Help to understand that $xleft( zright) ={Re}int_{z_{0}}^{z}phi left( xi right) dxi $ is well defined











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Let $phi :Omega subset
%TCIMACRO{U{2102} }%
%BeginExpansion
mathbb{C}
%EndExpansion
rightarrow
%TCIMACRO{U{2102} }%
%BeginExpansion
mathbb{C}
%EndExpansion
$
holomorphic function. Define $x:Omega subset
rightarrow
mathbb{C}
$
by $xleft( zright) ={Re}int_{z_{0}}^{z}phi left( xi right)
dxi $
.



The author states:
"It is not necessary that the domain $Omega$ is simply connected, but only that the periods (that is, that the integral along any closed curve) $phi_k$ are pure imaginary so that the functions $x_k$ are well defined."



Because for $x_k$ is well defined, just enough?
I can not understand. For me if the integral is pure imaginary, then the function $x_k$ will be identically zero.



Could someone help me to understand this better?










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    up vote
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    down vote

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    Let $phi :Omega subset
    %TCIMACRO{U{2102} }%
    %BeginExpansion
    mathbb{C}
    %EndExpansion
    rightarrow
    %TCIMACRO{U{2102} }%
    %BeginExpansion
    mathbb{C}
    %EndExpansion
    $
    holomorphic function. Define $x:Omega subset
    rightarrow
    mathbb{C}
    $
    by $xleft( zright) ={Re}int_{z_{0}}^{z}phi left( xi right)
    dxi $
    .



    The author states:
    "It is not necessary that the domain $Omega$ is simply connected, but only that the periods (that is, that the integral along any closed curve) $phi_k$ are pure imaginary so that the functions $x_k$ are well defined."



    Because for $x_k$ is well defined, just enough?
    I can not understand. For me if the integral is pure imaginary, then the function $x_k$ will be identically zero.



    Could someone help me to understand this better?










    share|cite|improve this question
























      up vote
      1
      down vote

      favorite
      1









      up vote
      1
      down vote

      favorite
      1






      1





      Let $phi :Omega subset
      %TCIMACRO{U{2102} }%
      %BeginExpansion
      mathbb{C}
      %EndExpansion
      rightarrow
      %TCIMACRO{U{2102} }%
      %BeginExpansion
      mathbb{C}
      %EndExpansion
      $
      holomorphic function. Define $x:Omega subset
      rightarrow
      mathbb{C}
      $
      by $xleft( zright) ={Re}int_{z_{0}}^{z}phi left( xi right)
      dxi $
      .



      The author states:
      "It is not necessary that the domain $Omega$ is simply connected, but only that the periods (that is, that the integral along any closed curve) $phi_k$ are pure imaginary so that the functions $x_k$ are well defined."



      Because for $x_k$ is well defined, just enough?
      I can not understand. For me if the integral is pure imaginary, then the function $x_k$ will be identically zero.



      Could someone help me to understand this better?










      share|cite|improve this question













      Let $phi :Omega subset
      %TCIMACRO{U{2102} }%
      %BeginExpansion
      mathbb{C}
      %EndExpansion
      rightarrow
      %TCIMACRO{U{2102} }%
      %BeginExpansion
      mathbb{C}
      %EndExpansion
      $
      holomorphic function. Define $x:Omega subset
      rightarrow
      mathbb{C}
      $
      by $xleft( zright) ={Re}int_{z_{0}}^{z}phi left( xi right)
      dxi $
      .



      The author states:
      "It is not necessary that the domain $Omega$ is simply connected, but only that the periods (that is, that the integral along any closed curve) $phi_k$ are pure imaginary so that the functions $x_k$ are well defined."



      Because for $x_k$ is well defined, just enough?
      I can not understand. For me if the integral is pure imaginary, then the function $x_k$ will be identically zero.



      Could someone help me to understand this better?







      complex-analysis analysis differential-geometry riemannian-geometry






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      asked Nov 26 at 1:35









      Takashi

      1946




      1946






















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          It doesn't say that the integral from $z_0$ to $z$ is purely imaginary, but that the integral along a closed curve must be purely imaginary.



          This means that if you have two different curves that both go from $z_0$ to $z$, the difference of their integrals will equal the integral around a closed curve (namely, out from $z_0$ to $z$ along one curve and back again along the other) will be purely imaginary, and therefore the two integrals have the same real part.






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            1 Answer
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            1 Answer
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            active

            oldest

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            active

            oldest

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            active

            oldest

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            up vote
            2
            down vote



            accepted










            It doesn't say that the integral from $z_0$ to $z$ is purely imaginary, but that the integral along a closed curve must be purely imaginary.



            This means that if you have two different curves that both go from $z_0$ to $z$, the difference of their integrals will equal the integral around a closed curve (namely, out from $z_0$ to $z$ along one curve and back again along the other) will be purely imaginary, and therefore the two integrals have the same real part.






            share|cite|improve this answer

























              up vote
              2
              down vote



              accepted










              It doesn't say that the integral from $z_0$ to $z$ is purely imaginary, but that the integral along a closed curve must be purely imaginary.



              This means that if you have two different curves that both go from $z_0$ to $z$, the difference of their integrals will equal the integral around a closed curve (namely, out from $z_0$ to $z$ along one curve and back again along the other) will be purely imaginary, and therefore the two integrals have the same real part.






              share|cite|improve this answer























                up vote
                2
                down vote



                accepted







                up vote
                2
                down vote



                accepted






                It doesn't say that the integral from $z_0$ to $z$ is purely imaginary, but that the integral along a closed curve must be purely imaginary.



                This means that if you have two different curves that both go from $z_0$ to $z$, the difference of their integrals will equal the integral around a closed curve (namely, out from $z_0$ to $z$ along one curve and back again along the other) will be purely imaginary, and therefore the two integrals have the same real part.






                share|cite|improve this answer












                It doesn't say that the integral from $z_0$ to $z$ is purely imaginary, but that the integral along a closed curve must be purely imaginary.



                This means that if you have two different curves that both go from $z_0$ to $z$, the difference of their integrals will equal the integral around a closed curve (namely, out from $z_0$ to $z$ along one curve and back again along the other) will be purely imaginary, and therefore the two integrals have the same real part.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 26 at 1:44









                Henning Makholm

                236k16300534




                236k16300534






























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