Is this true? $Reint{f(z)dz}=int{Re(f(z))dz}$ [closed]












1












$begingroup$


I have to say if this is true or not and why.



Let f a complex function, then $$Reint_{gamma}{f(z)dz}=int_{gamma}{Re(f(z))dz}$$










share|cite|improve this question











$endgroup$



closed as off-topic by T. Bongers, Henrik, Kavi Rama Murthy, Cesareo, Chinnapparaj R Dec 8 '18 at 4:56


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – T. Bongers, Henrik, Kavi Rama Murthy, Cesareo, Chinnapparaj R

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 2




    $begingroup$
    No. ${}{}{}{}{}$
    $endgroup$
    – T. Bongers
    Dec 7 '18 at 22:17






  • 2




    $begingroup$
    How about $dz$? Try integrating along an arc of the unit cicle $z=e^{i theta}$.
    $endgroup$
    – mlerma54
    Dec 7 '18 at 22:32










  • $begingroup$
    Consider that the imaginary part of $f(z)$ could integrate to a real value, or vice versa. It depends a lot on the path $gamma$.
    $endgroup$
    – zahbaz
    Dec 7 '18 at 23:13
















1












$begingroup$


I have to say if this is true or not and why.



Let f a complex function, then $$Reint_{gamma}{f(z)dz}=int_{gamma}{Re(f(z))dz}$$










share|cite|improve this question











$endgroup$



closed as off-topic by T. Bongers, Henrik, Kavi Rama Murthy, Cesareo, Chinnapparaj R Dec 8 '18 at 4:56


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – T. Bongers, Henrik, Kavi Rama Murthy, Cesareo, Chinnapparaj R

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 2




    $begingroup$
    No. ${}{}{}{}{}$
    $endgroup$
    – T. Bongers
    Dec 7 '18 at 22:17






  • 2




    $begingroup$
    How about $dz$? Try integrating along an arc of the unit cicle $z=e^{i theta}$.
    $endgroup$
    – mlerma54
    Dec 7 '18 at 22:32










  • $begingroup$
    Consider that the imaginary part of $f(z)$ could integrate to a real value, or vice versa. It depends a lot on the path $gamma$.
    $endgroup$
    – zahbaz
    Dec 7 '18 at 23:13














1












1








1


0



$begingroup$


I have to say if this is true or not and why.



Let f a complex function, then $$Reint_{gamma}{f(z)dz}=int_{gamma}{Re(f(z))dz}$$










share|cite|improve this question











$endgroup$




I have to say if this is true or not and why.



Let f a complex function, then $$Reint_{gamma}{f(z)dz}=int_{gamma}{Re(f(z))dz}$$







integration complex-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 '18 at 23:07









Jean-Claude Arbaut

14.8k63464




14.8k63464










asked Dec 7 '18 at 22:13









Juan De Dios RojasJuan De Dios Rojas

222




222




closed as off-topic by T. Bongers, Henrik, Kavi Rama Murthy, Cesareo, Chinnapparaj R Dec 8 '18 at 4:56


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – T. Bongers, Henrik, Kavi Rama Murthy, Cesareo, Chinnapparaj R

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by T. Bongers, Henrik, Kavi Rama Murthy, Cesareo, Chinnapparaj R Dec 8 '18 at 4:56


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – T. Bongers, Henrik, Kavi Rama Murthy, Cesareo, Chinnapparaj R

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 2




    $begingroup$
    No. ${}{}{}{}{}$
    $endgroup$
    – T. Bongers
    Dec 7 '18 at 22:17






  • 2




    $begingroup$
    How about $dz$? Try integrating along an arc of the unit cicle $z=e^{i theta}$.
    $endgroup$
    – mlerma54
    Dec 7 '18 at 22:32










  • $begingroup$
    Consider that the imaginary part of $f(z)$ could integrate to a real value, or vice versa. It depends a lot on the path $gamma$.
    $endgroup$
    – zahbaz
    Dec 7 '18 at 23:13














  • 2




    $begingroup$
    No. ${}{}{}{}{}$
    $endgroup$
    – T. Bongers
    Dec 7 '18 at 22:17






  • 2




    $begingroup$
    How about $dz$? Try integrating along an arc of the unit cicle $z=e^{i theta}$.
    $endgroup$
    – mlerma54
    Dec 7 '18 at 22:32










