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Unit 4: Integration




          Next, let’s integrate along the straight line segment C  joining 0 and 1 + i.         Notes
                                                     2

















          Here we have  (t) = t + it, 0  t  1. Thus,   ' 2 (t)  = 1 + i, and our integral looks like
                      2
                              1
                              
                       f(z)dz  = f( (t)) (t)dt 2   ' 2
                     C  2     0
                              1
                                        2
                                 2
                              
                            = [(t  t) it ](1 i) dt
                                     
                                           
                              0
                              1
                              
                            = [t i(t 2t )]dt    2
                              0
                              1  7
                            =     i
                              2  6
          Finally, let’s integrate along C , the path consisting of the line segment from 0 to 1 together with
                                  3
          the segment from 1 to 1 + i.


















          We shall do this in two parts: C , the line from 0 to 1 ; and C , the line from 1 to 1 + i. Then we
                                                           32
                                   31
          have
                                       f(z)dz     f(z)dz     f(z)dz.
                                    C  3     C  31   C  32











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