Cauchy's Formula vs. Cauchy's Theorem confusion

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I'm using Donald Sarason's book as a reference.



(One version of) Cauchy's theorem states that for a holomorphic function defined on an open, convex set, $int_gamma f(z)mathrmdz$ = 0 for any piecewise-$C^1$ closed curve $gamma$ in this set.



On the other hand, Cauchy's formula for a circle gives one a method to compute the value of a holomorphic $f$ for any point $z$ in the interior of the circle:
$$
f(z) = frac12 pi i int_C fracf(xi)xi - z mathrmdxi .
$$



My question is the following: why doesn't the integral on the right hand side of the above question vanish by Cauchy's theorem? To be painfully explicit, if we consider a circle centered around the origin, why is it not true that
$$
f(0) = frac12 pi i int_C fracf(xi)xi mathrmdxi = 0
$$
for all holomorphic functions $f$?



I know I'm missing something elementary; I just can't see what it is.




complex-analysis




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    up vote
    2
    down vote

    favorite












    I'm using Donald Sarason's book as a reference.



    (One version of) Cauchy's theorem states that for a holomorphic function defined on an open, convex set, $int_gamma f(z)mathrmdz$ = 0 for any piecewise-$C^1$ closed curve $gamma$ in this set.



    On the other hand, Cauchy's formula for a circle gives one a method to compute the value of a holomorphic $f$ for any point $z$ in the interior of the circle:
    $$
    f(z) = frac12 pi i int_C fracf(xi)xi - z mathrmdxi .
    $$



    My question is the following: why doesn't the integral on the right hand side of the above question vanish by Cauchy's theorem? To be painfully explicit, if we consider a circle centered around the origin, why is it not true that
    $$
    f(0) = frac12 pi i int_C fracf(xi)xi mathrmdxi = 0
    $$
    for all holomorphic functions $f$?



    I know I'm missing something elementary; I just can't see what it is.




    complex-analysis




    share|cite|improve this question
























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      I'm using Donald Sarason's book as a reference.



      (One version of) Cauchy's theorem states that for a holomorphic function defined on an open, convex set, $int_gamma f(z)mathrmdz$ = 0 for any piecewise-$C^1$ closed curve $gamma$ in this set.



      On the other hand, Cauchy's formula for a circle gives one a method to compute the value of a holomorphic $f$ for any point $z$ in the interior of the circle:
      $$
      f(z) = frac12 pi i int_C fracf(xi)xi - z mathrmdxi .
      $$



      My question is the following: why doesn't the integral on the right hand side of the above question vanish by Cauchy's theorem? To be painfully explicit, if we consider a circle centered around the origin, why is it not true that
      $$
      f(0) = frac12 pi i int_C fracf(xi)xi mathrmdxi = 0
      $$
      for all holomorphic functions $f$?



      I know I'm missing something elementary; I just can't see what it is.




      complex-analysis




      share|cite|improve this question














      I'm using Donald Sarason's book as a reference.



      (One version of) Cauchy's theorem states that for a holomorphic function defined on an open, convex set, $int_gamma f(z)mathrmdz$ = 0 for any piecewise-$C^1$ closed curve $gamma$ in this set.



      On the other hand, Cauchy's formula for a circle gives one a method to compute the value of a holomorphic $f$ for any point $z$ in the interior of the circle:
      $$
      f(z) = frac12 pi i int_C fracf(xi)xi - z mathrmdxi .
      $$



      My question is the following: why doesn't the integral on the right hand side of the above question vanish by Cauchy's theorem? To be painfully explicit, if we consider a circle centered around the origin, why is it not true that
      $$
      f(0) = frac12 pi i int_C fracf(xi)xi mathrmdxi = 0
      $$
      for all holomorphic functions $f$?



