Stuck on contour integral, can I use Cauchy's theorem instead?












2












$begingroup$


I am trying to calculate



$$int_{|z-2| = 3} e^{1/z}$$



I parametric the circle of radius three centered at 2 by $gamma(t) = 3e^{it} +2$ and so I can instead evaluate



$$int_gamma f(gamma(t))gamma'(t) = int_0^{2pi} exp(frac{1}{3exp(itheta)+2})*3iexp(itheta)$$ and I am stuck here because I do not know how to compute the antiderivate of this. This makes me think there must be an easier way to do this



I am aware that the integrand is not defined at z = 0 is this what is causing problems? Since $gamma$ is a closed path can I just say that the integral is zero, or does the fact that the integrand is not defined at 0 mean that it is not analytic and hence I can not use Cauchy's theorem?






complex-analysis contour-integration







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    I am trying to calculate



    $$int_{|z-2| = 3} e^{1/z}$$



    I parametric the circle of radius three centered at 2 by $gamma(t) = 3e^{it} +2$ and so I can instead evaluate



    $$int_gamma f(gamma(t))gamma'(t) = int_0^{2pi} exp(frac{1}{3exp(itheta)+2})*3iexp(itheta)$$ and I am stuck here because I do not know how to compute the antiderivate of this. This makes me think there must be an easier way to do this



    I am aware that the integrand is not defined at z = 0 is this what is causing problems? Since $gamma$ is a closed path can I just say that the integral is zero, or does the fact that the integrand is not defined at 0 mean that it is not analytic and hence I can not use Cauchy's theorem?






    complex-analysis contour-integration







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      I am trying to calculate



      $$int_{|z-2| = 3} e^{1/z}$$



      I parametric the circle of radius three centered at 2 by $gamma(t) = 3e^{it} +2$ and so I can instead evaluate



      $$int_gamma f(gamma(t))gamma'(t) = int_0^{2pi} exp(frac{1}{3exp(itheta)+2})*3iexp(itheta)$$ and I am stuck here because I do not know how to compute the antiderivate of this. This makes me think there must be an easier way to do this



      I am aware that the integrand is not defined at z = 0 is this what is causing problems? Since $gamma$ is a closed path can I just say that the integral is zero, or does the fact that the integrand is not defined at 0 mean that it is not analytic and hence I can not use Cauchy's theorem?






      complex-analysis contour-integration







      share|cite|improve this question









      $endgroup$




      I am trying to calculate



      $$int_{|z-2| = 3} e^{1/z}$$



      I parametric the circle of radius three centered at 2 by $gamma(t) = 3e^{it} +2$ and so I can instead evaluate



      $$int_gamma f(gamma(t))gamma'(t) = int_0^{2pi} exp(frac{1}{3exp(itheta)+2})*3iexp(itheta)$$ and I am stuck here because I do not know how to compute the antiderivate of this. This makes me think there must be an easier way to do this



      I am aware that the integrand is not defined at z = 0 is this what is causing problems? Since $gamma$ is a closed path can I just say that the integral is zero, or does the fact that the integrand is not defined at 0 mean that it is not analytic and hence I can not use Cauchy's theorem?






      complex-analysis contour-integration




      complex-analysis contour-integration






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 13 '18 at 0:45









      Richard VillalobosRichard Villalobos

      1757




      1757






















          2 Answers
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          active

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          6












          $begingroup$

          You can apply the residue theorem here. Since$$e^{frac1z}=1+frac1z+frac1{2z^2}+cdots,$$you know that $operatorname{res}_{z=0}left(e^{frac1z}right)=1$. Therefore, your integral is equal to $2pi itimes1=2pi i$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            ah yes! Thank you, I had completely forgotten about the residue theorem.
            $endgroup$
            – Richard Villalobos
            Dec 13 '18 at 0:53










          • $begingroup$
            Jose Carlos Santos, may I ask what is that hat which appeared on your avatar?
            $endgroup$
            – Mark
            Dec 13 '18 at 0:54








          • 1




            $begingroup$
            It is called “The Merlin”.
            $endgroup$
            – José Carlos Santos
            Dec 13 '18 at 1:04



















          2












          $begingroup$

          I'm going to assume you aren't versed in the residue theorem yet. (Well, you implied that you were in a comment, so oh well. You can take this as an alternate answer, or this might be helpful to those in the future looking at this who aren't as well versed. As you will.)




