Compute $int_{|z|=R} frac{|dz|}{|z-a|^4}$ when $R>0, |a|neq R$












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I want to compute $int_{|z|=R} frac{|dz|}{|z-a|^4}$ when $R>0, |a|neq R$.
I did so by parametrizing the circle, but this way required many pages of tedious calculations.



Do you have a more straightforward way to do it?



Let me mention that my issue here is that we integrate by $|dz|$ not by $dz$, so all the known theorems I know fail to be applied -at least in this form of the integral.



Thanks






complex-integration







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    0












    $begingroup$


    I want to compute $int_{|z|=R} frac{|dz|}{|z-a|^4}$ when $R>0, |a|neq R$.
    I did so by parametrizing the circle, but this way required many pages of tedious calculations.



    Do you have a more straightforward way to do it?



    Let me mention that my issue here is that we integrate by $|dz|$ not by $dz$, so all the known theorems I know fail to be applied -at least in this form of the integral.



    Thanks






    complex-integration







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I want to compute $int_{|z|=R} frac{|dz|}{|z-a|^4}$ when $R>0, |a|neq R$.
      I did so by parametrizing the circle, but this way required many pages of tedious calculations.



      Do you have a more straightforward way to do it?



      Let me mention that my issue here is that we integrate by $|dz|$ not by $dz$, so all the known theorems I know fail to be applied -at least in this form of the integral.



      Thanks






      complex-integration







      share|cite|improve this question











      $endgroup$




      I want to compute $int_{|z|=R} frac{|dz|}{|z-a|^4}$ when $R>0, |a|neq R$.
      I did so by parametrizing the circle, but this way required many pages of tedious calculations.



      Do you have a more straightforward way to do it?



      Let me mention that my issue here is that we integrate by $|dz|$ not by $dz$, so all the known theorems I know fail to be applied -at least in this form of the integral.



      Thanks






      complex-integration




      complex-integration






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Oct 24 '17 at 5:28







      perlman

















      asked Oct 24 '17 at 5:22









      perlmanperlman

      8110




      8110






















          2 Answers
          2






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          0












          $begingroup$

          Try This



          $f^n(z)$ = $frac {n!}{2(pi)n}$ $int$ $frac {f(w) dw}{(w-z)^{n+1}}$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Can you make it a bit more clear? Specifically, how do I treat the $|dz|$?
            $endgroup$
            – perlman
            Oct 24 '17 at 5:47










          • $begingroup$
            I just assumed the absolute value sign was meant to be around the function. I don't think I've specifically dealt with a case you're talking about, because I tried googling it, and |dz| turns out to be some kind of arc length.
            $endgroup$
            – Math Model
            Oct 24 '17 at 5:57



















          0












          $begingroup$

          After parametrising $z= R e^{itheta}$ and after a lot of simplifications you will get
          $$I = int_{0}^{2pi} dtheta frac{R}{[(Rcostheta-Re(a))^2+(Rsintheta-Im(a))^2]^2_{}}.$$
          To proceed further : go back to complex variable $Z = e^{itheta}$, rewrite it in terms of $costheta=frac{1}{2}(z+frac{1}{z})$ and $sintheta=frac{1}{2i}(z-frac{1}{z})$. The resulting integral can be evaluated using complex integration methods.






          share|cite|improve this answer









          $endgroup$














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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            0












            $begingroup$

            Try This



            $f^n(z)$ = $frac {n!}{2(pi)n}$ $int$ $frac {f(w) dw}{(w-z)^{n+1}}$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Can you make it a bit more clear? Specifically, how do I treat the $|dz|$?
              $endgroup$
              – perlman
              Oct 24 '17 at 5:47










            • $begingroup$
              I just assumed the absolute value sign was meant to be around the function. I don't think I've specifically dealt with a case you're talking about, because I tried googling it, and |dz| turns out to be some kind of arc length.
              $endgroup$
              – Math Model
              Oct 24 '17 at 5:57
















            0












            $begingroup$

            Try This



            $f^n(z)$ = $frac {n!}{2(pi)n}$ $int$ $frac {f(w) dw}{(w-z)^{n+1}}$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Can you make it a bit more clear? Specifically, how do I treat the $|dz|$?
              $endgroup$
              – perlman
              Oct 24 '17 at 5:47










            • $begingroup$
              I just assumed the absolute value sign was meant to be around the function. I don't think I've specifically dealt with a case you're talking about, because I tried googling it, and |dz| turns out to be some kind of arc length.
              $endgroup$
              – Math Model
              Oct 24 '17 at 5:57














