Complex Integral $int_{C} frac{e^{iz}}{z^3} dz$ using Cauchy Integral Formula of Derivatives












2












$begingroup$


I am trying to find $int_{C} frac{e^{iz}}{z^3} dz$ on circle of $|z| = 2$ traversing once on positive direction.



My approach was using Cauchy derivative formula $f^{(n)}(a) = frac{n!}{2pi i} int_{C} frac{f(z)}{(z-a)^{n+1}} dz$.



This is a case with $a = 0$ and $n = 2$. Second derivative of $f(z) = e^{iz}$ is $-e^{iz}$ so our integral is simply $frac{2 pi i}{2} (-e^{iz})$ at $z=0$ which gives me $-pi i$.



However, I am not sure if this is correct mostly because of the fact that the circle is not a unit circle. Can somebody verify and if I am wrong, can somebody tell me which part I have done wrong? Thanks.






complex-analysis proof-verification complex-numbers complex-integration







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    2












    $begingroup$


    I am trying to find $int_{C} frac{e^{iz}}{z^3} dz$ on circle of $|z| = 2$ traversing once on positive direction.



    My approach was using Cauchy derivative formula $f^{(n)}(a) = frac{n!}{2pi i} int_{C} frac{f(z)}{(z-a)^{n+1}} dz$.



    This is a case with $a = 0$ and $n = 2$. Second derivative of $f(z) = e^{iz}$ is $-e^{iz}$ so our integral is simply $frac{2 pi i}{2} (-e^{iz})$ at $z=0$ which gives me $-pi i$.



    However, I am not sure if this is correct mostly because of the fact that the circle is not a unit circle. Can somebody verify and if I am wrong, can somebody tell me which part I have done wrong? Thanks.






    complex-analysis proof-verification complex-numbers complex-integration







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      I am trying to find $int_{C} frac{e^{iz}}{z^3} dz$ on circle of $|z| = 2$ traversing once on positive direction.



      My approach was using Cauchy derivative formula $f^{(n)}(a) = frac{n!}{2pi i} int_{C} frac{f(z)}{(z-a)^{n+1}} dz$.



      This is a case with $a = 0$ and $n = 2$. Second derivative of $f(z) = e^{iz}$ is $-e^{iz}$ so our integral is simply $frac{2 pi i}{2} (-e^{iz})$ at $z=0$ which gives me $-pi i$.



      However, I am not sure if this is correct mostly because of the fact that the circle is not a unit circle. Can somebody verify and if I am wrong, can somebody tell me which part I have done wrong? Thanks.






      complex-analysis proof-verification complex-numbers complex-integration







      share|cite|improve this question











      $endgroup$




      I am trying to find $int_{C} frac{e^{iz}}{z^3} dz$ on circle of $|z| = 2$ traversing once on positive direction.



      My approach was using Cauchy derivative formula $f^{(n)}(a) = frac{n!}{2pi i} int_{C} frac{f(z)}{(z-a)^{n+1}} dz$.



      This is a case with $a = 0$ and $n = 2$. Second derivative of $f(z) = e^{iz}$ is $-e^{iz}$ so our integral is simply $frac{2 pi i}{2} (-e^{iz})$ at $z=0$ which gives me $-pi i$.



      However, I am not sure if this is correct mostly because of the fact that the circle is not a unit circle. Can somebody verify and if I am wrong, can somebody tell me which part I have done wrong? Thanks.






      complex-analysis proof-verification complex-numbers complex-integration




      complex-analysis proof-verification complex-numbers complex-integration






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 12 '18 at 3:13









      Eevee Trainer

      5,7571936




      5,7571936










      asked Dec 12 '18 at 3:02









      dmsj djsldmsj djsl

      35517




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          2 Answers
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          $begingroup$

          As an addendum to the previous answer, you actually do not need to compute any derivative since the Maclaurin series of the exponential function is fairly well-known:
          $$ e^{iz} = 1 + iz -frac{z^2}{2} + O(z^3) $$
          directly leads to
          $$ oint_{|z|=rho}frac{e^{iz}}{z^3}= 2pi icdotoperatorname*{Res}_{z=0}left(frac{e^{iz}}{z^3}right)=2pi icdotleft(-frac{1}{2}right)=color{red}{-pi i} $$
          for any $rho>0$.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            The circle in question need not be a unit circle.