  • $begingroup$
    Consider that the imaginary part of $f(z)$ could integrate to a real value, or vice versa. It depends a lot on the path $gamma$.
    $endgroup$
    – zahbaz
    Dec 7 '18 at 23:13








2




2




$begingroup$
No. ${}{}{}{}{}$
$endgroup$
– T. Bongers
Dec 7 '18 at 22:17




$begingroup$
No. ${}{}{}{}{}$
$endgroup$
– T. Bongers
Dec 7 '18 at 22:17




2




2




$begingroup$
How about $dz$? Try integrating along an arc of the unit cicle $z=e^{i theta}$.
$endgroup$
– mlerma54
Dec 7 '18 at 22:32




$begingroup$
How about $dz$? Try integrating along an arc of the unit cicle $z=e^{i theta}$.
$endgroup$
– mlerma54
Dec 7 '18 at 22:32












$begingroup$
Consider that the imaginary part of $f(z)$ could integrate to a real value, or vice versa. It depends a lot on the path $gamma$.
$endgroup$
– zahbaz
Dec 7 '18 at 23:13




$begingroup$
Consider that the imaginary part of $f(z)$ could integrate to a real value, or vice versa. It depends a lot on the path $gamma$.
$endgroup$
– zahbaz
Dec 7 '18 at 23:13










2 Answers
2






active

oldest

votes


















3












$begingroup$

$$Reint_{gamma}{frac{i}{z}text{ } dz}=-2pi$$



$$int_{0}^{2pi}-Refrac{1}{e^{it} }dt=int_{0}^{2pi}-costdt=0$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    $dz$ is complex for a general path $gamma$. Your statement would be true if $gamma$ was on the real line. The point is that for two complex numbers $a$ and $b$, $Re(ab) neq Re(a)b$. Indeed, they are equal iff $b$ is real.






    share|cite|improve this answer









    $endgroup$




















      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      $$Reint_{gamma}{frac{i}{z}text{ } dz}=-2pi$$



      $$int_{0}^{2pi}-Refrac{1}{e^{it} }dt=int_{0}^{2pi}-costdt=0$$






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        $$Reint_{gamma}{frac{i}{z}text{ } dz}=-2pi$$



        $$int_{0}^{2pi}-Refrac{1}{e^{it} }dt=int_{0}^{2pi}-costdt=0$$






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          $$Reint_{gamma}{frac{i}{z}text{ } dz}=-2pi$$



          $$int_{0}^{2pi}-Refrac{1}{e^{it} }dt=int_{0}^{2pi}-costdt=0$$






          share|cite|improve this answer









          $endgroup$



          $$Reint_{gamma}{frac{i}{z}text{ } dz}=-2pi$$



          $$int_{0}^{2pi}-Refrac{1}{e^{it} }dt=int_{0}^{2pi}-costdt=0$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 7 '18 at 23:01









          ryszard egginkryszard eggink

          413110




          413110























              1












              $begingroup$

              $dz$ is complex for a general path $gamma$. Your statement would be true if $gamma$ was on the real line. The point is that for two complex numbers $a$ and $b$, $Re(ab) neq Re(a)b$. Indeed, they are equal iff $b$ is real.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                $dz$ is complex for a general path $gamma$. Your statement would be true if $gamma$ was on the real line. The point is that for two complex numbers $a$ and $b$, $Re(ab) neq Re(a)b$. Indeed, they are equal iff $b$ is real.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  $dz$ is complex for a general path $gamma$. Your statement would be true if $gamma$ was on the real line. The point is that for two complex numbers $a$ and $b$, $Re(ab) neq Re(a)b$. Indeed, they are equal iff $b$ is real.






                  share|cite|improve this answer









                  $endgroup$



                  $dz$ is complex for a general path $gamma$. Your statement would be true if $gamma$ was on the real line. The point is that for two complex numbers $a$ and $b$, $Re(ab) neq Re(a)b$. Indeed, they are equal iff $b$ is real.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 8 '18 at 0:22









                  zoidbergzoidberg

                  1,080113




                  1,080113















                      Popular posts from this blog

                      mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924

                      How to change which sound is reproduced for terminal bell?

                      Can I use Tabulator js library in my java Spring + Thymeleaf project?