A line integral equals zero implies a real integral also is zero












1












$begingroup$


I'm asked to check that the following line integral is zero:



$$int_{C(0,r)} frac {log(1+z)}z dz=0$$



(where $C(0,r)$ is the circle of radius $r$ centered at $0$) and then to conclude that for every $rin (0,1)$ the following integral is also $0$:



$$int_0^{pi}log(1+r^2+2rcos t)dt=0$$



I've managed to calculated the first integral expanding in Taylor series (we didn't do any residue yet in class) and I can imagine that it is connected in some way with the second, however I cannot find out exactly how.






complex-analysis complex-numbers complex-integration line-integrals







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$endgroup$












  • $begingroup$
    Write $z=re^{itheta}$ and remember how log works.
    $endgroup$
    – user10354138
    Dec 7 '18 at 18:15










  • $begingroup$
    $log(1+z)/z$ is holomorphic in the region bounded by $C(0,r)$, so en.wikipedia.org/wiki/Cauchy%27s_integral_theorem
    $endgroup$
    – Federico
    Dec 7 '18 at 18:31
















1












$begingroup$


I'm asked to check that the following line integral is zero:



$$int_{C(0,r)} frac {log(1+z)}z dz=0$$



(where $C(0,r)$ is the circle of radius $r$ centered at $0$) and then to conclude that for every $rin (0,1)$ the following integral is also $0$:



$$int_0^{pi}log(1+r^2+2rcos t)dt=0$$



I've managed to calculated the first integral expanding in Taylor series (we didn't do any residue yet in class) and I can imagine that it is connected in some way with the second, however I cannot find out exactly how.






complex-analysis complex-numbers complex-integration line-integrals







share|cite|improve this question









$endgroup$












  • $begingroup$
    Write $z=re^{itheta}$ and remember how log works.
    $endgroup$
    – user10354138
    Dec 7 '18 at 18:15










  • $begingroup$
    $log(1+z)/z$ is holomorphic in the region bounded by $C(0,r)$, so en.wikipedia.org/wiki/Cauchy%27s_integral_theorem
    $endgroup$
    – Federico
    Dec 7 '18 at 18:31














1












1








1





$begingroup$


I'm asked to check that the following line integral is zero:



$$int_{C(0,r)} frac {log(1+z)}z dz=0$$



(where $C(0,r)$ is the circle of radius $r$ centered at $0$) and then to conclude that for every $rin (0,1)$ the following integral is also $0$:



$$int_0^{pi}log(1+r^2+2rcos t)dt=0$$



I've managed to calculated the first integral expanding in Taylor series (we didn't do any residue yet in class) and I can imagine that it is connected in some way with the second, however I cannot find out exactly how.






complex-analysis complex-numbers complex-integration line-integrals







share|cite|improve this question









$endgroup$




I'm asked to check that the following line integral is zero:



$$int_{C(0,r)} frac {log(1+z)}z dz=0$$



(where $C(0,r)$ is the circle of radius $r$ centered at $0$) and then to conclude that for every $rin (0,1)$ the following integral is also $0$:



$$int_0^{pi}log(1+r^2+2rcos t)dt=0$$



I've managed to calculated the first integral expanding in Taylor series (we didn't do any residue yet in class) and I can imagine that it is connected in some way with the second, however I cannot find out exactly how.






complex-analysis complex-numbers complex-integration line-integrals




complex-analysis complex-numbers complex-integration line-integrals






share|cite|improve this question













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asked Dec 7 '18 at 18:10









Renato FaraoneRenato Faraone

2,33111627




2,33111627












  • $begingroup$
    Write $z=re^{itheta}$ and remember how log works.
    $endgroup$
    – user10354138
    Dec 7 '18 at 18:15










  • $begingroup$
    $log(1+z)/z$ is holomorphic in the region bounded by $C(0,r)$, so en.wikipedia.org/wiki/Cauchy%27s_integral_theorem
    $endgroup$
    – Federico
    Dec 7 '18 at 18:31


















  • $begingroup$
    Write $z=re^{itheta}$ and remember how log works.
    $endgroup$
    – user10354138
    Dec 7 '18 at 18:15










  • $begingroup$
    $log(1+z)/z$ is holomorphic in the region bounded by $C(0,r)$, so en.wikipedia.org/wiki/Cauchy%27s_integral_theorem
    $endgroup$
    – Federico
    Dec 7 '18 at 18:31
















$begingroup$
Write $z=re^{itheta}$ and remember how log works.
$endgroup$
– user10354138
Dec 7 '18 at 18:15




$begingroup$
Write $z=re^{itheta}$ and remember how log works.
$endgroup$
– user10354138
Dec 7 '18 at 18:15












$begingroup$
$log(1+z)/z$ is holomorphic in the region bounded by $C(0,r)$, so en.wikipedia.org/wiki/Cauchy%27s_integral_theorem
$endgroup$
– Federico
Dec 7 '18 at 18:31




$begingroup$
$log(1+z)/z$ is holomorphic in the region bounded by $C(0,r)$, so en.wikipedia.org/wiki/Cauchy%27s_integral_theorem
$endgroup$
– Federico
Dec 7 '18 at 18:31










2 Answers
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$begingroup$

Parametrizing the contour integral by setting $z = re^{it}$, $-pi le t le pi$, we write
$$0 =int_{C(0,r)} frac{log(1 + z)}{z}, dz = int_{-pi}^pi frac{log(1 + re^{it})}{re^{it}} ire^{it}, dt = i int_{-pi}^pi log(1 + re^{it}), dt$$
Taking imaginary parts, $0 = int_{-pi}^pi log|1 + re^{it}|, dt$, or
$$0 = frac{1}{2}int_{-pi}^pi log|1 + re^{it}|^2, dt$$ Expanding $|1 + re^{it}|^2$ and using symmetry, the result is obtained.






