If $intlimits_gamma f(z)dzinBbb R$ does it imply $f(z)in Bbb R~forall zingamma$?












0












$begingroup$


If $intlimits_gamma f(z)dzinBbb R$ does it imply $f(z)in Bbb R~forall zingamma$?



This is from a proof where $gamma$ is a circle and we have $1={1over 2pi}intlimits_gamma f(z)dz={1over 2pi}intlimits_gamma Re~f(z)dz$ and I'm unsure how the second equality is justified.



Clearily $0=intlimits_0^{2pi}ie^{it}dt$ and $ie^{it}notin Bbb R$...






complex-analysis complex-integration







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$endgroup$








  • 5




    $begingroup$
    The answer is no. For example, if $gamma$ is a closed path and $f(z)$ is holomorphic, then the integral is $0$, and $0 in Bbb R$. This does not mean that $f$ is real on $gamma$.
    $endgroup$
    – Crostul
    Dec 31 '18 at 13:52












  • $begingroup$
    Further specialized example. $f(z) = i$ the constant, and $gamma$ is the unit circle. Then $int_gamma f(z);dz = 0$ but $f(z)$ is never real.
    $endgroup$
    – GEdgar
    Dec 31 '18 at 13:54










  • $begingroup$
    Also $int_gamma f(z) dz in mathbb{C}setminusmathbb{R}$ does not imply that there is a $z$ with $f(z) in mathbb{C}$. Take $f equiv 1$ and $gamma: [0, pi] to mathbb{C}$ with $gamma(t) := e^{it}$. (Naively one could see curve integration as just summation of the function values over the curve, but this severely fails!)
    $endgroup$
    – ComFreek
    Dec 31 '18 at 14:29


















0












$begingroup$


If $intlimits_gamma f(z)dzinBbb R$ does it imply $f(z)in Bbb R~forall zingamma$?



This is from a proof where $gamma$ is a circle and we have $1={1over 2pi}intlimits_gamma f(z)dz={1over 2pi}intlimits_gamma Re~f(z)dz$ and I'm unsure how the second equality is justified.



Clearily $0=intlimits_0^{2pi}ie^{it}dt$ and $ie^{it}notin Bbb R$...






complex-analysis complex-integration







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    The answer is no. For example, if $gamma$ is a closed path and $f(z)$ is holomorphic, then the integral is $0$, and $0 in Bbb R$. This does not mean that $f$ is real on $gamma$.
    $endgroup$
    – Crostul
    Dec 31 '18 at 13:52












  • $begingroup$
    Further specialized example. $f(z) = i$ the constant, and $gamma$ is the unit circle. Then $int_gamma f(z);dz = 0$ but $f(z)$ is never real.
    $endgroup$
    – GEdgar
    Dec 31 '18 at 13:54










  • $begingroup$
    Also $int_gamma f(z) dz in mathbb{C}setminusmathbb{R}$ does not imply that there is a $z$ with $f(z) in mathbb{C}$. Take $f equiv 1$ and $gamma: [0, pi] to mathbb{C}$ with $gamma(t) := e^{it}$. (Naively one could see curve integration as just summation of the function values over the curve, but this severely fails!)
    $endgroup$
    – ComFreek
    Dec 31 '18 at 14:29
















0












0








0





$begingroup$


If $intlimits_gamma f(z)dzinBbb R$ does it imply $f(z)in Bbb R~forall zingamma$?



This is from a proof where $gamma$ is a circle and we have $1={1over 2pi}intlimits_gamma f(z)dz={1over 2pi}intlimits_gamma Re~f(z)dz$ and I'm unsure how the second equality is justified.



Clearily $0=intlimits_0^{2pi}ie^{it}dt$ and $ie^{it}notin Bbb R$...






complex-analysis complex-integration







share|cite|improve this question











$endgroup$




If $intlimits_gamma f(z)dzinBbb R$ does it imply $f(z)in Bbb R~forall zingamma$?



This is from a proof where $gamma$ is a circle and we have $1={1over 2pi}intlimits_gamma f(z)dz={1over 2pi}intlimits_gamma Re~f(z)dz$ and I'm unsure how the second equality is justified.



Clearily $0=intlimits_0^{2pi}ie^{it}dt$ and $ie^{it}notin Bbb R$...






