Using the Cauchy integral formula show that ∮ | z | = 2 e z d z ( z − 1 ) 2 ( z − 3 ) = − 3 2 j e π .

Question

Using the Cauchy integral formula show that      ∮                  |            z              |            =      2                                    e          z                d        z                    (        z        −        1                  )          2                (        z        −        3        )              =  −      3    2    j  e  π  .

Rhett Pruitt · Accepted Answer

You have      1          (      z      −      1              )        2            (      z      −      3      )        =  −      1    4        1          z      −      1        −      1    2        1          (      z      −      1              )        2              +      1    4        1          z      −      3      Now, you can use Cauchy&#039;s formula directly omitting the on       |    z      |    ≤  2 holomorphic part belonging to       1          z      −      3      :                              ∮                                    |                        z                          |                        =            2                                                e            z                                (            z            −            1                          )              2                        (            z            −            3            )                          d        z                            =                    −                  1          4                          ∮                                    |                        z                          |                        =            2                                                e            z                                z            −            1                          d        z        −                  1          2                          ∮                                    |                        z                          |                        =            2                                                e            z                                (            z            −            1                          )              2                                      d        z                                          =                    −                  1          4                ⋅        2        π        i        ⋅                  e          1                −                  1          2                ⋅        2        π        i        ⋅                  e          1                                                  =                    −                  3          2                π        i        e

Using the Cauchy integral formula show that ∮ <mrow class="MJX-TeXAtom

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