Assume that $z_1$, $z_2$, $z_3$, and $z_4$ are all distinct complex numbers and $f(z)=\frac{az+b}{cz+d}$ is defined to be a Möbius transformation. If $f(z_1)=0$, $f(z_2)=1$ and $f(z_3)=∞$, then show that $f(z_4)=(z_1,z_2;z_3,z_4)$.
I have tried using two different $S_4$ permutations of cross ratios: $(z_1,z_2;z_3,z_4)=\frac{(z_1-z_3)(z_2-z_4)}{(z_1-z_4)(z_2-z_3)} \rightarrow f(z_4)=\frac{1}{1-\lambda}$, $(z_1,z_2;z_3,z_4)=\frac{(z_1-z_2)(z_3-z_4)}{(z_1-z_4)(z_3-z_2)} \rightarrow f(z_4)=\frac{1}{λ}$,
where $\lambda = (z_1,z_2;z_3,z_4)$.
Why is $f(z_4)$ not equal to $\lambda$?
Also, there seems to be more than one definition of $\lambda$ on the web, one for the reals and another for the complex, which I don't know why BTW.
I'm starting to think that maybe there's something wrong with the question itself, as appeared in exercise 2.10. Treisman 2009. A typo maybe? Or am I missing something subtle here?
Consult the Wikipedia page, for instance. "Möbius transformation - Wikipedia" https://en.m.wikipedia.org/wiki/M%C3%B6bius_transformation
That cross-ratios are invariant under Mobius transformations doesn't imply that $f(z_4)=\lambda$.
Furthermore, the definition of cross-ratio that I found there was the first one.