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Contour integral of square root on its Riemann surface
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)contour integral with singularity on the contourUsing the Residue Theorem for a contour integral along the Riemann sphereIntegration of $z^1/2$ along a contour winding the origin twice, without introducing Riemann surfaceAnalytic continuation of the Riemann zeta function using contour integrationHow to evaluate the contour integral $int_C(0,1) fracz e^z tan^2 zdz$ over the unit circle?Contour integral with a logarithm squaredGive an example of a function with singularity at $0$ and whose contour integral is zero.Showing that parts of a contour integral vanishAbout closed curves in Riemann surfaceResidue and Laurent series for an integral involving exponential and square root.
$begingroup$
Consider a branch of the square root function $f(z)=z^1/2, zinmathbbC$ with $Im thinspace f(z)>=0$, i.e.
$f(re^iphi):=sqrtre^i(phi thinspace mod thinspace 2pi)/2.$
The function is not holomorphic at the origin and has a discontinuity on the non-negative real axis. Consequently, the contour integral of the function along the unit circle doesn't vanish.
If we extend the function to it's Riemann surface then we'll get a single-valued continuous function, (informally $F(re^iphi)=sqrtre^iphi/2$).
The contour integral along the "double unit circle" $z(phi)=e^iphi, phiin[0,4pi]$ on the Riemann surface now vanishes.
I suspect this will be true for any other closed curve on the Riemann surface.
Is there a way to understand this in context of the residue theorem, suitably generalized to Riemann surfaces? Is it correct to say that the origin has now become an isolated (essential) singularity whose residue is zero and hence the integral of the function along any closed curve on the Riemann surface must always vanish?
complex-analysis residue-calculus riemann-surfaces
asked Mar 31 at 19:48
user1552913user1552913
62
$endgroup$
add a comment |
$begingroup$
Consider a branch of the square root function $f(z)=z^1/2, zinmathbbC$ with $Im thinspace f(z)>=0$, i.e.
$f(re^iphi):=sqrtre^i(phi thinspace mod thinspace 2pi)/2.$
The function is not holomorphic at the origin and has a discontinuity on the non-negative real axis. Consequently, the contour integral of the function along the unit circle doesn't vanish.
If we extend the function to it's Riemann surface then we'll get a single-valued continuous function, (informally $F(re^iphi)=sqrtre^iphi/2$).
The contour integral along the "double unit circle" $z(phi)=e^iphi, phiin[0,4pi]$ on the Riemann surface now vanishes.
I suspect this will be true for any other closed curve on the Riemann surface.
Is there a way to understand this in context of the residue theorem, suitably generalized to Riemann surfaces? Is it correct to say that the origin has now become an isolated (essential) singularity whose residue is zero and hence the integral of the function along any closed curve on the Riemann surface must always vanish?
complex-analysis residue-calculus riemann-surfaces
asked Mar 31 at 19:48
user1552913user1552913
62
$endgroup$
1
$begingroup$
Its Riemann surface is $X =(z,z^1/2), z in BbbC^*$ isomorphic to $BbbC^*$ and the image of $z^1/2$ is $z$ having a removable singularity at $z=0$. The function field of $X$ becomes the field of meromorphic functions on $BbbC^*$. It is their derivatives having vanishing "residue" $int_z dF = 0$ corresponding to $int_ = r^1/2 in X dF(z^1/2) = 0$ on $X$. Note $F$ can have infinitely many poles around $z=0$ (so infinitely many different $r$) to disallow that you need to add the $z=0$ point to obtain $BbbC$ instead of $BbbC^*$.
$endgroup$
– reuns
Mar 31 at 20:03
$begingroup$
Why is the integral $int_=r^1/2∈X dF(z^1/2)=0$ ? Is $dF(z^1/2)$ a closed form on the Riemann surface?
$endgroup$
– user1552913
Apr 1 at 16:29
$begingroup$
$F(z)$ is meromorphic on $BbbC^*$ iff $f = F(z^1/2)$ is meromorphic on $X$ and $df$ is exact holomorphic one-form on $X$ minus the poles of $f$. Of course $z^1/2$ is not the principal branch of the squareroot, it is the multivalued one corresponding to the charts of $X$.
$endgroup$
– reuns
Apr 1 at 19:56
$begingroup$
Concretely $int_=r^1/2∈X dF(z^1/2) = int_0^4pi dF(r^1/2 e^it/2) =F( r^1/2 e^it/2)|_0^4pi$
$endgroup$
– reuns
Apr 1 at 20:02
add a comment |
$begingroup$
Consider a branch of the square root function $f(z)=z^1/2, zinmathbbC$ with $Im thinspace f(z)>=0$, i.e.
$f(re^iphi):=sqrtre^i(phi thinspace mod thinspace 2pi)/2.$
The function is not holomorphic at the origin and has a discontinuity on the non-negative real axis. Consequently, the contour integral of the function along the unit circle doesn't vanish.
If we extend the function to it's Riemann surface then we'll get a single-valued continuous function, (informally $F(re^iphi)=sqrtre^iphi/2$).