          Hint #1:



          Since the integrand is not defined at $z=0$, and the contour encloses that singularity, you cannot use the Cauchy Integral Theorem, at least not immediately.



          My recommendation would be to use the power series expansion of $e^z$ (plugging in $1/z$ for $z$). You can then express the integral by



          $$int_{|z-2|=3} e^{1/z}dz = int_{|z-2|=3} sum_{k=0}^infty frac{1}{k! cdot z^k}= sum_{k=0}^infty int_{|z-2|=3} frac{1}{k! cdot z^k}$$



          You should find some pleasant surprises in that summation that simplify the process a bit.




          Hint #2:



          For any contour of positive orientation which is a circle of radius $r$ centered at the complex number $z_star$, you can show



          $$int_{|z-z_star|=r} frac{1}{z^n} dz = left{begin{matrix}
          0 & forall n neq 1 \
          2pi i & n = 1
          end{matrix}right.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you, I appreciate this answer very much as well because its gives me another insight into how to solve similar problems!
            $endgroup$
            – Richard Villalobos
            Dec 13 '18 at 0:57











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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

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          6












          $begingroup$

          You can apply the residue theorem here. Since$$e^{frac1z}=1+frac1z+frac1{2z^2}+cdots,$$you know that $operatorname{res}_{z=0}left(e^{frac1z}right)=1$. Therefore, your integral is equal to $2pi itimes1=2pi i$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            ah yes! Thank you, I had completely forgotten about the residue theorem.
            $endgroup$
            – Richard Villalobos
            Dec 13 '18 at 0:53










          • $begingroup$
            Jose Carlos Santos, may I ask what is that hat which appeared on your avatar?
            $endgroup$
            – Mark
            Dec 13 '18 at 0:54








          • 1




            $begingroup$
            It is called “The Merlin”.
            $endgroup$
            – José Carlos Santos
            Dec 13 '18 at 1:04
















          6












          $begingroup$

          You can apply the residue theorem here. Since$$e^{frac1z}=1+frac1z+frac1{2z^2}+cdots,$$you know that $operatorname{res}_{z=0}left(e^{frac1z}right)=1$. Therefore, your integral is equal to $2pi itimes1=2pi i$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            ah yes! Thank you, I had completely forgotten about the residue theorem.
            $endgroup$
            – Richard Villalobos
            Dec 13 '18 at 0:53










          • $begingroup$
            Jose Carlos Santos, may I ask what is that hat which appeared on your avatar?
            $endgroup$
            – Mark
            Dec 13 '18 at 0:54








          • 1




            $begingroup$
            It is called “The Merlin”.
            $endgroup$
            – José Carlos Santos
            Dec 13 '18 at 1:04














          6












          6








          6





          $begingroup$

          You can apply the residue theorem here. Since$$e^{frac1z}=1+frac1z+frac1{2z^2}+cdots,$$you know that $operatorname{res}_{z=0}left(e^{frac1z}right)=1$. Therefore, your integral is equal to $2pi itimes1=2pi i$.






          share|cite|improve this answer











          $endgroup$



          You can apply the residue theorem here. Since$$e^{frac1z}=1+frac1z+frac1{2z^2}+cdots,$$you know that $operatorname{res}_{z=0}left(e^{frac1z}right)=1$. Therefore, your integral is equal to $2pi itimes1=2pi i$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 13 '18 at 1:06

























          answered Dec 13 '18 at 0:52









          José Carlos SantosJosé Carlos Santos

          159k22126231




          159k22126231












          • $begingroup$
            ah yes! Thank you, I had completely forgotten about the residue theorem.
            $endgroup$
            – Richard Villalobos
            Dec 13 '18 at 0:53










