Proving that the cross ratio is a Möbius transformation I'm trying to show that given three distinct points $z_1,z_2,z_3\in\mathbb C$, the rational function
$$
f(z) = \frac{(z-z_1)(z_2 - z_3)}{(z - z_3)(z_2 - z_1)} = \frac{(z_2 - z_3)z + (z_1z_3 - z_1z_2)}{(z_2 - z_1)z + (z_1z_3 - z_2z_3)} = \frac{az + b}{cz + d}
$$
is a Möbius transformation. That is, I must show that $ad - bc \neq 0$. I worked out that
        \begin{align*}
  ad - bc & = (z_1z_2z_3 - z_2^2z_3 - z_1z_3^2 + z_2z_3^2) - (z_1z_2z_3 - z_1^2z_3 - z_1z_2^2 + z_1^2z_2) \\
  & = - z_2^2z_3 - z_1z_3^2 + z_2z_3^2 + z_1^2z_3 + z_1z_2^2 - z_1^2z_2 \\
  & = (z_2z_3^2 - z_2^2z_3) + (z_1^2z_3 - z_1z_3^2) + (z_1z_2^2 - z_1^2z_2) \\
  & = z_1^2(z_3 - z_2) + z_2^2(z_1 - z_3) + z_3^2(z_2 - z_1),
  \end{align*}
but I'm unsure of where to go from here to show that this expression is nonzero.
 A: To show that your final expression is non-zero, note that
\begin{align*}
(z_3 - z_2)(z_3 - z_1)(z_2 - z_1) &= (z_3^2 - z_1z_3 - z_2z_3 + z_1z_2)(z_2 - z_1)\\
&= z_2z_3^2 - z_1z_2z_3 - z_2^2z_3 + z_1z_2^2 - z_1z_3^2 + z_1^2z_3 + z_1z_2z_3 - z_1^2z_2\\
&= z_2z_3^2 - z_2^2z_3 + z_1z_2^2 - z_1z_3^2 + z_1^2z_3 - z_1^2z_2\\
&= z_1^2(z_3 - z_2) + z_2^2(z_1 - z_3) + z_3^2(z_2 - z_1).
\end{align*}
As $z_1, z_2, z_3$ are all distinct, the expression is non-zero.

Here is an alternative approach to the initial problem.
If $g(z) = \frac{az + b}{cz + d}$ is a Möbius transformation ($ad - bc \neq 0$), then $f(z) = kg(z) = \frac{kaz + kb}{cz + d}$ is also a Möbius transformation for $k \neq 0$; note that $(ka)d - (kb)c = k(ad - bc) \neq 0$. 
The cross-ratio can be written as
$$f(z) = \frac{(z - z_1)(z_2 - z_3)}{(z - z_3)(z_2 - z_1)} = \left(\frac{z_2 - z_3}{z_2 - z_1}\right)\frac{z - z_1}{z - z_3} = kg(z)$$
where $k = \frac{z_2 - z_3}{z_2 - z_1}$ and $g(z) = \frac{z - z_1}{z - z_3}$. As $g$ is a Möbius transformation ($z_1 - z_3 \neq 0$), and $k \neq 0$, $f$ is a Möbius transformation by the above.
A: We can factor $z_3$ from the expression for $d$ and $z_1$ from the expression for $c$ and then group:
\begin{align}
ad-bc
&= (z_2 - z_3) (z_1 z_3 - z_2 z_3) - (z_2 - z_1) (z_1 z_3 - z_1 z_2) \\
&= (z_1 - z_2) (z_2 - z_3) z_3 - (z_1 - z_2) (z_2 - z_3) z_1 \\
&= (z_1 - z_2) (z_2 - z_3) (z_3 - z_1) .
\end{align}
Since $z_1, z_2, z_3$ are distinct, $ad - bc$ is a product of three nonzero expressions and hence nonzero.
