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Complex Analysis and Differential Geometry




                    Notes
                                                                                         z 4
                                                                                   
                                                                                or (z) 
                                                                           4
                                                                     4
                                                                                         
                                                                  =  lim  z    a       z ia
                                                                                        4
                                                                                            
                                                                z ia z ia   2ia    z /(z ia)
                                                                                as f(z) 
                                                                                         (z ia)
                                                                                           
                                                               a i  ia 3
                                                                4
                                                                  =   
                                                               2a   2
                                   (b)  Find the residues of e  z  at its poles.
                                                         iz
                                                           -4
                                                   e iz
                                   Solution. Let f(z) =
                                                   z 4
                                   f(z) has pole of order 4 at z = 0, so
                                                                   iz 
                                                 Res (z = 0) =   1    d 3 3 (e )      i           | (z) = e iz
                                                            3 dz
                                                           |        z 0  6
                                                                      
                                   Alternatively, by the Laurent’s expansion
                                                 e iz  1  i   1    i   ...
                                                  z 4   z 4    z 3   | 2z 2   | 3z  

                                   we find that

                                                                  1
                                            Res (z = 0) = co-efficient of
                                                                  z
                                                        i
                                                     = –
                                                        6

                                                         z 3
                                   (c)  Find the residue of   at  z = .
                                                        2
                                                        z  1
                                                                    
                                   Solution. Let f(z) =   z 3    z 3    z 1   1   1
                                                                    
                                                                         2 
                                                    2
                                                   z  1  2   1      z 
                                                         z  1   2 
                                                             z 
                                                          1  1
                                                      =  z(1      ....)
                                                         z 2  z 4
                                                        1  1
                                                      =  z      ...
                                                        z  z 3
                                          Therefore,

                                                                         1
                                                 Res (z = ) = –(co-efficient of  )   1
                                                                         z
                                   (d)  Find the residues of at its poles.

                                                         z 3
                                   Solution. Let f(z) =
                                                   (z 1) (z 2)(z 3)
                                                       4
                                                          
                                                               
                                                     
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