          • $begingroup$
            Jose Carlos Santos, may I ask what is that hat which appeared on your avatar?
            $endgroup$
            – Mark
            Dec 13 '18 at 0:54








          • 1




            $begingroup$
            It is called “The Merlin”.
            $endgroup$
            – José Carlos Santos
            Dec 13 '18 at 1:04


















          • $begingroup$
            ah yes! Thank you, I had completely forgotten about the residue theorem.
            $endgroup$
            – Richard Villalobos
            Dec 13 '18 at 0:53










          • $begingroup$
            Jose Carlos Santos, may I ask what is that hat which appeared on your avatar?
            $endgroup$
            – Mark
            Dec 13 '18 at 0:54








          • 1




            $begingroup$
            It is called “The Merlin”.
            $endgroup$
            – José Carlos Santos
            Dec 13 '18 at 1:04
















          $begingroup$
          ah yes! Thank you, I had completely forgotten about the residue theorem.
          $endgroup$
          – Richard Villalobos
          Dec 13 '18 at 0:53




          $begingroup$
          ah yes! Thank you, I had completely forgotten about the residue theorem.
          $endgroup$
          – Richard Villalobos
          Dec 13 '18 at 0:53












          $begingroup$
          Jose Carlos Santos, may I ask what is that hat which appeared on your avatar?
          $endgroup$
          – Mark
          Dec 13 '18 at 0:54






          $begingroup$
          Jose Carlos Santos, may I ask what is that hat which appeared on your avatar?
          $endgroup$
          – Mark
          Dec 13 '18 at 0:54






          1




          1




          $begingroup$
          It is called “The Merlin”.
          $endgroup$
          – José Carlos Santos
          Dec 13 '18 at 1:04




          $begingroup$
          It is called “The Merlin”.
          $endgroup$
          – José Carlos Santos
          Dec 13 '18 at 1:04











          2












          $begingroup$

          I'm going to assume you aren't versed in the residue theorem yet. (Well, you implied that you were in a comment, so oh well. You can take this as an alternate answer, or this might be helpful to those in the future looking at this who aren't as well versed. As you will.)




          Hint #1:



          Since the integrand is not defined at $z=0$, and the contour encloses that singularity, you cannot use the Cauchy Integral Theorem, at least not immediately.



          My recommendation would be to use the power series expansion of $e^z$ (plugging in $1/z$ for $z$). You can then express the integral by



          $$int_{|z-2|=3} e^{1/z}dz = int_{|z-2|=3} sum_{k=0}^infty frac{1}{k! cdot z^k}= sum_{k=0}^infty int_{|z-2|=3} frac{1}{k! cdot z^k}$$



          You should find some pleasant surprises in that summation that simplify the process a bit.




          Hint #2:



          For any contour of positive orientation which is a circle of radius $r$ centered at the complex number $z_star$, you can show



          $$int_{|z-z_star|=r} frac{1}{z^n} dz = left{begin{matrix}
          0 & forall n neq 1 \
          2pi i & n = 1
          end{matrix}right.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you, I appreciate this answer very much as well because its gives me another insight into how to solve similar problems!
            $endgroup$
            – Richard Villalobos
            Dec 13 '18 at 0:57
















          2












          $begingroup$

          I'm going to assume you aren't versed in the residue theorem yet. (Well, you implied that you were in a comment, so oh well. You can take this as an alternate answer, or this might be helpful to those in the future looking at this who aren't as well versed. As you will.)




          Hint #1:



          Since the integrand is not defined at $z=0$, and the contour encloses that singularity, you cannot use the Cauchy Integral Theorem, at least not immediately.



          My recommendation would be to use the power series expansion of $e^z$ (plugging in $1/z$ for $z$). You can then express the integral by



          $$int_{|z-2|=3} e^{1/z}dz = int_{|z-2|=3} sum_{k=0}^infty frac{1}{k! cdot z^k}= sum_{k=0}^infty int_{|z-2|=3} frac{1}{k! cdot z^k}$$



          You should find some pleasant surprises in that summation that simplify the process a bit.