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








edited Dec 31 '18 at 13:40







John Cataldo

















asked Dec 31 '18 at 13:38









John CataldoJohn Cataldo

1,1371316




1,1371316








  • 5




    $begingroup$
    The answer is no. For example, if $gamma$ is a closed path and $f(z)$ is holomorphic, then the integral is $0$, and $0 in Bbb R$. This does not mean that $f$ is real on $gamma$.
    $endgroup$
    – Crostul
    Dec 31 '18 at 13:52












  • $begingroup$
    Further specialized example. $f(z) = i$ the constant, and $gamma$ is the unit circle. Then $int_gamma f(z);dz = 0$ but $f(z)$ is never real.
    $endgroup$
    – GEdgar
    Dec 31 '18 at 13:54










  • $begingroup$
    Also $int_gamma f(z) dz in mathbb{C}setminusmathbb{R}$ does not imply that there is a $z$ with $f(z) in mathbb{C}$. Take $f equiv 1$ and $gamma: [0, pi] to mathbb{C}$ with $gamma(t) := e^{it}$. (Naively one could see curve integration as just summation of the function values over the curve, but this severely fails!)
    $endgroup$
    – ComFreek
    Dec 31 '18 at 14:29
















  • 5




    $begingroup$
    The answer is no. For example, if $gamma$ is a closed path and $f(z)$ is holomorphic, then the integral is $0$, and $0 in Bbb R$. This does not mean that $f$ is real on $gamma$.
    $endgroup$
    – Crostul
    Dec 31 '18 at 13:52












  • $begingroup$
    Further specialized example. $f(z) = i$ the constant, and $gamma$ is the unit circle. Then $int_gamma f(z);dz = 0$ but $f(z)$ is never real.
    $endgroup$
    – GEdgar
    Dec 31 '18 at 13:54










  • $begingroup$
    Also $int_gamma f(z) dz in mathbb{C}setminusmathbb{R}$ does not imply that there is a $z$ with $f(z) in mathbb{C}$. Take $f equiv 1$ and $gamma: [0, pi] to mathbb{C}$ with $gamma(t) := e^{it}$. (Naively one could see curve integration as just summation of the function values over the curve, but this severely fails!)
    $endgroup$
    – ComFreek
    Dec 31 '18 at 14:29










5




5




$begingroup$
The answer is no. For example, if $gamma$ is a closed path and $f(z)$ is holomorphic, then the integral is $0$, and $0 in Bbb R$. This does not mean that $f$ is real on $gamma$.
$endgroup$
– Crostul
Dec 31 '18 at 13:52






$begingroup$
The answer is no. For example, if $gamma$ is a closed path and $f(z)$ is holomorphic, then the integral is $0$, and $0 in Bbb R$. This does not mean that $f$ is real on $gamma$.
$endgroup$
– Crostul
Dec 31 '18 at 13:52














$begingroup$
Further specialized example. $f(z) = i$ the constant, and $gamma$ is the unit circle. Then $int_gamma f(z);dz = 0$ but $f(z)$ is never real.
$endgroup$
– GEdgar
Dec 31 '18 at 13:54




$begingroup$
Further specialized example. $f(z) = i$ the constant, and $gamma$ is the unit circle. Then $int_gamma f(z);dz = 0$ but $f(z)$ is never real.
$endgroup$
– GEdgar
Dec 31 '18 at 13:54












$begingroup$
Also $int_gamma f(z) dz in mathbb{C}setminusmathbb{R}$ does not imply that there is a $z$ with $f(z) in mathbb{C}$. Take $f equiv 1$ and $gamma: [0, pi] to mathbb{C}$ with $gamma(t) := e^{it}$. (Naively one could see curve integration as just summation of the function values over the curve, but this severely fails!)
$endgroup$
– ComFreek
Dec 31 '18 at 14:29






$begingroup$
Also $int_gamma f(z) dz in mathbb{C}setminusmathbb{R}$ does not imply that there is a $z$ with $f(z) in mathbb{C}$. Take $f equiv 1$ and $gamma: [0, pi] to mathbb{C}$ with $gamma(t) := e^{it}$. (Naively one could see curve integration as just summation of the function values over the curve, but this severely fails!)
$endgroup$
– ComFreek
Dec 31 '18 at 14:29