            Indeed, the key fact is that it's a closed curve of counterclockwise orientation, enclosing the one singularity of the integrand, $z = 0$, which is being traversed one time. Strictly speaking, it could be any sort of crazy curve so long as those facts are maintained, even if it's not even circle! (If you can continuously deform/reshape the curve, the integral remains the same.)



            That nuance aside - yup, your solution is correct!






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Well, not too crazy: better for such a curve to be closed, simple and rectifiable.
              $endgroup$
              – Jack D'Aurizio
              Dec 12 '18 at 3:31










            • $begingroup$
              Fair point. No sense in burdening oneself unnecessarily in the computations.
              $endgroup$
              – Eevee Trainer
              Dec 12 '18 at 3:33











            Your Answer





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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            As an addendum to the previous answer, you actually do not need to compute any derivative since the Maclaurin series of the exponential function is fairly well-known:
            $$ e^{iz} = 1 + iz -frac{z^2}{2} + O(z^3) $$
            directly leads to
            $$ oint_{|z|=rho}frac{e^{iz}}{z^3}= 2pi icdotoperatorname*{Res}_{z=0}left(frac{e^{iz}}{z^3}right)=2pi icdotleft(-frac{1}{2}right)=color{red}{-pi i} $$
            for any $rho>0$.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              As an addendum to the previous answer, you actually do not need to compute any derivative since the Maclaurin series of the exponential function is fairly well-known:
              $$ e^{iz} = 1 + iz -frac{z^2}{2} + O(z^3) $$
              directly leads to
              $$ oint_{|z|=rho}frac{e^{iz}}{z^3}= 2pi icdotoperatorname*{Res}_{z=0}left(frac{e^{iz}}{z^3}right)=2pi icdotleft(-frac{1}{2}right)=color{red}{-pi i} $$
              for any $rho>0$.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                As an addendum to the previous answer, you actually do not need to compute any derivative since the Maclaurin series of the exponential function is fairly well-known:
                $$ e^{iz} = 1 + iz -frac{z^2}{2} + O(z^3) $$
                directly leads to
                $$ oint_{|z|=rho}frac{e^{iz}}{z^3}= 2pi icdotoperatorname*{Res}_{z=0}left(frac{e^{iz}}{z^3}right)=2pi icdotleft(-frac{1}{2}right)=color{red}{-pi i} $$
                for any $rho>0$.






                share|cite|improve this answer









                $endgroup$



                As an addendum to the previous answer, you actually do not need to compute any derivative since the Maclaurin series of the exponential function is fairly well-known:
                $$ e^{iz} = 1 + iz -frac{z^2}{2} + O(z^3) $$
                directly leads to
                $$ oint_{|z|=rho}frac{e^{iz}}{z^3}= 2pi icdotoperatorname*{Res}_{z=0}left(frac{e^{iz}}{z^3}right)=2pi icdotleft(-frac{1}{2}right)=color{red}{-pi i} $$
                for any $rho>0$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 12 '18 at 3:17









                Jack D'AurizioJack D'Aurizio

                289k33281661




                289k33281661























                    1












                    $begingroup$

                    The circle in question need not be a unit circle.



                    Indeed, the key fact is that it's a closed curve of counterclockwise orientation, enclosing the one singularity of the integrand, $z = 0$, which is being traversed one time. Strictly speaking, it could be any sort of crazy curve so long as those facts are maintained, even if it's not even circle! (If you can continuously deform/reshape the curve, the integral remains the same.)



