📂Complex Anaylsis

Cross Ratio in Complex Analysis

Definition ¹

For four distinct points $ z_{1} , z_{2} , z_{3} , z_{4} \in \overline{ \mathbb{C} }$ on the extended complex plane, the following is defined as the cross ratio. $$ (z_{1} , z_{2} , z_{3} , z_{4} ) = {{( z_{1} - z_{4})( z_{3} - z_{2})} \over {(z_{1} - z_{2}) ( z_{3} - z_{4}) } } $$

Explanation

Rewriting it slightly as $\displaystyle (z_{1} , z_{2} , z_{3} , z ) = {{( z_{3} - z_{2}) } \over {(z_{1} - z_{2})} } \cdot {{ ( z - z_{1}) } \over { ( z - z_{3}) } }$, we find that $$ (z_{1} , z_{2} , z_{3} , z_{1} ) = 0 \\ (z_{1} , z_{2} , z_{3} , z_{2} ) = 1 \\ (z_{1} , z_{2} , z_{3} , z_{3} ) = \infty $$ holds, and since at least three points appear, one can readily guess that it is useful for dealing with circles or lines.

The key here is the following property.

Theorem

The cross ratio is invariant under bilinear transformations.

Proof

Setting the bilinear transformation $f$ as $w_{k} = f ( z_{k} )$ and the cross ratio as $g(z) = (z_{1} , z_{2} , z_{3} , z )$, then $$ g ( f^{-1} (w_{1}) ) = g (z_{1}) = 0 \\ g ( f^{-1} (w_{2}) ) = g (z_{2}) = 1 \\ g ( f^{-1} (w_{3}) ) = g (z_{3}) = \infty $$ that is, $g \circ f^{-1}$ becomes the cross ratio for $w_{1} , w_{2} , w_{3} , w_{4} \in \overline{ \mathbb{C} }$. Therefore $$ ( z_{1} , z_{2} , z_{3} , z_{4} ) = g(z_{4}) = g ( f^{-1} (w_{4} ) ) = ( w_{1} , w_{2} , w_{3} , w_{4} ) $$

■