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Complex Analysis and Differential Geometry




                    Notes
                                          Example:
                                   We shall find all the residues of the function

                                                                          e z
                                                                  f(z)        .
                                                                        2
                                                                          2
                                                                       z (z  1)
                                   First, observe that f has isolated singularities at 0, and ± i. Let’s see about the residue at 0.
                                   Here, we have,

                                                                             e z
                                                                     2
                                                                (z)   z f(z)   .
                                                                              2
                                                                           2
                                                                          z (z  1)
                                   The residue is simply ’(0) :
                                                                           z
                                                                       2
                                                                     (z   1)e  2ze z
                                                                '(z)            .
                                                                          2
                                                                        (z  1) 2
                                   Hence,
                                                                            
                                                                  Resf   '(0) 1.
                                                                   z 0
                                                                    
                                   Next, let’s see what we have at z = i:
                                                                              e z
                                                             f(z) = (z – i)f(z) =   ,
                                                                             2
                                                                               2
                                                                            z (z  1)
                                   and so
                                                                              e i
                                                                 Resf(z)   (i)    .
                                                                 z 0          2i
                                                                  
                                   In the same way, we see that

                                                                          e  i 
                                                                    Resf    .
                                                                     z i  2i


                                   Let’s find the integral   2   e 2 z  dz,  where C is the contour pictured:
                                                     C z (z  1)
                                                                    Figure  10.2






















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