Substitution in complex integral and the Argument Principle.












2












$begingroup$


Let's say $C$ is a simple closed curve in the complex plane and $f(z)$ is holomorphic and doesn't vanish on $C$.



According to wikipedia, one can make the following change of variables:
$$omega = fleft( z right)$$



and get:
$$ointlimits_C {{{f'left( z right)} over {fleft( z right)}}dz} = ointlimits_{fleft( C right)} {{1 over omega }domega } $$



I would like to inquire about what theorem was used to justify this change.



I've searched online and I've found several mentions of the change of variables theorem for complex integrals but in all of them (for example here) it was required that $omega = fleft( z right)$ will be a biholorphic mapping.



So my question is, how can the change of variables be rigorously justified?






complex-analysis contour-integration substitution change-of-variable







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    Let's say $C$ is a simple closed curve in the complex plane and $f(z)$ is holomorphic and doesn't vanish on $C$.



    According to wikipedia, one can make the following change of variables:
    $$omega = fleft( z right)$$



    and get:
    $$ointlimits_C {{{f'left( z right)} over {fleft( z right)}}dz} = ointlimits_{fleft( C right)} {{1 over omega }domega } $$



    I would like to inquire about what theorem was used to justify this change.



    I've searched online and I've found several mentions of the change of variables theorem for complex integrals but in all of them (for example here) it was required that $omega = fleft( z right)$ will be a biholorphic mapping.



    So my question is, how can the change of variables be rigorously justified?






    complex-analysis contour-integration substitution change-of-variable







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      Let's say $C$ is a simple closed curve in the complex plane and $f(z)$ is holomorphic and doesn't vanish on $C$.



      According to wikipedia, one can make the following change of variables:
      $$omega = fleft( z right)$$



      and get:
      $$ointlimits_C {{{f'left( z right)} over {fleft( z right)}}dz} = ointlimits_{fleft( C right)} {{1 over omega }domega } $$



      I would like to inquire about what theorem was used to justify this change.



      I've searched online and I've found several mentions of the change of variables theorem for complex integrals but in all of them (for example here) it was required that $omega = fleft( z right)$ will be a biholorphic mapping.



      So my question is, how can the change of variables be rigorously justified?






      complex-analysis contour-integration substitution change-of-variable







      share|cite|improve this question











      $endgroup$




      Let's say $C$ is a simple closed curve in the complex plane and $f(z)$ is holomorphic and doesn't vanish on $C$.



      According to wikipedia, one can make the following change of variables:
      $$omega = fleft( z right)$$



      and get:
      $$ointlimits_C {{{f'left( z right)} over {fleft( z right)}}dz} = ointlimits_{fleft( C right)} {{1 over omega }domega } $$



      I would like to inquire about what theorem was used to justify this change.



      I've searched online and I've found several mentions of the change of variables theorem for complex integrals but in all of them (for example here) it was required that $omega = fleft( z right)$ will be a biholorphic mapping.



      So my question is, how can the change of variables be rigorously justified?






      complex-analysis contour-integration substitution change-of-variable




      complex-analysis contour-integration substitution change-of-variable






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Feb 2 at 18:20









      José Carlos Santos

      174k23134243




      174k23134243










      asked Feb 2 at 18:07









      zokomokozokomoko

      169214




      169214






















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          $begingroup$

          Let $[a,b]$ be the domain of $C$. Thenbegin{align}oint_{f(C)}frac1omega,mathrm domega&=int_a^bfrac{(fcirc C)'(t)}{(fcirc C)(t)},mathrm dt\&=int_a^bfrac{f'bigl(C(t)bigr)C'(t)}{fbigl(C(t)bigr)},mathrm dt\&=oint_Cfrac{f'(z)}{f(z)},mathrm dz.end{align}






          share|cite|improve this answer









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          • $begingroup$
            Simple and clear - thank you! :)
            $endgroup$
            – zokomoko
            Feb 2 at 18:31












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          $begingroup$

          Let $[a,b]$ be the domain of $C$. Thenbegin{align}oint_{f(C)}frac1omega,mathrm domega&=int_a^bfrac{(fcirc C)'(t)}{(fcirc C)(t)},mathrm dt\&=int_a^bfrac{f'bigl(C(t)bigr)C'(t)}{fbigl(C(t)bigr)},mathrm dt\&=oint_Cfrac{f'(z)}{f(z)},mathrm dz.end{align}






          share|cite|improve this answer









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          • $begingroup$
            Simple and clear - thank you! :)
            $endgroup$
            – zokomoko
            Feb 2 at 18:31
















          1












          $begingroup$

          Let $[a,b]$ be the domain of $C$. Thenbegin{align}oint_{f(C)}frac1omega,mathrm domega&=int_a^bfrac{(fcirc C)'(t)}{(fcirc C)(t)},mathrm dt\&=int_a^bfrac{f'bigl(C(t)bigr)C'(t)}{fbigl(C(t)bigr)},mathrm dt\&=oint_Cfrac{f'(z)}{f(z)},mathrm dz.end{align}






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Simple and clear - thank you! :)
            $endgroup$
            – zokomoko
            Feb 2 at 18:31














          1












          1








          1





          $begingroup$

          Let $[a,b]$ be the domain of $C$. Thenbegin{align}oint_{f(C)}frac1omega,mathrm domega&=int_a^bfrac{(fcirc C)'(t)}{(fcirc C)(t)},mathrm dt\&=int_a^bfrac{f'bigl(C(t)bigr)C'(t)}{fbigl(C(t)bigr)},mathrm dt\&=oint_Cfrac{f'(z)}{f(z)},mathrm dz.end{align}






          share|cite|improve this answer









          $endgroup$



          Let $[a,b]$ be the domain of $C$. Thenbegin{align}oint_{f(C)}frac1omega,mathrm domega&=int_a^bfrac{(fcirc C)'(t)}{(fcirc C)(t)},mathrm dt\&=int_a^bfrac{f'bigl(C(t)bigr)C'(t)}{fbigl(C(t)bigr)},mathrm dt\&=oint_Cfrac{f'(z)}{f(z)},mathrm dz.end{align}







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Feb 2 at 18:19









          José Carlos SantosJosé Carlos Santos

          174k23134243




          174k23134243












          • $begingroup$
            Simple and clear - thank you! :)
            $endgroup$
            – zokomoko
            Feb 2 at 18:31


















          • $begingroup$
            Simple and clear - thank you! :)
            $endgroup$
            – zokomoko
            Feb 2 at 18:31
















          $begingroup$
          Simple and clear - thank you! :)
          $endgroup$
          – zokomoko
          Feb 2 at 18:31




          $begingroup$
          Simple and clear - thank you! :)
          $endgroup$
          – zokomoko
          Feb 2 at 18:31


















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