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Computation of integral $int_rho fracdz(z-a)(z-b)$ [duplicate]



The 2019 Stack Overflow Developer Survey Results Are InComplex integral over a circleintegral of complex logarithmResidue Calculus Integral computationHow to calculate $ int_0^infty frac x^2 log(x) 1 + x^4 $?complex integration, how to evaluate it?Evaluate $I = frac 12pi i int_gamma_ rho frac 1z^n , dz$Explain me the formula of the inverse of a complex number, or how does $frac1rho cdot cis(theta)$ become $frac1rhocdot cis(-theta)$Complex Integrals to find infinite integral on the real line.Compute the integral $int_z|z-a|^-4|dz|$ with $|a|neq rho$Stuck on contour integral, can I use Cauchy's theorem instead?Finding contour integral $ int_gamma fracmathrmIm (z)z - alpha dz $










0












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This question already has an answer here:



  • Complex integral over a circle

    3 answers



Let $a,b$ be complex number and $|a| < r < |b|$, compute.



$int_rho fracdz(z-a)(z-b)$



where $rho$ is the circle with radius $r$ and the usual orientation.



I've tried the common path $int_rho fracdz(z-a)(z-b) = int_0^2pifracrie^i theta dtheta(re^i theta -a)(re^i theta-b)$ but in this point I don't know how use the $r$










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marked as duplicate by Arnaud D., Lord Shark the Unknown, user477343, Cesareo, Joel Reyes Noche 2 days ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






















    0












    $begingroup$



    This question already has an answer here:



    • Complex integral over a circle

      3 answers



    Let $a,b$ be complex number and $|a| < r < |b|$, compute.



    $int_rho fracdz(z-a)(z-b)$



    where $rho$ is the circle with radius $r$ and the usual orientation.



    I've tried the common path $int_rho fracdz(z-a)(z-b) = int_0^2pifracrie^i theta dtheta(re^i theta -a)(re^i theta-b)$ but in this point I don't know how use the $r$










    share|cite|improve this question











    $endgroup$



    marked as duplicate by Arnaud D., Lord Shark the Unknown, user477343, Cesareo, Joel Reyes Noche 2 days ago


    This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.




















      0












      0








      0





      $begingroup$



      This question already has an answer here:



      • Complex integral over a circle

        3 answers



      Let $a,b$ be complex number and $|a| < r < |b|$, compute.



      $int_rho fracdz(z-a)(z-b)$



      where $rho$ is the circle with radius $r$ and the usual orientation.



      I've tried the common path $int_rho fracdz(z-a)(z-b) = int_0^2pifracrie^i theta dtheta(re^i theta -a)(re^i theta-b)$ but in this point I don't know how use the $r$










      share|cite|improve this question











      $endgroup$





      This question already has an answer here:



      • Complex integral over a circle

        3 answers



      Let $a,b$ be complex number and $|a| < r < |b|$, compute.



      $int_rho fracdz(z-a)(z-b)$



      where $rho$ is the circle with radius $r$ and the usual orientation.



      I've tried the common path $int_rho fracdz(z-a)(z-b) = int_0^2pifracrie^i theta dtheta(re^i theta -a)(re^i theta-b)$ but in this point I don't know how use the $r$





      This question already has an answer here:



      • Complex integral over a circle

        3 answers







      complex-analysis complex-numbers contour-integration residue-calculus






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Apr 7 at 19:31









      Arnaud D.

      16.2k52445




      16.2k52445










      asked Apr 7 at 19:29









      jfruizc273jfruizc273

      126




      126




      marked as duplicate by Arnaud D., Lord Shark the Unknown, user477343, Cesareo, Joel Reyes Noche 2 days ago


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









      marked as duplicate by Arnaud D., Lord Shark the Unknown, user477343, Cesareo, Joel Reyes Noche 2 days ago


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






















          1 Answer
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          active

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          0












          $begingroup$

          By residue theorem the integral is
          $$
          2pi itext Res_z=afrac1 (z-a)(z-b)=frac 2pi ia-b,
          $$

          since only the pole $z=a $ is inside the integration contour.






          share|cite|improve this answer









          $endgroup$



















            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            0












            $begingroup$

            By residue theorem the integral is
            $$
            2pi itext Res_z=afrac1 (z-a)(z-b)=frac 2pi ia-b,
            $$

            since only the pole $z=a $ is inside the integration contour.






            share|cite|improve this answer









            $endgroup$

















              0












              $begingroup$

              By residue theorem the integral is
              $$
              2pi itext Res_z=afrac1 (z-a)(z-b)=frac 2pi ia-b,
              $$

              since only the pole $z=a $ is inside the integration contour.






              share|cite|improve this answer









              $endgroup$















                0












                0








                0





                $begingroup$

                By residue theorem the integral is
                $$
                2pi itext Res_z=afrac1 (z-a)(z-b)=frac 2pi ia-b,
                $$

                since only the pole $z=a $ is inside the integration contour.






                share|cite|improve this answer









                $endgroup$



                By residue theorem the integral is
                $$
                2pi itext Res_z=afrac1 (z-a)(z-b)=frac 2pi ia-b,
                $$

                since only the pole $z=a $ is inside the integration contour.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Apr 7 at 20:19









                useruser

                6,47811031




                6,47811031













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