If $z_{1}, z_{2}, z_{3}$ are distinct points in $C \cup \infty$ then there is only one Möbius Transformation which sends these points to $1,0,\infty$ Write out the formula for this Möbius Transformation. You will have to consider the cases when one of the $z_{j}'s $ is $\infty$
The first part of this question I have worked out. The second part is what I have trouble with. I think this has to do with the cross ratio in projective geometry. If $A,B,C,D$ are points that are collinear. then their cross ratio is
$$(A,B,C,D) = \frac{\bar{AC} * \bar{BD}}{\bar{AD} * \bar{BC}} .$$ If $A,B,C,D$ were points on the real numbers then the ratio would be $$(A,B,C,D) = \frac{(C-A) * (D-B)}{(D-A) * (C-B)} .$$
Therefore I think the cross ratio of the three points would be
$$(z_{1}, z_{2} , z_{3}, z_{4}) = \frac{(z_{1} - z_{3})(z_{2} - z_{4})}{(z_1- z_{4}) (z_{3} - z_{2})}$$ The last formula is sort of what I want but the problem is I don't understand the part where we have to let one of the z_{j}'s equal to infinity. What do you think? Thanks for the help.
For the first part of the question, if all of the $z_i$ are in $\mathbb{C}$, you should have found that
$\displaystyle f(z)= \frac{(z-z_2)(z_1-z_3)}{(z-z_3)(z_1-z_2)}$
maps $z_1$ to $1$, $z_2$ to $0$ and $z_3$ to $\infty$. To see what happens if one of the $z_i$ is at $\infty$, consider what happens as it approaches $\infty$ and take the limit.
For example, as $z_1 \rightarrow \infty$, then $\frac{z_1-z_3}{z_1-z_2} \rightarrow 1$, so
$\displaystyle\lim_{z_1 \rightarrow \infty} f(z) = \frac{z-z_2}{z-z_3}$
and this function now maps $\infty$ to $1$, $z_2$ to $0$ and $z_3$ to $\infty$ as required.