How to evaluate a complex integral? [closed]












1














Show that |$int_{C} frac{e^z}{bar z + 1} dz$| ≤ $2pi e^2$ where C is the circle |z-1| = 1.



I think C can be parametrized by z = $1 + e^{it}$ and $z' = ie^{it}dt$



with 0 ≤ t ≤ $2 pi$



How do i have to continue this proof?






complex-analysis complex-integration







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closed as off-topic by Brahadeesh, Did, Alexander Gruber♦ Nov 30 '18 at 3:18


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." – Brahadeesh, Did, Alexander Gruber

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


















    1














    Show that |$int_{C} frac{e^z}{bar z + 1} dz$| ≤ $2pi e^2$ where C is the circle |z-1| = 1.



    I think C can be parametrized by z = $1 + e^{it}$ and $z' = ie^{it}dt$



    with 0 ≤ t ≤ $2 pi$



    How do i have to continue this proof?






    complex-analysis complex-integration







    share|cite|improve this question













    closed as off-topic by Brahadeesh, Did, Alexander Gruber♦ Nov 30 '18 at 3:18


    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." – Brahadeesh, Did, Alexander Gruber

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
















      1












      1








      1







      Show that |$int_{C} frac{e^z}{bar z + 1} dz$| ≤ $2pi e^2$ where C is the circle |z-1| = 1.



      I think C can be parametrized by z = $1 + e^{it}$ and $z' = ie^{it}dt$



      with 0 ≤ t ≤ $2 pi$



      How do i have to continue this proof?






      complex-analysis complex-integration







      share|cite|improve this question













      Show that |$int_{C} frac{e^z}{bar z + 1} dz$| ≤ $2pi e^2$ where C is the circle |z-1| = 1.



      I think C can be parametrized by z = $1 + e^{it}$ and $z' = ie^{it}dt$



      with 0 ≤ t ≤ $2 pi$



      How do i have to continue this proof?






      complex-analysis complex-integration




      complex-analysis complex-integration






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 29 '18 at 12:13









      Peter van de Berg

      198




      198




      closed as off-topic by Brahadeesh, Did, Alexander Gruber♦ Nov 30 '18 at 3:18


      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." – Brahadeesh, Did, Alexander Gruber

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




      closed as off-topic by Brahadeesh, Did, Alexander Gruber♦ Nov 30 '18 at 3:18


      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." – Brahadeesh, Did, Alexander Gruber

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






















          1 Answer
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          By the ML inequality, we have that



          $$Bigg|int_Gamma f(z)mathrm{d}zBigg| leq max_{z in Gamma}f(z) cdot l(Gamma)$$



          where $l(Gamma)$ is the arc length of $Gamma$.



          Thus, for your given integral, it is :



          $$Bigg|int_C frac{e^z}{bar{z}+1}mathrm{d}zBigg| leq max_{z in Gamma}Bigg{frac{e^z}{bar{z}+1}Bigg}cdot l(Gamma)$$



          But $C$ is the circle $|z-1| = 1$, which means that it has radius $r=1$ and thus its length is $l(Gamma) = 2pi$. Can you use complex algebra now and make a case for the maximum of the integrand function ?






          share|cite|improve this answer





















          • Thanks, i already see what i have to do!
            – Peter van de Berg
            Nov 29 '18 at 12:30










          • @PetervandeBerg Glad I could help !
            – Rebellos
            Nov 29 '18 at 12:33


















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1














          By the ML inequality, we have that



          $$Bigg|int_Gamma f(z)mathrm{d}zBigg| leq max_{z in Gamma}f(z) cdot l(Gamma)$$



          where $l(Gamma)$ is the arc length of $Gamma$.