The contour integral along the "double unit circle" $z(phi)=e^iphi, phiin[0,4pi]$ on the Riemann surface now vanishes.
I suspect this will be true for any other closed curve on the Riemann surface.
Is there a way to understand this in context of the residue theorem, suitably generalized to Riemann surfaces? Is it correct to say that the origin has now become an isolated (essential) singularity whose residue is zero and hence the integral of the function along any closed curve on the Riemann surface must always vanish?
complex-analysis residue-calculus riemann-surfaces
asked Mar 31 at 19:48
user1552913user1552913
62
$endgroup$
Consider a branch of the square root function $f(z)=z^1/2, zinmathbbC$ with $Im thinspace f(z)>=0$, i.e.
$f(re^iphi):=sqrtre^i(phi thinspace mod thinspace 2pi)/2.$
The function is not holomorphic at the origin and has a discontinuity on the non-negative real axis. Consequently, the contour integral of the function along the unit circle doesn't vanish.
If we extend the function to it's Riemann surface then we'll get a single-valued continuous function, (informally $F(re^iphi)=sqrtre^iphi/2$).
The contour integral along the "double unit circle" $z(phi)=e^iphi, phiin[0,4pi]$ on the Riemann surface now vanishes.
I suspect this will be true for any other closed curve on the Riemann surface.
Is there a way to understand this in context of the residue theorem, suitably generalized to Riemann surfaces? Is it correct to say that the origin has now become an isolated (essential) singularity whose residue is zero and hence the integral of the function along any closed curve on the Riemann surface must always vanish?
complex-analysis residue-calculus riemann-surfaces
complex-analysis residue-calculus riemann-surfaces
asked Mar 31 at 19:48
user1552913user1552913
62
asked Mar 31 at 19:48
user1552913user1552913
62
edited Apr 1 at 0:56
user1552913
asked Mar 31 at 19:48
user1552913user1552913
62
asked Mar 31 at 19:48
user1552913user1552913
62
asked Mar 31 at 19:48
user1552913user1552913
62
62
1
$begingroup$
Its Riemann surface is $X =(z,z^1/2), z in BbbC^*$ isomorphic to $BbbC^*$ and the image of $z^1/2$ is $z$ having a removable singularity at $z=0$. The function field of $X$ becomes the field of meromorphic functions on $BbbC^*$. It is their derivatives having vanishing "residue" $int_z dF = 0$ corresponding to $int_ = r^1/2 in X dF(z^1/2) = 0$ on $X$. Note $F$ can have infinitely many poles around $z=0$ (so infinitely many different $r$) to disallow that you need to add the $z=0$ point to obtain $BbbC$ instead of $BbbC^*$.
$endgroup$
– reuns
Mar 31 at 20:03
$begingroup$
Why is the integral $int_=r^1/2∈X dF(z^1/2)=0$ ? Is $dF(z^1/2)$ a closed form on the Riemann surface?
$endgroup$
– user1552913
Apr 1 at 16:29
$begingroup$
$F(z)$ is meromorphic on $BbbC^*$ iff $f = F(z^1/2)$ is meromorphic on $X$ and $df$ is exact holomorphic one-form on $X$ minus the poles of $f$. Of course $z^1/2$ is not the principal branch of the squareroot, it is the multivalued one corresponding to the charts of $X$.
$endgroup$
– reuns
Apr 1 at 19:56
$begingroup$
Concretely $int_=r^1/2∈X dF(z^1/2) = int_0^4pi dF(r^1/2 e^it/2) =F( r^1/2 e^it/2)|_0^4pi$
$endgroup$
– reuns
Apr 1 at 20:02
add a comment |
1
$begingroup$
Its Riemann surface is $X =(z,z^1/2), z in BbbC^*$ isomorphic to $BbbC^*$ and the image of $z^1/2$ is $z$ having a removable singularity at $z=0$. The function field of $X$ becomes the field of meromorphic functions on $BbbC^*$. It is their derivatives having vanishing "residue" $int_z dF = 0$ corresponding to $int_ = r^1/2 in X dF(z^1/2) = 0$ on $X$. Note $F$ can have infinitely many poles around $z=0$ (so infinitely many different $r$) to disallow that you need to add the $z=0$ point to obtain $BbbC$ instead of $BbbC^*$.
$endgroup$
– reuns
Mar 31 at 20:03
$begingroup$
Why is the integral $int_=r^1/2∈X dF(z^1/2)=0$ ? Is $dF(z^1/2)$ a closed form on the Riemann surface?
$endgroup$
– user1552913
Apr 1 at 16:29
$begingroup$
$F(z)$ is meromorphic on $BbbC^*$ iff $f = F(z^1/2)$ is meromorphic on $X$ and $df$ is exact holomorphic one-form on $X$ minus the poles of $f$. Of course $z^1/2$ is not the principal branch of the squareroot, it is the multivalued one corresponding to the charts of $X$.