            0












            0








            0





            $begingroup$

            Try This



            $f^n(z)$ = $frac {n!}{2(pi)n}$ $int$ $frac {f(w) dw}{(w-z)^{n+1}}$






            share|cite|improve this answer











            $endgroup$



            Try This



            $f^n(z)$ = $frac {n!}{2(pi)n}$ $int$ $frac {f(w) dw}{(w-z)^{n+1}}$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Oct 24 '17 at 5:55

























            answered Oct 24 '17 at 5:41









            Math ModelMath Model

            445




            445












            • $begingroup$
              Can you make it a bit more clear? Specifically, how do I treat the $|dz|$?
              $endgroup$
              – perlman
              Oct 24 '17 at 5:47










            • $begingroup$
              I just assumed the absolute value sign was meant to be around the function. I don't think I've specifically dealt with a case you're talking about, because I tried googling it, and |dz| turns out to be some kind of arc length.
              $endgroup$
              – Math Model
              Oct 24 '17 at 5:57


















            • $begingroup$
              Can you make it a bit more clear? Specifically, how do I treat the $|dz|$?
              $endgroup$
              – perlman
              Oct 24 '17 at 5:47










            • $begingroup$
              I just assumed the absolute value sign was meant to be around the function. I don't think I've specifically dealt with a case you're talking about, because I tried googling it, and |dz| turns out to be some kind of arc length.
              $endgroup$
              – Math Model
              Oct 24 '17 at 5:57
















            $begingroup$
            Can you make it a bit more clear? Specifically, how do I treat the $|dz|$?
            $endgroup$
            – perlman
            Oct 24 '17 at 5:47




            $begingroup$
            Can you make it a bit more clear? Specifically, how do I treat the $|dz|$?
            $endgroup$
            – perlman
            Oct 24 '17 at 5:47












            $begingroup$
            I just assumed the absolute value sign was meant to be around the function. I don't think I've specifically dealt with a case you're talking about, because I tried googling it, and |dz| turns out to be some kind of arc length.
            $endgroup$
            – Math Model
            Oct 24 '17 at 5:57




            $begingroup$
            I just assumed the absolute value sign was meant to be around the function. I don't think I've specifically dealt with a case you're talking about, because I tried googling it, and |dz| turns out to be some kind of arc length.
            $endgroup$
            – Math Model
            Oct 24 '17 at 5:57











            0












            $begingroup$

            After parametrising $z= R e^{itheta}$ and after a lot of simplifications you will get
            $$I = int_{0}^{2pi} dtheta frac{R}{[(Rcostheta-Re(a))^2+(Rsintheta-Im(a))^2]^2_{}}.$$
            To proceed further : go back to complex variable $Z = e^{itheta}$, rewrite it in terms of $costheta=frac{1}{2}(z+frac{1}{z})$ and $sintheta=frac{1}{2i}(z-frac{1}{z})$. The resulting integral can be evaluated using complex integration methods.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              After parametrising $z= R e^{itheta}$ and after a lot of simplifications you will get
              $$I = int_{0}^{2pi} dtheta frac{R}{[(Rcostheta-Re(a))^2+(Rsintheta-Im(a))^2]^2_{}}.$$
              To proceed further : go back to complex variable $Z = e^{itheta}$, rewrite it in terms of $costheta=frac{1}{2}(z+frac{1}{z})$ and $sintheta=frac{1}{2i}(z-frac{1}{z})$. The resulting integral can be evaluated using complex integration methods.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                After parametrising $z= R e^{itheta}$ and after a lot of simplifications you will get
                $$I = int_{0}^{2pi} dtheta frac{R}{[(Rcostheta-Re(a))^2+(Rsintheta-Im(a))^2]^2_{}}.$$
                To proceed further : go back to complex variable $Z = e^{itheta}$, rewrite it in terms of $costheta=frac{1}{2}(z+frac{1}{z})$ and $sintheta=frac{1}{2i}(z-frac{1}{z})$. The resulting integral can be evaluated using complex integration methods.






                share|cite|improve this answer









                $endgroup$



                After parametrising $z= R e^{itheta}$ and after a lot of simplifications you will get
                $$I = int_{0}^{2pi} dtheta frac{R}{[(Rcostheta-Re(a))^2+(Rsintheta-Im(a))^2]^2_{}}.$$
                To proceed further : go back to complex variable $Z = e^{itheta}$, rewrite it in terms of $costheta=frac{1}{2}(z+frac{1}{z})$ and $sintheta=frac{1}{2i}(z-frac{1}{z})$. The resulting integral can be evaluated using complex integration methods.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Oct 24 '17 at 6:34









                SunyamSunyam

                167214




                167214






























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