          Thus, for your given integral, it is :



          $$Bigg|int_C frac{e^z}{bar{z}+1}mathrm{d}zBigg| leq max_{z in Gamma}Bigg{frac{e^z}{bar{z}+1}Bigg}cdot l(Gamma)$$



          But $C$ is the circle $|z-1| = 1$, which means that it has radius $r=1$ and thus its length is $l(Gamma) = 2pi$. Can you use complex algebra now and make a case for the maximum of the integrand function ?






          share|cite|improve this answer





















          • Thanks, i already see what i have to do!
            – Peter van de Berg
            Nov 29 '18 at 12:30










          • @PetervandeBerg Glad I could help !
            – Rebellos
            Nov 29 '18 at 12:33
















          1














          By the ML inequality, we have that



          $$Bigg|int_Gamma f(z)mathrm{d}zBigg| leq max_{z in Gamma}f(z) cdot l(Gamma)$$



          where $l(Gamma)$ is the arc length of $Gamma$.



          Thus, for your given integral, it is :



          $$Bigg|int_C frac{e^z}{bar{z}+1}mathrm{d}zBigg| leq max_{z in Gamma}Bigg{frac{e^z}{bar{z}+1}Bigg}cdot l(Gamma)$$



          But $C$ is the circle $|z-1| = 1$, which means that it has radius $r=1$ and thus its length is $l(Gamma) = 2pi$. Can you use complex algebra now and make a case for the maximum of the integrand function ?






          share|cite|improve this answer





















          • Thanks, i already see what i have to do!
            – Peter van de Berg
            Nov 29 '18 at 12:30










          • @PetervandeBerg Glad I could help !
            – Rebellos
            Nov 29 '18 at 12:33














          1












          1








          1






          By the ML inequality, we have that



          $$Bigg|int_Gamma f(z)mathrm{d}zBigg| leq max_{z in Gamma}f(z) cdot l(Gamma)$$



          where $l(Gamma)$ is the arc length of $Gamma$.



          Thus, for your given integral, it is :



          $$Bigg|int_C frac{e^z}{bar{z}+1}mathrm{d}zBigg| leq max_{z in Gamma}Bigg{frac{e^z}{bar{z}+1}Bigg}cdot l(Gamma)$$



          But $C$ is the circle $|z-1| = 1$, which means that it has radius $r=1$ and thus its length is $l(Gamma) = 2pi$. Can you use complex algebra now and make a case for the maximum of the integrand function ?






          share|cite|improve this answer












          By the ML inequality, we have that



          $$Bigg|int_Gamma f(z)mathrm{d}zBigg| leq max_{z in Gamma}f(z) cdot l(Gamma)$$



          where $l(Gamma)$ is the arc length of $Gamma$.



          Thus, for your given integral, it is :



          $$Bigg|int_C frac{e^z}{bar{z}+1}mathrm{d}zBigg| leq max_{z in Gamma}Bigg{frac{e^z}{bar{z}+1}Bigg}cdot l(Gamma)$$



          But $C$ is the circle $|z-1| = 1$, which means that it has radius $r=1$ and thus its length is $l(Gamma) = 2pi$. Can you use complex algebra now and make a case for the maximum of the integrand function ?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 29 '18 at 12:26









          Rebellos

          14.5k31245




          14.5k31245












          • Thanks, i already see what i have to do!
            – Peter van de Berg
            Nov 29 '18 at 12:30










          • @PetervandeBerg Glad I could help !
            – Rebellos
            Nov 29 '18 at 12:33


















          • Thanks, i already see what i have to do!
            – Peter van de Berg
            Nov 29 '18 at 12:30










          • @PetervandeBerg Glad I could help !
            – Rebellos
            Nov 29 '18 at 12:33
















          Thanks, i already see what i have to do!
          – Peter van de Berg
          Nov 29 '18 at 12:30




          Thanks, i already see what i have to do!
          – Peter van de Berg
          Nov 29 '18 at 12:30












          @PetervandeBerg Glad I could help !
          – Rebellos
          Nov 29 '18 at 12:33




          @PetervandeBerg Glad I could help !
          – Rebellos
          Nov 29 '18 at 12:33



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