$endgroup$
– reuns
Apr 1 at 19:56
$begingroup$
Concretely $int_=r^1/2∈X dF(z^1/2) = int_0^4pi dF(r^1/2 e^it/2) =F( r^1/2 e^it/2)|_0^4pi$
$endgroup$
– reuns
Apr 1 at 20:02
1
1
$begingroup$
Its Riemann surface is $X =(z,z^1/2), z in BbbC^*$ isomorphic to $BbbC^*$ and the image of $z^1/2$ is $z$ having a removable singularity at $z=0$. The function field of $X$ becomes the field of meromorphic functions on $BbbC^*$. It is their derivatives having vanishing "residue" $int_z dF = 0$ corresponding to $int_ = r^1/2 in X dF(z^1/2) = 0$ on $X$. Note $F$ can have infinitely many poles around $z=0$ (so infinitely many different $r$) to disallow that you need to add the $z=0$ point to obtain $BbbC$ instead of $BbbC^*$.
$endgroup$
– reuns
Mar 31 at 20:03
$begingroup$
Its Riemann surface is $X =(z,z^1/2), z in BbbC^*$ isomorphic to $BbbC^*$ and the image of $z^1/2$ is $z$ having a removable singularity at $z=0$. The function field of $X$ becomes the field of meromorphic functions on $BbbC^*$. It is their derivatives having vanishing "residue" $int_z dF = 0$ corresponding to $int_ = r^1/2 in X dF(z^1/2) = 0$ on $X$. Note $F$ can have infinitely many poles around $z=0$ (so infinitely many different $r$) to disallow that you need to add the $z=0$ point to obtain $BbbC$ instead of $BbbC^*$.
$endgroup$
– reuns
Mar 31 at 20:03
$begingroup$
Why is the integral $int_=r^1/2∈X dF(z^1/2)=0$ ? Is $dF(z^1/2)$ a closed form on the Riemann surface?
$endgroup$
– user1552913
Apr 1 at 16:29
$begingroup$
Why is the integral $int_=r^1/2∈X dF(z^1/2)=0$ ? Is $dF(z^1/2)$ a closed form on the Riemann surface?
$endgroup$
– user1552913
Apr 1 at 16:29
$begingroup$
$F(z)$ is meromorphic on $BbbC^*$ iff $f = F(z^1/2)$ is meromorphic on $X$ and $df$ is exact holomorphic one-form on $X$ minus the poles of $f$. Of course $z^1/2$ is not the principal branch of the squareroot, it is the multivalued one corresponding to the charts of $X$.
$endgroup$
– reuns
Apr 1 at 19:56
$begingroup$
$F(z)$ is meromorphic on $BbbC^*$ iff $f = F(z^1/2)$ is meromorphic on $X$ and $df$ is exact holomorphic one-form on $X$ minus the poles of $f$. Of course $z^1/2$ is not the principal branch of the squareroot, it is the multivalued one corresponding to the charts of $X$.
$endgroup$
– reuns
Apr 1 at 19:56
$begingroup$
Concretely $int_=r^1/2∈X dF(z^1/2) = int_0^4pi dF(r^1/2 e^it/2) =F( r^1/2 e^it/2)|_0^4pi$
$endgroup$
– reuns
Apr 1 at 20:02
$begingroup$
Concretely $int_=r^1/2∈X dF(z^1/2) = int_0^4pi dF(r^1/2 e^it/2) =F( r^1/2 e^it/2)|_0^4pi$
$endgroup$
– reuns
Apr 1 at 20:02
add a comment |
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$begingroup$
Its Riemann surface is $X =(z,z^1/2), z in BbbC^*$ isomorphic to $BbbC^*$ and the image of $z^1/2$ is $z$ having a removable singularity at $z=0$. The function field of $X$ becomes the field of meromorphic functions on $BbbC^*$. It is their derivatives having vanishing "residue" $int_z dF = 0$ corresponding to $int_ = r^1/2 in X dF(z^1/2) = 0$ on $X$. Note $F$ can have infinitely many poles around $z=0$ (so infinitely many different $r$) to disallow that you need to add the $z=0$ point to obtain $BbbC$ instead of $BbbC^*$.
$endgroup$
– reuns
Mar 31 at 20:03
$begingroup$
Why is the integral $int_=r^1/2∈X dF(z^1/2)=0$ ? Is $dF(z^1/2)$ a closed form on the Riemann surface?
$endgroup$
– user1552913
Apr 1 at 16:29
$begingroup$
$F(z)$ is meromorphic on $BbbC^*$ iff $f = F(z^1/2)$ is meromorphic on $X$ and $df$ is exact holomorphic one-form on $X$ minus the poles of $f$. Of course $z^1/2$ is not the principal branch of the squareroot, it is the multivalued one corresponding to the charts of $X$.
$endgroup$
– reuns
Apr 1 at 19:56
$begingroup$
Concretely $int_=r^1/2∈X dF(z^1/2) = int_0^4pi dF(r^1/2 e^it/2) =F( r^1/2 e^it/2)|_0^4pi$
$endgroup$
– reuns
Apr 1 at 20:02