Find a circle in which the integral sea different from zero












0














My function is $z^{2} hat z^{3}$
I don't know if this can be:
Taking the radio circle 2
begin{equation}
int z^{2} hat z^{3} dx = int frac{64}{z} dx
end{equation}
This because $z hat z= |z|^2 = 4$ but this happens twice, so $z hat z z hat z= |z|^2 = 16$ y then there is a $hat z$ without $z$ so $z hat z = 4 rightarrow hat z = frac{4}{z}$
but I do not know if this is valid






integration complex-analysis







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    0














    My function is $z^{2} hat z^{3}$
    I don't know if this can be:
    Taking the radio circle 2
    begin{equation}
    int z^{2} hat z^{3} dx = int frac{64}{z} dx
    end{equation}
    This because $z hat z= |z|^2 = 4$ but this happens twice, so $z hat z z hat z= |z|^2 = 16$ y then there is a $hat z$ without $z$ so $z hat z = 4 rightarrow hat z = frac{4}{z}$
    but I do not know if this is valid






    integration complex-analysis







    share|cite|improve this question

























      0












      0








      0







      My function is $z^{2} hat z^{3}$
      I don't know if this can be:
      Taking the radio circle 2
      begin{equation}
      int z^{2} hat z^{3} dx = int frac{64}{z} dx
      end{equation}
      This because $z hat z= |z|^2 = 4$ but this happens twice, so $z hat z z hat z= |z|^2 = 16$ y then there is a $hat z$ without $z$ so $z hat z = 4 rightarrow hat z = frac{4}{z}$
      but I do not know if this is valid






      integration complex-analysis







      share|cite|improve this question













      My function is $z^{2} hat z^{3}$
      I don't know if this can be:
      Taking the radio circle 2
      begin{equation}
      int z^{2} hat z^{3} dx = int frac{64}{z} dx
      end{equation}
      This because $z hat z= |z|^2 = 4$ but this happens twice, so $z hat z z hat z= |z|^2 = 16$ y then there is a $hat z$ without $z$ so $z hat z = 4 rightarrow hat z = frac{4}{z}$
      but I do not know if this is valid






      integration complex-analysis




      integration complex-analysis






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      asked Nov 25 at 7:23









      Jazmín Jones

      519




      519






















          2 Answers
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          2














          First, $z^2bar z^3=(zbar z)^2bar z=|z|^4bar z$ so using the unit circle $gamma=e^{it}$ gives
          $$ int_{gamma}|z|^4bar z dz=int_0^{2pi} e^{-it}ie^{it}dt=2pi i$$






          share|cite|improve this answer

















          • 2




            I guess @JazminJones mentioned "radius of circle being 2". So if thats the case $gamma=2e^{it}$ and the integral evaluates to $32pi i$.
            – Yadati Kiran
            Nov 25 at 7:47












          • $32cdot2pi i$ not $32pi i$.
            – Yadati Kiran
            Nov 25 at 8:37












          • can you explain the equality in more detail please
            – Jazmín Jones
            Nov 25 at 17:09










          • @JazminJones which one?
            – Guacho Perez
            Nov 25 at 18:06










          • Forget it thanks haha
            – Jazmín Jones
            Nov 25 at 18:14



















          0














          In your case the integrand is



          $$(rho e^{itheta})^2(rho e^{-itheta})^3de^{itheta}=2^5e^{-itheta}de^{itheta}$$ giving the integral



          $$left.32iright|_0^{2pi}.$$






          share|cite|improve this answer





















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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2














            First, $z^2bar z^3=(zbar z)^2bar z=|z|^4bar z$ so using the unit circle $gamma=e^{it}$ gives
            $$ int_{gamma}|z|^4bar z dz=int_0^{2pi} e^{-it}ie^{it}dt=2pi i$$






            share|cite|improve this answer

















            • 2




              I guess @JazminJones mentioned "radius of circle being 2". So if thats the case $gamma=2e^{it}$ and the integral evaluates to $32pi i$.
              – Yadati Kiran
              Nov 25 at 7:47












            • $32cdot2pi i$ not $32pi i$.
              – Yadati Kiran
              Nov 25 at 8:37












            • can you explain the equality in more detail please
              – Jazmín Jones
              Nov 25 at 17:09










            • @JazminJones which one?
              – Guacho Perez
              Nov 25 at 18:06










            • Forget it thanks haha
              – Jazmín Jones
              Nov 25 at 18:14
















            2














            First, $z^2bar z^3=(zbar z)^2bar z=|z|^4bar z$ so using the unit circle $gamma=e^{it}$ gives
            $$ int_{gamma}|z|^4bar z dz=int_0^{2pi} e^{-it}ie^{it}dt=2pi i$$






            share|cite|improve this answer

















            • 2




              I guess @JazminJones mentioned "radius of circle being 2". So if thats the case $gamma=2e^{it}$ and the integral evaluates to $32pi i$.
              – Yadati Kiran
              Nov 25 at 7:47