2 Answers
2






active

oldest

votes


















0












$begingroup$

$$frac1{2pi} int_gamma f(z) mathrm dz = 1$$



Comparing real parts:
$$Re left( frac1{2pi} int_gamma f(z) mathrm dz right) = 1$$



Taking real parts commutes with everything:
$$1 = Re left( frac1{2pi} int_gamma f(z) mathrm dz right) = frac1{2pi} Re left( int_gamma f(z) mathrm dz right) = frac1{2pi} int_gamma Re left( f(z) right) mathrm dz$$




Notice, though, that $displaystyle int_gamma Re left( f(z) right) mathrm dz = int_gamma f(z) mathrm dz$ does not imply $Re left( f(z) right) = f(z)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    That's not justification, that's just saying it works because it works
    $endgroup$
    – Jakobian
    Dec 31 '18 at 13:44






  • 1




    $begingroup$
    @Jakobian but that's pointing out the author's mistake in thinking that we somehow deduced $f(z) in Bbb R$ when writing down that equality.
    $endgroup$
    – Kenny Lau
    Dec 31 '18 at 13:44



















0












$begingroup$

A you talking about a curve Integral?



If yes then f is probably a One-Form:
Integration over a curve $gamma : [a,b] to mathbb{R}^n$ is defined as follows:



$$
int_gamma f(z) dz=int_a^b f(gamma(z))gamma'(z) dz
$$



(I know this should be a comment rather than an answer but i don't have sufficient points jet)






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












    $begingroup$

    $$frac1{2pi} int_gamma f(z) mathrm dz = 1$$



    Comparing real parts:
    $$Re left( frac1{2pi} int_gamma f(z) mathrm dz right) = 1$$



    Taking real parts commutes with everything:
    $$1 = Re left( frac1{2pi} int_gamma f(z) mathrm dz right) = frac1{2pi} Re left( int_gamma f(z) mathrm dz right) = frac1{2pi} int_gamma Re left( f(z) right) mathrm dz$$




    Notice, though, that $displaystyle int_gamma Re left( f(z) right) mathrm dz = int_gamma f(z) mathrm dz$ does not imply $Re left( f(z) right) = f(z)$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      That's not justification, that's just saying it works because it works
      $endgroup$
      – Jakobian
      Dec 31 '18 at 13:44






    • 1




      $begingroup$
      @Jakobian but that's pointing out the author's mistake in thinking that we somehow deduced $f(z) in Bbb R$ when writing down that equality.
      $endgroup$
      – Kenny Lau
      Dec 31 '18 at 13:44
















    0












    $begingroup$

    $$frac1{2pi} int_gamma f(z) mathrm dz = 1$$



    Comparing real parts:
    $$Re left( frac1{2pi} int_gamma f(z) mathrm dz right) = 1$$



    Taking real parts commutes with everything:
    $$1 = Re left( frac1{2pi} int_gamma f(z) mathrm dz right) = frac1{2pi} Re left( int_gamma f(z) mathrm dz right) = frac1{2pi} int_gamma Re left( f(z) right) mathrm dz$$




    Notice, though, that $displaystyle int_gamma Re left( f(z) right) mathrm dz = int_gamma f(z) mathrm dz$ does not imply $Re left( f(z) right) = f(z)$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      That's not justification, that's just saying it works because it works
      $endgroup$
      – Jakobian
      Dec 31 '18 at 13:44






    • 1




      $begingroup$
      @Jakobian but that's pointing out the author's mistake in thinking that we somehow deduced $f(z) in Bbb R$ when writing down that equality.
      $endgroup$
      – Kenny Lau
      Dec 31 '18 at 13:44














    0












    0








    0





    $begingroup$

    $$frac1{2pi} int_gamma f(z) mathrm dz = 1$$



    Comparing real parts:
    $$Re left( frac1{2pi} int_gamma f(z) mathrm dz right) = 1$$



    Taking real parts commutes with everything:
    $$1 = Re left( frac1{2pi} int_gamma f(z) mathrm dz right) = frac1{2pi} Re left( int_gamma f(z) mathrm dz right) = frac1{2pi} int_gamma Re left( f(z) right) mathrm dz$$




