear fractional transformation.
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3.6 Proposition. If $S$ is a Möbius transformation then $S$ is the composition of translations, dilations, and the inversion.
Proof.
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Now let $c\neq 0$ and put $S_1(z)=z+d/c,\ S_2(z)=1/z,\ S_3(z)=\frac{bc-ad}{c^2}z,\ S_4(z)=z+a/c$. Then $S=S_4\circ S_3\circ S_2\circ S_1$.
I just can't verify $S=S_4\circ S_3\circ S_2\circ S_1$, so I'm not convinced.
$\begin{align}&S_4\circ S_3\left(\frac 1{z+d/c}\right)=S_4\left(\frac{c^2}{(bc-ad)z+cd} \right) = \frac{c^2}{(bc-ad)z+cd}+\frac a c\\ &=\frac{c^3}{c\left((bc-ad)z+cd\right)}+\frac{a((bc-ad)z+cd)}{c((bc-ad)z+cd)}\\ &=\frac{c^3+abcz-a^2dx+acd}{bc^2z-acdz+c^2d}\end{align}$
From which I have to get $\frac{az+b}{cz+d}$. In high school, questions involving simplification of fractions would have let me be able to take out a common factor of $(az+b)$ in the numerator, $(cz+d)$ in the denominator and there would be something left to cancel on top and bottom. Should I be approaching this challenge differently?
Reference: John Conway, Functions of One Complex Variable I, 2nd ed., page 47.
From астон вілла олоф мэллбэрг's suggestion, \begin{align}&S_4\circ S_3\left(\frac 1{z+d/c}\right)=S_4\left(\frac{bc-ad}{c^2z+cd} \right) = \frac{bc-ad}{c^2z+cd}+\frac ac = \frac{bc-ad+a(cz+d)}{c^2z+cd}\\ &= \frac{c(az+b)}{c(cz+d)}=\frac{az+b}{cz+d}.\end{align}