          Hint #2:



          For any contour of positive orientation which is a circle of radius $r$ centered at the complex number $z_star$, you can show



          $$int_{|z-z_star|=r} frac{1}{z^n} dz = left{begin{matrix}
          0 & forall n neq 1 \
          2pi i & n = 1
          end{matrix}right.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you, I appreciate this answer very much as well because its gives me another insight into how to solve similar problems!
            $endgroup$
            – Richard Villalobos
            Dec 13 '18 at 0:57














          2












          2








          2





          $begingroup$

          I'm going to assume you aren't versed in the residue theorem yet. (Well, you implied that you were in a comment, so oh well. You can take this as an alternate answer, or this might be helpful to those in the future looking at this who aren't as well versed. As you will.)




          Hint #1:



          Since the integrand is not defined at $z=0$, and the contour encloses that singularity, you cannot use the Cauchy Integral Theorem, at least not immediately.



          My recommendation would be to use the power series expansion of $e^z$ (plugging in $1/z$ for $z$). You can then express the integral by



          $$int_{|z-2|=3} e^{1/z}dz = int_{|z-2|=3} sum_{k=0}^infty frac{1}{k! cdot z^k}= sum_{k=0}^infty int_{|z-2|=3} frac{1}{k! cdot z^k}$$



          You should find some pleasant surprises in that summation that simplify the process a bit.




          Hint #2:



          For any contour of positive orientation which is a circle of radius $r$ centered at the complex number $z_star$, you can show



          $$int_{|z-z_star|=r} frac{1}{z^n} dz = left{begin{matrix}
          0 & forall n neq 1 \
          2pi i & n = 1
          end{matrix}right.$$






          share|cite|improve this answer









          $endgroup$



          I'm going to assume you aren't versed in the residue theorem yet. (Well, you implied that you were in a comment, so oh well. You can take this as an alternate answer, or this might be helpful to those in the future looking at this who aren't as well versed. As you will.)




          Hint #1:



          Since the integrand is not defined at $z=0$, and the contour encloses that singularity, you cannot use the Cauchy Integral Theorem, at least not immediately.



          My recommendation would be to use the power series expansion of $e^z$ (plugging in $1/z$ for $z$). You can then express the integral by



          $$int_{|z-2|=3} e^{1/z}dz = int_{|z-2|=3} sum_{k=0}^infty frac{1}{k! cdot z^k}= sum_{k=0}^infty int_{|z-2|=3} frac{1}{k! cdot z^k}$$



          You should find some pleasant surprises in that summation that simplify the process a bit.




          Hint #2:



          For any contour of positive orientation which is a circle of radius $r$ centered at the complex number $z_star$, you can show



          $$int_{|z-z_star|=r} frac{1}{z^n} dz = left{begin{matrix}
          0 & forall n neq 1 \
          2pi i & n = 1
          end{matrix}right.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 13 '18 at 0:56









          Eevee TrainerEevee Trainer

          5,8521936




          5,8521936












          • $begingroup$
            Thank you, I appreciate this answer very much as well because its gives me another insight into how to solve similar problems!
            $endgroup$
            – Richard Villalobos
            Dec 13 '18 at 0:57


















          • $begingroup$
            Thank you, I appreciate this answer very much as well because its gives me another insight into how to solve similar problems!
            $endgroup$
            – Richard Villalobos
            Dec 13 '18 at 0:57
















          $begingroup$
          Thank you, I appreciate this answer very much as well because its gives me another insight into how to solve similar problems!
          $endgroup$
          – Richard Villalobos
          Dec 13 '18 at 0:57




          $begingroup$
          Thank you, I appreciate this answer very much as well because its gives me another insight into how to solve similar problems!
          $endgroup$
          – Richard Villalobos
          Dec 13 '18 at 0:57


















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