Linear Transformations mapping four points Problem: Show that any four distinct points can be carried by a linear transformation to positions $1, -1, k, -k$, where the value of $k$ depends on the points. How many solutions are there, and how are they related?
Attempt at a solution:
So I know that given any three points $z_2, z_3, z_4$ I can find a linear transformation that carries these points to some other points $w_2, w_3, w_4$ this can be done since the ratio is preserved; that is $$(w, w_2, w_3, w_4)=(z, z_2, z_3,z_4).---(1)$$
So, in our case we have $w_2=-1, w_3=k$ and $w_4=-k$, and we want this transformation to be such that it will also take $z_1$ to $1$. 
After performing the necessary algebraic steps in $(1)$ I found 
$$w(z)=\frac{(1+k)(z-z_3)(z_2-z_4)+(1-k)(z-z_4)(z_2-z_3)}{(1-k)(z-z_4)(z_2-z_3)-(1+k)(z-z_3)(z_2-z_4)}k$$
This transformation takes the $z_2$ to $-1$, $z_3$ to $k$ and $z_4$ to $-k.$
So now, do I just force the transformation so that $w(z_1)=1$?
I am not sure how the whole $k$ being dependent on the points we choose is coming into play?
Any hints??
Thanks!
 A: Starting from scratch, let's consider the cross-ratio
$$(z_1,z_2;z_3,z_4) = \frac{z_1-z_3}{z_2-z_3}\frac{z_2-z_4}{z_1-z_4}$$
The map 
$$f(z)=\frac{z_1-z_3}{z_2-z_3}\frac{z_2-z }{z_1-z }$$
sends $z_1$ to $\infty$, $z_2$ to $0$,  $z_3$ to $1$, and $z_4$ to $(z_1,z_2;z_3,z_4)$. Therefore, two quadruples of points with the same cross-ratio  are related by a Möbius map. The converse is also true, because cross-ratio is preserved by $z\mapsto az$, $z\mapsto z+b$, and $z\mapsto 1/z$, which generate all Möbius maps. 
It remains to count the quadruples $1,-1;k,-k$ with given cross-ratio $r\in\mathbb C\setminus \{0,1\}$. We have
$$( 1,-1;k,-k) = \frac{1-k}{-1-k}\frac{-1+k}{1+k} = \left(\frac{1-k}{1+k}\right)^2$$
Here $k\mapsto \frac{1-k}{1+k}$ is a bijection from $\mathbb C\setminus \{-1,0,1\}$ onto $\mathbb C\setminus \{-1,0,1\}$. Therefore, $k\mapsto ( 1,-1;k,-k)$ is a 2-to-1 map from $\mathbb C\setminus \{-1,0,1\}$ onto $\mathbb C\setminus \{0,1\}$. Conclusion: there are precisely two suitable values of $k$ for every quadruple. (They are reciprocals of each other).