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    $begingroup$

    As long as $D(0,r)$ is a simply connected domain and the function $f(z)=1+zneq 0, forall z in D(0,r)$ then a holomorphic branch of $log f$ is defined well, and you proceed as the others said. We must be very careful with branches of logarithms and how we use them






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

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      active

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      2












      $begingroup$

      Parametrizing the contour integral by setting $z = re^{it}$, $-pi le t le pi$, we write
      $$0 =int_{C(0,r)} frac{log(1 + z)}{z}, dz = int_{-pi}^pi frac{log(1 + re^{it})}{re^{it}} ire^{it}, dt = i int_{-pi}^pi log(1 + re^{it}), dt$$
      Taking imaginary parts, $0 = int_{-pi}^pi log|1 + re^{it}|, dt$, or
      $$0 = frac{1}{2}int_{-pi}^pi log|1 + re^{it}|^2, dt$$ Expanding $|1 + re^{it}|^2$ and using symmetry, the result is obtained.






      share|cite|improve this answer









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        2












        $begingroup$

        Parametrizing the contour integral by setting $z = re^{it}$, $-pi le t le pi$, we write
        $$0 =int_{C(0,r)} frac{log(1 + z)}{z}, dz = int_{-pi}^pi frac{log(1 + re^{it})}{re^{it}} ire^{it}, dt = i int_{-pi}^pi log(1 + re^{it}), dt$$
        Taking imaginary parts, $0 = int_{-pi}^pi log|1 + re^{it}|, dt$, or
        $$0 = frac{1}{2}int_{-pi}^pi log|1 + re^{it}|^2, dt$$ Expanding $|1 + re^{it}|^2$ and using symmetry, the result is obtained.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Parametrizing the contour integral by setting $z = re^{it}$, $-pi le t le pi$, we write
          $$0 =int_{C(0,r)} frac{log(1 + z)}{z}, dz = int_{-pi}^pi frac{log(1 + re^{it})}{re^{it}} ire^{it}, dt = i int_{-pi}^pi log(1 + re^{it}), dt$$
          Taking imaginary parts, $0 = int_{-pi}^pi log|1 + re^{it}|, dt$, or
          $$0 = frac{1}{2}int_{-pi}^pi log|1 + re^{it}|^2, dt$$ Expanding $|1 + re^{it}|^2$ and using symmetry, the result is obtained.






          share|cite|improve this answer









          $endgroup$



          Parametrizing the contour integral by setting $z = re^{it}$, $-pi le t le pi$, we write
          $$0 =int_{C(0,r)} frac{log(1 + z)}{z}, dz = int_{-pi}^pi frac{log(1 + re^{it})}{re^{it}} ire^{it}, dt = i int_{-pi}^pi log(1 + re^{it}), dt$$
          Taking imaginary parts, $0 = int_{-pi}^pi log|1 + re^{it}|, dt$, or
          $$0 = frac{1}{2}int_{-pi}^pi log|1 + re^{it}|^2, dt$$ Expanding $|1 + re^{it}|^2$ and using symmetry, the result is obtained.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 7 '18 at 18:37









          kobekobe

          34.8k22248




          34.8k22248























              0












              $begingroup$

              As long as $D(0,r)$ is a simply connected domain and the function $f(z)=1+zneq 0, forall z in D(0,r)$ then a holomorphic branch of $log f$ is defined well, and you proceed as the others said. We must be very careful with branches of logarithms and how we use them






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                As long as $D(0,r)$ is a simply connected domain and the function $f(z)=1+zneq 0, forall z in D(0,r)$ then a holomorphic branch of $log f$ is defined well, and you proceed as the others said. We must be very careful with branches of logarithms and how we use them






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  As long as $D(0,r)$ is a simply connected domain and the function $f(z)=1+zneq 0, forall z in D(0,r)$ then a holomorphic branch of $log f$ is defined well, and you proceed as the others said. We must be very careful with branches of logarithms and how we use them






                  share|cite|improve this answer









                  $endgroup$



                  As long as $D(0,r)$ is a simply connected domain and the function $f(z)=1+zneq 0, forall z in D(0,r)$ then a holomorphic branch of $log f$ is defined well, and you proceed as the others said. We must be very careful with branches of logarithms and how we use them







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 8 '18 at 21:36









                  Beslikas ThanosBeslikas Thanos

                  758314




                  758314






























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