                    That nuance aside - yup, your solution is correct!






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Well, not too crazy: better for such a curve to be closed, simple and rectifiable.
                      $endgroup$
                      – Jack D'Aurizio
                      Dec 12 '18 at 3:31










                    • $begingroup$
                      Fair point. No sense in burdening oneself unnecessarily in the computations.
                      $endgroup$
                      – Eevee Trainer
                      Dec 12 '18 at 3:33
















                    1












                    $begingroup$

                    The circle in question need not be a unit circle.



                    Indeed, the key fact is that it's a closed curve of counterclockwise orientation, enclosing the one singularity of the integrand, $z = 0$, which is being traversed one time. Strictly speaking, it could be any sort of crazy curve so long as those facts are maintained, even if it's not even circle! (If you can continuously deform/reshape the curve, the integral remains the same.)



                    That nuance aside - yup, your solution is correct!






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Well, not too crazy: better for such a curve to be closed, simple and rectifiable.
                      $endgroup$
                      – Jack D'Aurizio
                      Dec 12 '18 at 3:31










                    • $begingroup$
                      Fair point. No sense in burdening oneself unnecessarily in the computations.
                      $endgroup$
                      – Eevee Trainer
                      Dec 12 '18 at 3:33














                    1












                    1








                    1





                    $begingroup$

                    The circle in question need not be a unit circle.



                    Indeed, the key fact is that it's a closed curve of counterclockwise orientation, enclosing the one singularity of the integrand, $z = 0$, which is being traversed one time. Strictly speaking, it could be any sort of crazy curve so long as those facts are maintained, even if it's not even circle! (If you can continuously deform/reshape the curve, the integral remains the same.)



                    That nuance aside - yup, your solution is correct!






                    share|cite|improve this answer









                    $endgroup$



                    The circle in question need not be a unit circle.



                    Indeed, the key fact is that it's a closed curve of counterclockwise orientation, enclosing the one singularity of the integrand, $z = 0$, which is being traversed one time. Strictly speaking, it could be any sort of crazy curve so long as those facts are maintained, even if it's not even circle! (If you can continuously deform/reshape the curve, the integral remains the same.)



                    That nuance aside - yup, your solution is correct!







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 12 '18 at 3:10









                    Eevee TrainerEevee Trainer

                    5,7571936




                    5,7571936












                    • $begingroup$
                      Well, not too crazy: better for such a curve to be closed, simple and rectifiable.
                      $endgroup$
                      – Jack D'Aurizio
                      Dec 12 '18 at 3:31










                    • $begingroup$
                      Fair point. No sense in burdening oneself unnecessarily in the computations.
                      $endgroup$
                      – Eevee Trainer
                      Dec 12 '18 at 3:33


















                    • $begingroup$
                      Well, not too crazy: better for such a curve to be closed, simple and rectifiable.
                      $endgroup$
                      – Jack D'Aurizio
                      Dec 12 '18 at 3:31










                    • $begingroup$
                      Fair point. No sense in burdening oneself unnecessarily in the computations.
                      $endgroup$
                      – Eevee Trainer
                      Dec 12 '18 at 3:33
















                    $begingroup$
                    Well, not too crazy: better for such a curve to be closed, simple and rectifiable.
                    $endgroup$
                    – Jack D'Aurizio
                    Dec 12 '18 at 3:31




                    $begingroup$
                    Well, not too crazy: better for such a curve to be closed, simple and rectifiable.
                    $endgroup$
                    – Jack D'Aurizio
                    Dec 12 '18 at 3:31












                    $begingroup$
                    Fair point. No sense in burdening oneself unnecessarily in the computations.
                    $endgroup$
                    – Eevee Trainer
                    Dec 12 '18 at 3:33




                    $begingroup$
                    Fair point. No sense in burdening oneself unnecessarily in the computations.
                    $endgroup$
                    – Eevee Trainer
                    Dec 12 '18 at 3:33


















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