# 交比

${\displaystyle (z_{1},z_{2};z_{3},z_{4})={\frac {(z_{1}-z_{3})(z_{2}-z_{4})}{(z_{1}-z_{4})(z_{2}-z_{3})}}}$

## 对称

${\displaystyle (z_{1},z_{2};z_{3},z_{4})=(z_{2},z_{1};z_{4},z_{3})=(z_{3},z_{4};z_{1},z_{2})=(z_{4},z_{3};z_{2},z_{1})\,}$

 ${\displaystyle (z_{1},z_{2};z_{3},z_{4})=\lambda \,}$ ${\displaystyle (z_{1},z_{2};z_{4},z_{3})={1 \over \lambda }}$ ${\displaystyle (z_{1},z_{3};z_{4},z_{2})={1 \over {1-\lambda }}}$ ${\displaystyle (z_{1},z_{3};z_{2},z_{4})=1-\lambda \,}$ ${\displaystyle (z_{1},z_{4};z_{3},z_{2})={\lambda \over {\lambda -1}}}$ ${\displaystyle (z_{1},z_{4};z_{2},z_{3})={{\lambda -1} \over \lambda }}$

${\displaystyle (z,z_{2};z,z_{4})=(z_{1},z;z_{3},z)=0\,}$
${\displaystyle (z,z;z_{3},z_{4})=(z_{1},z_{2};z,z)=1\,}$
${\displaystyle (z,z_{2};z_{3},z)=(z_{1},z;z,z_{4})=\infty \,}$

## 从变换出发

${\displaystyle f(z)={\frac {az+b}{cz+d}}\;,\quad ad-bc\neq 0}$

${\displaystyle (f(z_{1}),f(z_{2});f(z_{3}),f(z_{4}))=(z_{1},z_{2};z_{3},z_{4})\,}$

${\displaystyle (z,1;0,\infty )=\lim _{w\to \infty }{\frac {z(1-w)}{z-w}}=z\,}$

${\displaystyle z_{2}\to 1,\;z_{3}\to 0,\;z_{4}\to \infty }$

${\displaystyle z_{1}}$就被映射到${\displaystyle (z_{1},z_{2};z_{3},z_{4})=f(z_{1})}$。换个角度看，若把交比看为${\displaystyle z_{1}}$的函数，交比是唯一的变换把点${\displaystyle (z_{2},z_{3},z_{4})}$映射到${\displaystyle (1,0,\infty )}$