Help to understand that $xleft( zright) ={Re}int_{z_{0}}^{z}phi left( xi right) dxi $ is well defined
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
asked Nov 26 at 1:35
Takashi
1946
add a comment |
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
asked Nov 26 at 1:35
Takashi
1946
add a comment |
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
asked Nov 26 at 1:35
Takashi
1946
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
complex-analysis analysis differential-geometry riemannian-geometry
asked Nov 26 at 1:35
Takashi
1946
asked Nov 26 at 1:35
Takashi
1946
asked Nov 26 at 1:35
Takashi
1946
asked Nov 26 at 1:35
Takashi
1946
asked Nov 26 at 1:35
Takashi
1946
1946
add a comment |
add a comment |
1 Answer
1
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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.
answered Nov 26 at 1:44
Henning Makholm
237k16301536
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
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.
answered Nov 26 at 1:44
Henning Makholm
237k16301536
add a comment |
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.
answered Nov 26 at 1:44
Henning Makholm
237k16301536
add a comment |
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.
answered Nov 26 at 1:44
Henning Makholm
237k16301536
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.
answered Nov 26 at 1:44
Henning Makholm
237k16301536
answered Nov 26 at 1:44
Henning Makholm
237k16301536
answered Nov 26 at 1:44
Henning Makholm
237k16301536
answered Nov 26 at 1:44
Henning Makholm
237k16301536
237k16301536
add a comment |
add a comment |
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