            • $32cdot2pi i$ not $32pi i$.
              – Yadati Kiran
              Nov 25 at 8:37












            • can you explain the equality in more detail please
              – Jazmín Jones
              Nov 25 at 17:09










            • @JazminJones which one?
              – Guacho Perez
              Nov 25 at 18:06










            • Forget it thanks haha
              – Jazmín Jones
              Nov 25 at 18:14














            2












            2








            2






            First, $z^2bar z^3=(zbar z)^2bar z=|z|^4bar z$ so using the unit circle $gamma=e^{it}$ gives
            $$ int_{gamma}|z|^4bar z dz=int_0^{2pi} e^{-it}ie^{it}dt=2pi i$$






            share|cite|improve this answer












            First, $z^2bar z^3=(zbar z)^2bar z=|z|^4bar z$ so using the unit circle $gamma=e^{it}$ gives
            $$ int_{gamma}|z|^4bar z dz=int_0^{2pi} e^{-it}ie^{it}dt=2pi i$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 25 at 7:38









            Guacho Perez

            3,80411131




            3,80411131








            • 2




              I guess @JazminJones mentioned "radius of circle being 2". So if thats the case $gamma=2e^{it}$ and the integral evaluates to $32pi i$.
              – Yadati Kiran
              Nov 25 at 7:47












            • $32cdot2pi i$ not $32pi i$.
              – Yadati Kiran
              Nov 25 at 8:37












            • can you explain the equality in more detail please
              – Jazmín Jones
              Nov 25 at 17:09










            • @JazminJones which one?
              – Guacho Perez
              Nov 25 at 18:06










            • Forget it thanks haha
              – Jazmín Jones
              Nov 25 at 18:14














            • 2




              I guess @JazminJones mentioned "radius of circle being 2". So if thats the case $gamma=2e^{it}$ and the integral evaluates to $32pi i$.
              – Yadati Kiran
              Nov 25 at 7:47












            • $32cdot2pi i$ not $32pi i$.
              – Yadati Kiran
              Nov 25 at 8:37












            • can you explain the equality in more detail please
              – Jazmín Jones
              Nov 25 at 17:09










            • @JazminJones which one?
              – Guacho Perez
              Nov 25 at 18:06










            • Forget it thanks haha
              – Jazmín Jones
              Nov 25 at 18:14








            2




            2




            I guess @JazminJones mentioned "radius of circle being 2". So if thats the case $gamma=2e^{it}$ and the integral evaluates to $32pi i$.
            – Yadati Kiran
            Nov 25 at 7:47






            I guess @JazminJones mentioned "radius of circle being 2". So if thats the case $gamma=2e^{it}$ and the integral evaluates to $32pi i$.
            – Yadati Kiran
            Nov 25 at 7:47














            $32cdot2pi i$ not $32pi i$.
            – Yadati Kiran
            Nov 25 at 8:37






            $32cdot2pi i$ not $32pi i$.
            – Yadati Kiran
            Nov 25 at 8:37














            can you explain the equality in more detail please
            – Jazmín Jones
            Nov 25 at 17:09




            can you explain the equality in more detail please
            – Jazmín Jones
            Nov 25 at 17:09












            @JazminJones which one?
            – Guacho Perez
            Nov 25 at 18:06




            @JazminJones which one?
            – Guacho Perez
            Nov 25 at 18:06












            Forget it thanks haha
            – Jazmín Jones
            Nov 25 at 18:14




            Forget it thanks haha
            – Jazmín Jones
            Nov 25 at 18:14











            0














            In your case the integrand is



            $$(rho e^{itheta})^2(rho e^{-itheta})^3de^{itheta}=2^5e^{-itheta}de^{itheta}$$ giving the integral



            $$left.32iright|_0^{2pi}.$$






            share|cite|improve this answer


























              0














              In your case the integrand is



              $$(rho e^{itheta})^2(rho e^{-itheta})^3de^{itheta}=2^5e^{-itheta}de^{itheta}$$ giving the integral



              $$left.32iright|_0^{2pi}.$$






              share|cite|improve this answer
























                0












                0








                0






                In your case the integrand is



                $$(rho e^{itheta})^2(rho e^{-itheta})^3de^{itheta}=2^5e^{-itheta}de^{itheta}$$ giving the integral



                $$left.32iright|_0^{2pi}.$$






                share|cite|improve this answer












                In your case the integrand is



                $$(rho e^{itheta})^2(rho e^{-itheta})^3de^{itheta}=2^5e^{-itheta}de^{itheta}$$ giving the integral



                $$left.32iright|_0^{2pi}.$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 25 at 8:25









                Yves Daoust

                124k671221




                124k671221






























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