      I know I'm missing something elementary; I just can't see what it is.





      complex-analysis





      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Aug 27 at 17:13









      Bernard

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      111k635102










      asked Aug 27 at 16:54









      Anthony

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          2 Answers
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          Because the denominator of the integrand gives rise to a singularity, so the integrand is not holomorphic on any open convex set containing the circle you are integrating over.






          share|cite|improve this answer






















          • But the denominator is not singular on the circle itself, which is the contour along which we're integrating, correct?
            – Anthony
            Aug 27 at 16:57






          • 3




            Yes, but the Cauchy theorem requires the function to be holomorphic on an open convex set containing the contour, so the point $z = xi$ will always be contained there. More particularly, in your example, if you integrate around any circle centered at the origin, the open convex set that is required to apply Cauchy's theorem will always contain the origin, since it would not be convex otherwise. I can try to elaborate this further in my answer, if you need me to.
            – Sobi
            Aug 27 at 16:58











          • No, that's clear now, thanks! I just got hung up thinking about the contour itself and forgot that the region in which it's contained matters.
            – Anthony
            Aug 27 at 17:05






          • 1




            @Anthony If I remember correctly, I had a similar problem when I was learning this. :)
            – Sobi
            Aug 27 at 17:07

















          up vote
          1
          down vote













          the integrand in $$f(z) = frac12 pi i int_C fracf(xi)xi - z mathrmdxi $$



          is $$fracf(xi)xi - z$$ which is not holomorphic within the region bounded by the boundary.



          Consider the point $z=xi $ where your function is not defined.






          share|cite|improve this answer




















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            2 Answers
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            active

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            2 Answers
            2






            active

            oldest

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            active

            oldest

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            active

            oldest

            votes








            up vote
            3
            down vote



            accepted










            Because the denominator of the integrand gives rise to a singularity, so the integrand is not holomorphic on any open convex set containing the circle you are integrating over.






            share|cite|improve this answer






















            • But the denominator is not singular on the circle itself, which is the contour along which we're integrating, correct?
              – Anthony
              Aug 27 at 16:57






            • 3




              Yes, but the Cauchy theorem requires the function to be holomorphic on an open convex set containing the contour, so the point $z = xi$ will always be contained there. More particularly, in your example, if you integrate around any circle centered at the origin, the open convex set that is required to apply Cauchy's theorem will always contain the origin, since it would not be convex otherwise. I can try to elaborate this further in my answer, if you need me to.
              – Sobi
              Aug 27 at 16:58











            • No, that's clear now, thanks! I just got hung up thinking about the contour itself and forgot that the region in which it's contained matters.
              – Anthony
              Aug 27 at 17:05






            • 1




              @Anthony If I remember correctly, I had a similar problem when I was learning this. :)
              – Sobi
              Aug 27 at 17:07














            up vote
            3
            down vote



            accepted










            Because the denominator of the integrand gives rise to a singularity, so the integrand is not holomorphic on any open convex set containing the circle you are integrating over.






            share|cite|improve this answer






















            • But the denominator is not singular on the circle itself, which is the contour along which we're integrating, correct?
              – Anthony
              Aug 27 at 16:57






            • 3




              Yes, but the Cauchy theorem requires the function to be holomorphic on an open convex set containing the contour, so the point $z = xi$ will always be contained there. More particularly, in your example, if you integrate around any circle centered at the origin, the open convex set that is required to apply Cauchy's theorem will always contain the origin, since it would not be convex otherwise. I can try to elaborate this further in my answer, if you need me to.
              – Sobi
              Aug 27 at 16:58











            • No, that's clear now, thanks! I just got hung up thinking about the contour itself and forgot that the region in which it's contained matters.
              – Anthony
              Aug 27 at 17:05






            • 1




              @Anthony If I remember correctly, I had a similar problem when I was learning this. :)
              – Sobi
              Aug 27 at 17:07












            up vote
            3
            down vote



            accepted







            up vote
            3
            down vote



            accepted






            Because the denominator of the integrand gives rise to a singularity, so the integrand is not holomorphic on any open convex set containing the circle you are integrating over.






            share|cite|improve this answer














            Because the denominator of the integrand gives rise to a singularity, so the integrand is not holomorphic on any open convex set containing the circle you are integrating over.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Aug 27 at 17:03

























            answered Aug 27 at 16:55









            Sobi

            2,418315




            2,418315











            • But the denominator is not singular on the circle itself, which is the contour along which we're integrating, correct?
              – Anthony
              Aug 27 at 16:57






            • 3




              Yes, but the Cauchy theorem requires the function to be holomorphic on an open convex set containing the contour, so the point $z = xi$ will always be contained there. More particularly, in your example, if you integrate around any circle centered at the origin, the open convex set that is required to apply Cauchy's theorem will always contain the origin, since it would not be convex otherwise. I can try to elaborate this further in my answer, if you need me to.
              – Sobi
              Aug 27 at 16:58











