Value of a complex function inside an analytic region












1












$begingroup$


I encountered the following question which I am struggling with:




Suppose $f$ is analytic on ${z in mathbb{C} : |z| le 2 }$ and $f(z) = 3$ everywhere on the circle $|z|=2$. What is the value of $f$ at $z=1$?




All I can think of is about the Cauchy's integral formula which states that for a contour $gamma$ and $f$ analytic in a simply connected domain containing $gamma$:
$$f(1) = frac{1}{2pi i}int_{gamma}{frac{f(z)}{z-1}} dz$$



Then I can use as a contour the circle $|z|=2$ and
$$f(1) = frac{1}{2pi i}int_{|z|=2}{frac{3}{z-1}dz} = frac{1}{2pi i} [3ln(e^{itheta}-1)]^{2pi}_{0}$$



but this doesn't work. What is the correct/immediate way of deducing the statement above?






complex-analysis







share|cite|improve this question









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    1












    $begingroup$


    I encountered the following question which I am struggling with:




    Suppose $f$ is analytic on ${z in mathbb{C} : |z| le 2 }$ and $f(z) = 3$ everywhere on the circle $|z|=2$. What is the value of $f$ at $z=1$?




    All I can think of is about the Cauchy's integral formula which states that for a contour $gamma$ and $f$ analytic in a simply connected domain containing $gamma$:
    $$f(1) = frac{1}{2pi i}int_{gamma}{frac{f(z)}{z-1}} dz$$



    Then I can use as a contour the circle $|z|=2$ and
    $$f(1) = frac{1}{2pi i}int_{|z|=2}{frac{3}{z-1}dz} = frac{1}{2pi i} [3ln(e^{itheta}-1)]^{2pi}_{0}$$



    but this doesn't work. What is the correct/immediate way of deducing the statement above?






    complex-analysis







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I encountered the following question which I am struggling with:




      Suppose $f$ is analytic on ${z in mathbb{C} : |z| le 2 }$ and $f(z) = 3$ everywhere on the circle $|z|=2$. What is the value of $f$ at $z=1$?




      All I can think of is about the Cauchy's integral formula which states that for a contour $gamma$ and $f$ analytic in a simply connected domain containing $gamma$:
      $$f(1) = frac{1}{2pi i}int_{gamma}{frac{f(z)}{z-1}} dz$$



      Then I can use as a contour the circle $|z|=2$ and
      $$f(1) = frac{1}{2pi i}int_{|z|=2}{frac{3}{z-1}dz} = frac{1}{2pi i} [3ln(e^{itheta}-1)]^{2pi}_{0}$$



      but this doesn't work. What is the correct/immediate way of deducing the statement above?






      complex-analysis







      share|cite|improve this question









      $endgroup$




      I encountered the following question which I am struggling with:




      Suppose $f$ is analytic on ${z in mathbb{C} : |z| le 2 }$ and $f(z) = 3$ everywhere on the circle $|z|=2$. What is the value of $f$ at $z=1$?




      All I can think of is about the Cauchy's integral formula which states that for a contour $gamma$ and $f$ analytic in a simply connected domain containing $gamma$:
      $$f(1) = frac{1}{2pi i}int_{gamma}{frac{f(z)}{z-1}} dz$$



      Then I can use as a contour the circle $|z|=2$ and
      $$f(1) = frac{1}{2pi i}int_{|z|=2}{frac{3}{z-1}dz} = frac{1}{2pi i} [3ln(e^{itheta}-1)]^{2pi}_{0}$$



      but this doesn't work. What is the correct/immediate way of deducing the statement above?






      complex-analysis




      complex-analysis






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 12 at 20:15









      daljit97daljit97

      178111




      178111






















          1 Answer
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          $begingroup$

          The residue is $3$, so we should get $3$.



          That is, you have applied Cauchy's integral formula. Next apply the residue theorem.



          So $frac1{2pi i}cdot 6pi i=3$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            But isn't Cauchy's residue theorem valid only for a region where there is a singularity? In my question the function is defined to be analytic inside the region $|z| <2$
            $endgroup$
            – daljit97
            Jan 12 at 20:34










          • $begingroup$
            Yes, but let $g(z)=frac3{z-1}$.
            $endgroup$
            – Chris Custer
            Jan 12 at 20:36










          • $begingroup$
            Alright so $f(z)$ is analytic but $f(z)/z-1$ is not, correct?
            $endgroup$
            – daljit97
            Jan 12 at 20:39










          • $begingroup$
            Yes. That's what I'm saying. You can just replace $f(z)$ by $3$, in the residue theorem.
            $endgroup$
            – Chris Custer
            Jan 12 at 20:40













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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

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          active

          oldest

          votes









          1












          $begingroup$

          The residue is $3$, so we should get $3$.



          That is, you have applied Cauchy's integral formula. Next apply the residue theorem.