    Notice, though, that $displaystyle int_gamma Re left( f(z) right) mathrm dz = int_gamma f(z) mathrm dz$ does not imply $Re left( f(z) right) = f(z)$.






    share|cite|improve this answer









    $endgroup$



    $$frac1{2pi} int_gamma f(z) mathrm dz = 1$$



    Comparing real parts:
    $$Re left( frac1{2pi} int_gamma f(z) mathrm dz right) = 1$$



    Taking real parts commutes with everything:
    $$1 = Re left( frac1{2pi} int_gamma f(z) mathrm dz right) = frac1{2pi} Re left( int_gamma f(z) mathrm dz right) = frac1{2pi} int_gamma Re left( f(z) right) mathrm dz$$




    Notice, though, that $displaystyle int_gamma Re left( f(z) right) mathrm dz = int_gamma f(z) mathrm dz$ does not imply $Re left( f(z) right) = f(z)$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 31 '18 at 13:41









    Kenny LauKenny Lau

    19.9k2159




    19.9k2159












    • $begingroup$
      That's not justification, that's just saying it works because it works
      $endgroup$
      – Jakobian
      Dec 31 '18 at 13:44






    • 1




      $begingroup$
      @Jakobian but that's pointing out the author's mistake in thinking that we somehow deduced $f(z) in Bbb R$ when writing down that equality.
      $endgroup$
      – Kenny Lau
      Dec 31 '18 at 13:44


















    • $begingroup$
      That's not justification, that's just saying it works because it works
      $endgroup$
      – Jakobian
      Dec 31 '18 at 13:44






    • 1




      $begingroup$
      @Jakobian but that's pointing out the author's mistake in thinking that we somehow deduced $f(z) in Bbb R$ when writing down that equality.
      $endgroup$
      – Kenny Lau
      Dec 31 '18 at 13:44
















    $begingroup$
    That's not justification, that's just saying it works because it works
    $endgroup$
    – Jakobian
    Dec 31 '18 at 13:44




    $begingroup$
    That's not justification, that's just saying it works because it works
    $endgroup$
    – Jakobian
    Dec 31 '18 at 13:44




    1




    1




    $begingroup$
    @Jakobian but that's pointing out the author's mistake in thinking that we somehow deduced $f(z) in Bbb R$ when writing down that equality.
    $endgroup$
    – Kenny Lau
    Dec 31 '18 at 13:44




    $begingroup$
    @Jakobian but that's pointing out the author's mistake in thinking that we somehow deduced $f(z) in Bbb R$ when writing down that equality.
    $endgroup$
    – Kenny Lau
    Dec 31 '18 at 13:44











    0












    $begingroup$

    A you talking about a curve Integral?



    If yes then f is probably a One-Form:
    Integration over a curve $gamma : [a,b] to mathbb{R}^n$ is defined as follows:



    $$
    int_gamma f(z) dz=int_a^b f(gamma(z))gamma'(z) dz
    $$



    (I know this should be a comment rather than an answer but i don't have sufficient points jet)






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      A you talking about a curve Integral?



      If yes then f is probably a One-Form:
      Integration over a curve $gamma : [a,b] to mathbb{R}^n$ is defined as follows:



      $$
      int_gamma f(z) dz=int_a^b f(gamma(z))gamma'(z) dz
      $$



      (I know this should be a comment rather than an answer but i don't have sufficient points jet)






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        A you talking about a curve Integral?



        If yes then f is probably a One-Form:
        Integration over a curve $gamma : [a,b] to mathbb{R}^n$ is defined as follows:



        $$
        int_gamma f(z) dz=int_a^b f(gamma(z))gamma'(z) dz
        $$



        (I know this should be a comment rather than an answer but i don't have sufficient points jet)






        share|cite|improve this answer











        $endgroup$



        A you talking about a curve Integral?



        If yes then f is probably a One-Form:
        Integration over a curve $gamma : [a,b] to mathbb{R}^n$ is defined as follows:



        $$
        int_gamma f(z) dz=int_a^b f(gamma(z))gamma'(z) dz
        $$



        (I know this should be a comment rather than an answer but i don't have sufficient points jet)







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 31 '18 at 13:59

























        answered Dec 31 '18 at 13:49









        A. PA. P

        1085




        1085






























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            張江高科駅

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            張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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