Complex functions average [closed]

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If $f(z)$ is analitic inside and over C, a circle centered in $a$. Show that the average of $f(z)$ in C is $f(a)$.



$barf(z)=fracint_C f(z) dzint_C dz$



But if I use Cauchy-integral-formula $int_C f(z) dz=0$, so I do not know what I have to do.







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closed as off-topic by Greg Martin, Matthew Leingang, José Carlos Santos, T. Bongers, Leucippus Aug 29 at 0:10


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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Greg Martin, Matthew Leingang, José Carlos Santos, T. Bongers, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.












  • You should be able to find this in basically any complex analysis textbook.
    – Sobi
    Aug 28 at 17:18










  • What are your thoughts? What have you tried? Where are you stuck? You need to provide context for your question. Right now, it just looks like you want somebody to do your homework for you; that's not what this site is for. If you add some appropriate context, we will be happy to help.
    – Greg Martin
    Aug 28 at 17:20










  • You should show your steps for arriving at $int_C f(z)dz=0$.
    – Dave
    Aug 28 at 17:35










  • Because in Cauchy integral formula, the closed integral of an analitic function in this region is zero.
    – user557651
    Aug 28 at 17:40














up vote
-2
down vote

favorite












If $f(z)$ is analitic inside and over C, a circle centered in $a$. Show that the average of $f(z)$ in C is $f(a)$.



$barf(z)=fracint_C f(z) dzint_C dz$



But if I use Cauchy-integral-formula $int_C f(z) dz=0$, so I do not know what I have to do.







share|cite|improve this question














closed as off-topic by Greg Martin, Matthew Leingang, José Carlos Santos, T. Bongers, Leucippus Aug 29 at 0:10


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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Greg Martin, Matthew Leingang, José Carlos Santos, T. Bongers, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.












  • You should be able to find this in basically any complex analysis textbook.
    – Sobi
    Aug 28 at 17:18










  • What are your thoughts? What have you tried? Where are you stuck? You need to provide context for your question. Right now, it just looks like you want somebody to do your homework for you; that's not what this site is for. If you add some appropriate context, we will be happy to help.
    – Greg Martin
    Aug 28 at 17:20










  • You should show your steps for arriving at $int_C f(z)dz=0$.
    – Dave
    Aug 28 at 17:35










  • Because in Cauchy integral formula, the closed integral of an analitic function in this region is zero.
    – user557651
    Aug 28 at 17:40












up vote
-2
down vote

favorite









up vote
-2
down vote

favorite











If $f(z)$ is analitic inside and over C, a circle centered in $a$. Show that the average of $f(z)$ in C is $f(a)$.



$barf(z)=fracint_C f(z) dzint_C dz$



But if I use Cauchy-integral-formula $int_C f(z) dz=0$, so I do not know what I have to do.







share|cite|improve this question














If $f(z)$ is analitic inside and over C, a circle centered in $a$. Show that the average of $f(z)$ in C is $f(a)$.



$barf(z)=fracint_C f(z) dzint_C dz$



But if I use Cauchy-integral-formula $int_C f(z) dz=0$, so I do not know what I have to do.









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 28 at 17:32

























asked Aug 28 at 17:10









user557651

373




373




closed as off-topic by Greg Martin, Matthew Leingang, José Carlos Santos, T. Bongers, Leucippus Aug 29 at 0:10


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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Greg Martin, Matthew Leingang, José Carlos Santos, T. Bongers, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Greg Martin, Matthew Leingang, José Carlos Santos, T. Bongers, Leucippus Aug 29 at 0:10


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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Greg Martin, Matthew Leingang, José Carlos Santos, T. Bongers, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.











  • You should be able to find this in basically any complex analysis textbook.
    – Sobi
    Aug 28 at 17:18










  • What are your thoughts? What have you tried? Where are you stuck? You need to provide context for your question. Right now, it just looks like you want somebody to do your homework for you; that's not what this site is for. If you add some appropriate context, we will be happy to help.
    – Greg Martin
    Aug 28 at 17:20










  • You should show your steps for arriving at $int_C f(z)dz=0$.
    – Dave
    Aug 28 at 17:35










  • Because in Cauchy integral formula, the closed integral of an analitic function in this region is zero.
    – user557651
    Aug 28 at 17:40
















  • You should be able to find this in basically any complex analysis textbook.
    – Sobi
    Aug 28 at 17:18










  • What are your thoughts? What have you tried? Where are you stuck? You need to provide context for your question. Right now, it just looks like you want somebody to do your homework for you; that's not what this site is for. If you add some appropriate context, we will be happy to help.
    – Greg Martin
    Aug 28 at 17:20










  • You should show your steps for arriving at $int_C f(z)dz=0$.
    – Dave
    Aug 28 at 17:35










  • Because in Cauchy integral formula, the closed integral of an analitic function in this region is zero.
    – user557651
    Aug 28 at 17:40















You should be able to find this in basically any complex analysis textbook.
– Sobi
Aug 28 at 17:18




You should be able to find this in basically any complex analysis textbook.
– Sobi
Aug 28 at 17:18












What are your thoughts? What have you tried? Where are you stuck? You need to provide context for your question. Right now, it just looks like you want somebody to do your homework for you; that's not what this site is for. If you add some appropriate context, we will be happy to help.
– Greg Martin
Aug 28 at 17:20




What are your thoughts? What have you tried? Where are you stuck? You need to provide context for your question. Right now, it just looks like you want somebody to do your homework for you; that's not what this site is for. If you add some appropriate context, we will be happy to help.
– Greg Martin
Aug 28 at 17:20












You should show your steps for arriving at $int_C f(z)dz=0$.
– Dave
Aug 28 at 17:35




You should show your steps for arriving at $int_C f(z)dz=0$.
– Dave
Aug 28 at 17:35












Because in Cauchy integral formula, the closed integral of an analitic function in this region is zero.
– user557651
Aug 28 at 17:40




Because in Cauchy integral formula, the closed integral of an analitic function in this region is zero.
– user557651
Aug 28 at 17:40










1 Answer
1






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1
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Hint: Cauchy integral theorem.






share|cite|improve this answer



























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote













    Hint: Cauchy integral theorem.






    share|cite|improve this answer
























      up vote
      1
      down vote













      Hint: Cauchy integral theorem.






      share|cite|improve this answer






















        up vote
        1
        down vote










        up vote
        1
        down vote









        Hint: Cauchy integral theorem.






        share|cite|improve this answer












        Hint: Cauchy integral theorem.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 28 at 17:17









        Robert Israel

        306k22201443




        306k22201443












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