          So $frac1{2pi i}cdot 6pi i=3$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            But isn't Cauchy's residue theorem valid only for a region where there is a singularity? In my question the function is defined to be analytic inside the region $|z| <2$
            $endgroup$
            – daljit97
            Jan 12 at 20:34










          • $begingroup$
            Yes, but let $g(z)=frac3{z-1}$.
            $endgroup$
            – Chris Custer
            Jan 12 at 20:36










          • $begingroup$
            Alright so $f(z)$ is analytic but $f(z)/z-1$ is not, correct?
            $endgroup$
            – daljit97
            Jan 12 at 20:39










          • $begingroup$
            Yes. That's what I'm saying. You can just replace $f(z)$ by $3$, in the residue theorem.
            $endgroup$
            – Chris Custer
            Jan 12 at 20:40


















          1












          $begingroup$

          The residue is $3$, so we should get $3$.



          That is, you have applied Cauchy's integral formula. Next apply the residue theorem.



          So $frac1{2pi i}cdot 6pi i=3$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            But isn't Cauchy's residue theorem valid only for a region where there is a singularity? In my question the function is defined to be analytic inside the region $|z| <2$
            $endgroup$
            – daljit97
            Jan 12 at 20:34










          • $begingroup$
            Yes, but let $g(z)=frac3{z-1}$.
            $endgroup$
            – Chris Custer
            Jan 12 at 20:36










          • $begingroup$
            Alright so $f(z)$ is analytic but $f(z)/z-1$ is not, correct?
            $endgroup$
            – daljit97
            Jan 12 at 20:39










          • $begingroup$
            Yes. That's what I'm saying. You can just replace $f(z)$ by $3$, in the residue theorem.
            $endgroup$
            – Chris Custer
            Jan 12 at 20:40
















          1












          1








          1





          $begingroup$

          The residue is $3$, so we should get $3$.



          That is, you have applied Cauchy's integral formula. Next apply the residue theorem.



          So $frac1{2pi i}cdot 6pi i=3$.






          share|cite|improve this answer











          $endgroup$



          The residue is $3$, so we should get $3$.



          That is, you have applied Cauchy's integral formula. Next apply the residue theorem.



          So $frac1{2pi i}cdot 6pi i=3$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 12 at 20:32

























          answered Jan 12 at 20:21









          Chris CusterChris Custer

          14.2k3827




          14.2k3827












          • $begingroup$
            But isn't Cauchy's residue theorem valid only for a region where there is a singularity? In my question the function is defined to be analytic inside the region $|z| <2$
            $endgroup$
            – daljit97
            Jan 12 at 20:34










          • $begingroup$
            Yes, but let $g(z)=frac3{z-1}$.
            $endgroup$
            – Chris Custer
            Jan 12 at 20:36










          • $begingroup$
            Alright so $f(z)$ is analytic but $f(z)/z-1$ is not, correct?
            $endgroup$
            – daljit97
            Jan 12 at 20:39










          • $begingroup$
            Yes. That's what I'm saying. You can just replace $f(z)$ by $3$, in the residue theorem.
            $endgroup$
            – Chris Custer
            Jan 12 at 20:40




















          • $begingroup$
            But isn't Cauchy's residue theorem valid only for a region where there is a singularity? In my question the function is defined to be analytic inside the region $|z| <2$
            $endgroup$
            – daljit97
            Jan 12 at 20:34










          • $begingroup$
            Yes, but let $g(z)=frac3{z-1}$.
            $endgroup$
            – Chris Custer
            Jan 12 at 20:36










          • $begingroup$
            Alright so $f(z)$ is analytic but $f(z)/z-1$ is not, correct?
            $endgroup$
            – daljit97
            Jan 12 at 20:39










          • $begingroup$
            Yes. That's what I'm saying. You can just replace $f(z)$ by $3$, in the residue theorem.
            $endgroup$
            – Chris Custer
            Jan 12 at 20:40


















          $begingroup$
          But isn't Cauchy's residue theorem valid only for a region where there is a singularity? In my question the function is defined to be analytic inside the region $|z| <2$
          $endgroup$
          – daljit97
          Jan 12 at 20:34




          $begingroup$
          But isn't Cauchy's residue theorem valid only for a region where there is a singularity? In my question the function is defined to be analytic inside the region $|z| <2$
          $endgroup$
          – daljit97
          Jan 12 at 20:34












          $begingroup$
          Yes, but let $g(z)=frac3{z-1}$.
          $endgroup$
          – Chris Custer
          Jan 12 at 20:36




          $begingroup$
          Yes, but let $g(z)=frac3{z-1}$.
          $endgroup$
          – Chris Custer
          Jan 12 at 20:36












          $begingroup$
          Alright so $f(z)$ is analytic but $f(z)/z-1$ is not, correct?
          $endgroup$
          – daljit97
          Jan 12 at 20:39




          $begingroup$
          Alright so $f(z)$ is analytic but $f(z)/z-1$ is not, correct?
          $endgroup$
          – daljit97
          Jan 12 at 20:39












          $begingroup$
          Yes. That's what I'm saying. You can just replace $f(z)$ by $3$, in the residue theorem.
          $endgroup$
          – Chris Custer
          Jan 12 at 20:40






          $begingroup$
          Yes. That's what I'm saying. You can just replace $f(z)$ by $3$, in the residue theorem.
          $endgroup$
          – Chris Custer
          Jan 12 at 20:40




















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