= Solution
<Cross-ratio> conventions differ by permutations. Use the following ordering, which gives the signs requested in part (c):
$$
\boxed{r=[z_1,z_2,z_3,z_4]
=\frac{(z_1-z_4)(z_2-z_3)}{(z_1-z_2)(z_4-z_3)}.}
$$
Values involving infinity are defined by limits. This is the <cross-ratio with second point mapped to zero>: the <Möbius transformation>
$$
T(z)=\frac{(z_1-z_4)(z-z_2)}{(z_1-z_2)(z-z_4)}
$$
sends $z_1,z_2,z_4$ to $1,0,\infty$ and sends $z_3$ to $r$.
For $M(z)=(az+b)/(cz+d)$, $ad-bc\ne0$, the identity
$$
M(u)-M(v)=\frac{(ad-bc)(u-v)}{(cu+d)(cv+d)}
$$
shows that all factors cancel in the cross-ratio. This proves <Möbius invariance of the cross-ratio>, with the cases involving infinity following by limits. Algebraic subtraction also gives
$$
\boxed{[z_2,z_1,z_3,z_4]=1-r
=\frac{(z_1-z_3)(z_2-z_4)}{(z_1-z_2)(z_3-z_4)}.}
$$
The latter expression is another common ordering for the cross-ratio; keeping the convention explicit prevents an incorrect reciprocal or sign.
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