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Complex Analysis and Differential Geometry




                    Notes                            a            d
                                   and           T   = ,  T  () =   c  , if c  0
                                                             1
                                                 1
                                                   
                                                    c 
                                   From the above discussion, we conclude that inverse of a bilinear transformation is bilinear.
                                                d                    a
                                   The points z =   (w = ) and z = (w =   ) are called critical points.
                                                c                    c
                                   Theorem


                                   Composition (or resultant or product) of two bilinear transformations is a bilinear transformation.
                                   Proof. We consider the bilinear transformations

                                                        az b
                                                          
                                                    w =   cz d  ,  ad  bc  0                             …(1)
                                                          
                                                        a w b
                                                            
                                   and              w =   c w d 1 1  ,  a d   b c   0                    …(2)
                                                         1
                                                     1
                                                                    1
                                                                        1 1
                                                                   1
                                                            
                                                         1
                                   Putting the value of w from (1) in (2), we get
                                                             
                                                        a 1    az b    b 1
                                                                                    
                                                                          
                                                                               
                                                             
                                                    w =     cz d     (a a b c)z (b d a b)
                                                                                      1
                                                                        1
                                                                                  1
                                                                            1
                                                             
                                                     1
                                                                               
                                                                          
                                                                                    
                                                        c 1     az b    d 1  (c a d c)z (d d c b)
                                                                            1
                                                                        1
                                                                                  1
                                                                                      1
                                                           cz d 
                                                             
                                   Taking            A = a a + b c,  B = b d + a b,
                                                             1
                                                                          1
                                                         1
                                                                     1
                                                     C = c a + d c,  D = d d + c b, we get
                                                         1
                                                                          1
                                                                     1
                                                             1
                                                        Az B
                                                           
                                                    w =   Cz D
                                                           
                                                     1
                                   Also        AD  BC = (a a + b c) (d d + c b)  (b d + a b) (c a + d c)
                                                         1
                                                                  1
                                                                                         1
                                                                                     1
                                                                      1
                                                                            1
                                                              1
                                                                                 1
                                                      = (a ad d + a ac b + b cd d + b cc b)  (b dc a + b d d c + a bc a + a bd c)
                                                                       1
                                                                              1
                                                                         1
                                                            1
                                                                                                             1
                                                                                                          1
                                                                                        1
                                                                                            1
                                                                                                1
                                                         1
                                                                                      1
                                                                   1
                                                                                                    1
                                                                                                      1
                                                                1
                                                                                1
                                                      = a ad d + b bc c  b dc a  a bd c
                                                                              1
                                                                                1
                                                                  1
                                                                       1
                                                                1
                                                         1
                                                           1
                                                                         1
                                                      = ad(a d   b c )  bc(a d   b c )
                                                           1
                                                                          1
                                                             1
                                                                        1
                                                                              1 1
                                                                 1 1
                                                      = (ad  bc) (a d   b c )  0
                                                                   1
                                                                 1
                                                                      1 1
                                                        Az B
                                                           
                                   Thus             w =   Cz D  ,  AD  BC  0
                                                     1
                                                           
                                   is a bilinear transformation.
                                   This bilinear transformation is called the resultant (or product or composition) of the bilinear
                                   transformations (1) and (2).
                                   The above property is also expressed by saying that bilinear transformations form a group.
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