            • No, that's clear now, thanks! I just got hung up thinking about the contour itself and forgot that the region in which it's contained matters.
              – Anthony
              Aug 27 at 17:05






            • 1




              @Anthony If I remember correctly, I had a similar problem when I was learning this. :)
              – Sobi
              Aug 27 at 17:07
















            • But the denominator is not singular on the circle itself, which is the contour along which we're integrating, correct?
              – Anthony
              Aug 27 at 16:57






            • 3




              Yes, but the Cauchy theorem requires the function to be holomorphic on an open convex set containing the contour, so the point $z = xi$ will always be contained there. More particularly, in your example, if you integrate around any circle centered at the origin, the open convex set that is required to apply Cauchy's theorem will always contain the origin, since it would not be convex otherwise. I can try to elaborate this further in my answer, if you need me to.
              – Sobi
              Aug 27 at 16:58











            • No, that's clear now, thanks! I just got hung up thinking about the contour itself and forgot that the region in which it's contained matters.
              – Anthony
              Aug 27 at 17:05






            • 1




              @Anthony If I remember correctly, I had a similar problem when I was learning this. :)
              – Sobi
              Aug 27 at 17:07















            But the denominator is not singular on the circle itself, which is the contour along which we're integrating, correct?
            – Anthony
            Aug 27 at 16:57




            But the denominator is not singular on the circle itself, which is the contour along which we're integrating, correct?
            – Anthony
            Aug 27 at 16:57




            3




            3




            Yes, but the Cauchy theorem requires the function to be holomorphic on an open convex set containing the contour, so the point $z = xi$ will always be contained there. More particularly, in your example, if you integrate around any circle centered at the origin, the open convex set that is required to apply Cauchy's theorem will always contain the origin, since it would not be convex otherwise. I can try to elaborate this further in my answer, if you need me to.
            – Sobi
            Aug 27 at 16:58





            Yes, but the Cauchy theorem requires the function to be holomorphic on an open convex set containing the contour, so the point $z = xi$ will always be contained there. More particularly, in your example, if you integrate around any circle centered at the origin, the open convex set that is required to apply Cauchy's theorem will always contain the origin, since it would not be convex otherwise. I can try to elaborate this further in my answer, if you need me to.
            – Sobi
            Aug 27 at 16:58













            No, that's clear now, thanks! I just got hung up thinking about the contour itself and forgot that the region in which it's contained matters.
            – Anthony
            Aug 27 at 17:05




            No, that's clear now, thanks! I just got hung up thinking about the contour itself and forgot that the region in which it's contained matters.
            – Anthony
            Aug 27 at 17:05




            1




            1




            @Anthony If I remember correctly, I had a similar problem when I was learning this. :)
            – Sobi
            Aug 27 at 17:07




            @Anthony If I remember correctly, I had a similar problem when I was learning this. :)
            – Sobi
            Aug 27 at 17:07










            up vote
            1
            down vote













            the integrand in $$f(z) = frac12 pi i int_C fracf(xi)xi - z mathrmdxi $$



            is $$fracf(xi)xi - z$$ which is not holomorphic within the region bounded by the boundary.



            Consider the point $z=xi $ where your function is not defined.






            share|cite|improve this answer
























              up vote
              1
              down vote













              the integrand in $$f(z) = frac12 pi i int_C fracf(xi)xi - z mathrmdxi $$



              is $$fracf(xi)xi - z$$ which is not holomorphic within the region bounded by the boundary.



              Consider the point $z=xi $ where your function is not defined.






              share|cite|improve this answer






















                up vote
                1
                down vote










                up vote
                1
                down vote









                the integrand in $$f(z) = frac12 pi i int_C fracf(xi)xi - z mathrmdxi $$



                is $$fracf(xi)xi - z$$ which is not holomorphic within the region bounded by the boundary.



                Consider the point $z=xi $ where your function is not defined.






                share|cite|improve this answer












                the integrand in $$f(z) = frac12 pi i int_C fracf(xi)xi - z mathrmdxi $$



                is $$fracf(xi)xi - z$$ which is not holomorphic within the region bounded by the boundary.



                Consider the point $z=xi $ where your function is not defined.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Aug 27 at 17:10









                Mohammad Riazi-Kermani

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