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712    CHAPTER 20  Complex Integration


                                                                   y


                                                                    4i
                                                                        2


                                                                                     x
                                                               –2
                                                                1



                                                        FIGURE 20.12  , γ 1 , and γ 2 in Exam-
                                                        ple 20.15.


                                                  z                   z                   z

                                                         dz =                dz +                dz.
                                             (z + 2)(z − 4i)   γ 1  (z + 2)(z − 4i)  γ 2  (z + 2)(z − 4i)
                                 Use a partial fractions decomposition to write
                                                          z        1 − 2i  1   4 + 2i  1
                                                                =            +           .
                                                    (z + 2)(z − 4i)  5  z + 2    5  z − 4i
                                 Putting the last two equations together with Cauchy’s theorem and the conclusion of Exam-
                                 ple 20.11, we have

                                                    z          1 − 2i    1         4 + 2i  1
                                                          dz =              dz +              dz
                                               (z + 2)(z − 4i)   5   γ 1  z + 2  γ 1  5  z − 4i
                                                                 1 − 2i     1     4 + 2i     1
                                                               +              dz +              dz
                                                                   5     z + 2      5      z − 4i
                                                                       γ 2               γ 2
                                                               1 − 2i      4 + 2i
                                                             =      (2πi) +     (2πi)
                                                                 5           5
                                                             =2πi.

                                    A proof of the extended deformation theorem can be modeled after that of the deformation
                                 theorem, except now we draw a line segment from   to γ 1 , from γ 1 to γ 2 and so on until we come
                                 to γ n−1 to γ n , and finally from γ n to  , as in Figure 20.13 for n = 3.


                                                          y
                                                                             L 4


                                                                                3
                                                                  1      L 3
                                                     L 1

                                                                L 2          2

                                                                                         x



                                                    FIGURE 20.13 Argument for the extended deforma-
                                                    tion theorem.



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