Antiderivative 1/z on C Let Ω ⊆ C be open and γ : [ α , β ] → Ω be a piecewise continuously differentiable and closed path.Why does z − 1 have no antiderivative on C ∖ 0?Why is ∫ γ z − 1 d z = 2 π i ⋅ ind γ ( 0 ), where ind γ denotes the winding number.

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Antiderivative 1/z on       C  Let   Ω  ⊆      C   be open and   γ  :  [  α  ,  β  ]  →  Ω be a piecewise continuously differentiable and closed path.Why does       z          −      1       have no antiderivative on       C    ∖  0?Why is       ∫    γ        z          −      1        d  z  =  2  π  i  ⋅      ind    γ    (  0  ), where       ind    γ   denotes the winding number.

Louis Lawrence · Accepted Answer

Step 1The last line of the question does not feel entirely valid, as it basically questions the definition of the winding number. Here is how the classical argument about the direct computation goes. Suppose   z  =  z  (  t  ) is the equation of   γ on   [  α  ,  β  ]. Consider the function:                    (1)                    h        (        t        )        =                  ∫                      α                    t                                                    z              ′                        (            t            )                                z            (            t            )            −            a                          d        t            Then       h    ′    (  t  )  =                    z        ′            (      t      )              z      (      t      )      −      a      Step 2Now             (              e                  −          h                      (            t            )                                      (        z                  (          t          )                −        a        )            )        ′    =  −      h    ′        (    t    )        e          −      h              (        t        )                  (    z          (      t      )        −    a    )    +      e          −      h              (        t        )                  z    ′        (    t    )    =      e          −      h              (        t        )                  (          z      ′              (      t      )        −          h      ′              (      t      )              (      z              (        t        )            −      a      )        )    ≡  0Hence,       e          −      h              (        t        )                  (    z          (      t      )        −    a    )   reduces to a constant. By (1)   h  (  α  )  =  0    ⟹        e          h      (      α      )        =  1 and we can write      e          h      (      β      )        =            z      (      t      )      −      α              z      (      α      )      −      a      Since   z  (  β  )  =  z  (  α  ) we obtain       e          h      (      β      )        =  1 and so   h  (  β  ) must be a multiple of   2  π  i

Jordyn Calhoun · Accepted Answer

Step 1Suppose, for the sake of contradiction, that 1/z has an antiderivative F on       C    −  0. Then if   γ  :  [  a  ,  b  ]  →      C    −  0 is a path, the integral over this path is just   F  (  γ  (  a  )  )  −  F  (  γ  (  b  )  ), by the fundamental theorem of calculus.Step 2Now consider a closed loop that winds around 0. The integral should be   F  (  γ  (  a  )  )  −  F  (  γ  (  a  )  )  =  0 because the starting and ending points are the same. But if you directly compute the integral of 1/z over the unit circle, you get   2  π  i. This is a contradiction, and there is no antiderivative.Edit: I see you have edited your question, so I will edit my answer. Your second equation, about winding numbers, is true because of the generalized residue theorem. You can find this in any good book on complex analysis.

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