Mixing
Multivalued Functions for Dummies
$begingroup$
I have been studying complex analysis for a while, but I still cannot "get" how multivalued functions work. Despite having it explained to me many times, my brain cannot process it.
For example,
- I do not know / understand what $f(z) = sqrt{z^2}$, $g(z) = (sqrt{z})^2$ are.
- I don't know how to decide which branch to use when I see expressions like $f(z) = sqrt{z(z-1)(z-lambda)}$
- I don't understand how integrals involving roots work, especially when it seems like the integral is being taken along the branch cuts.
I feel like many books I have seen treats these as being obvious. Is there a reference that really drills on this topic? Kind of like 'multivalued functions for dummies'?
EDIT:
$g(z) = sqrt{z}^2 = z$ (with a removable singularity at $z=0$.)
$f(z) = sqrt{z^2}$ is a multivalued function.
EDIT2:
$f(z) = sqrt{z(z-1)(z-lambda)} = exp(frac12 log(g(z)))$
where
$g(z) = z(z-1)(z-lambda).$ Let $z_0$ some point that is not $0, 1, lambda$. Then, it is possible to define $L$ in a small disk around $z_0$, with
$$L(z) = ln|z| + i theta,$$
If $z_0$ is not a negative real number, $theta$ can be $[-pi,pi]$. If $z_0$ is a negative real number, then $theta$ can be $[0,2pi]$.
Then for $zne 0,1,lambda$, we can define
$$log z = int_{z_0}^z dfrac{g'(z)}{g(z)} dz + L(z_0),$$
which is multivalued.
Then,
$$sqrt{g(z)} =exp(dfrac{1}{2} left(int_{z_0}^z dfrac{g'(z)}{g(z)} dz + L(z_0)right))$$
By the argument principle, this is a single valued function on $D$ as long as $D$ does not allow the path from $z_0$ to $z$ to loop around an odd number of zeros of $g$. This can be prevented by pairing up the zeros with a branch cut. For an odd number of zeroes, the left over one can be connected to the point at infinity.
So any combination of branch cuts that connects two of the zeros, and the third zero to the point at infinity will make $sqrt{g(z)}$ single valued.
Similarly,
$sqrt[n]{g(z)}$ would require the branch cuts to group $n$-zeroes at a time, to prevent any loops around $k$-zeroes for $k<n$. If there is any shortage, $infty$ fire one branch cut off to infinity.
complex-analysis reference-request multivalued-functions
asked May 22 '15 at 16:12
BraindeadBraindead
3,4001739
$endgroup$
|
show 1 more comment
$begingroup$
I have been studying complex analysis for a while, but I still cannot "get" how multivalued functions work. Despite having it explained to me many times, my brain cannot process it.
For example,
- I do not know / understand what $f(z) = sqrt{z^2}$, $g(z) = (sqrt{z})^2$ are.
- I don't know how to decide which branch to use when I see expressions like $f(z) = sqrt{z(z-1)(z-lambda)}$
- I don't understand how integrals involving roots work, especially when it seems like the integral is being taken along the branch cuts.
I feel like many books I have seen treats these as being obvious. Is there a reference that really drills on this topic? Kind of like 'multivalued functions for dummies'?
EDIT:
$g(z) = sqrt{z}^2 = z$ (with a removable singularity at $z=0$.)
$f(z) = sqrt{z^2}$ is a multivalued function.
EDIT2:
$f(z) = sqrt{z(z-1)(z-lambda)} = exp(frac12 log(g(z)))$
where
$g(z) = z(z-1)(z-lambda).$ Let $z_0$ some point that is not $0, 1, lambda$. Then, it is possible to define $L$ in a small disk around $z_0$, with
$$L(z) = ln|z| + i theta,$$
If $z_0$ is not a negative real number, $theta$ can be $[-pi,pi]$. If $z_0$ is a negative real number, then $theta$ can be $[0,2pi]$.
Then for $zne 0,1,lambda$, we can define
$$log z = int_{z_0}^z dfrac{g'(z)}{g(z)} dz + L(z_0),$$
which is multivalued.
Then,
$$sqrt{g(z)} =exp(dfrac{1}{2} left(int_{z_0}^z dfrac{g'(z)}{g(z)} dz + L(z_0)right))$$
By the argument principle, this is a single valued function on $D$ as long as $D$ does not allow the path from $z_0$ to $z$ to loop around an odd number of zeros of $g$. This can be prevented by pairing up the zeros with a branch cut. For an odd number of zeroes, the left over one can be connected to the point at infinity.
So any combination of branch cuts that connects two of the zeros, and the third zero to the point at infinity will make $sqrt{g(z)}$ single valued.
Similarly,
$sqrt[n]{g(z)}$ would require the branch cuts to group $n$-zeroes at a time, to prevent any loops around $k$-zeroes for $k<n$. If there is any shortage, $infty$ fire one branch cut off to infinity.
complex-analysis reference-request multivalued-functions
asked May 22 '15 at 16:12
BraindeadBraindead
3,4001739
$endgroup$
1
$begingroup$
You're looking for algebraic functions. For each algebraic function, you can associate a Riemann surface such that the algebraic function becomes a holomorphic function. See Markushevich's book "Theory of Functions of a Complex Variable" for a practical procedure using triangulations. If you want something more abstract, there is Foster's "Lectures on Riemann Surfaces".
$endgroup$
– user40276
May 26 '15 at 16:14
$begingroup$
Thanks. Are there multiple volumes of Markushevich?
$endgroup$
– Braindead
May 26 '15 at 16:16
$begingroup$
And this is definitely not a trivial or obvious topic, since all the theory of elliptic curves and abelian varieties come from this (actually motives can be realized in this way!).
$endgroup$
– user40276
May 26 '15 at 16:17
$begingroup$
I think in the new edition all volumes were merged. For the old edition, I believe it's somewhere in the third volume.
$endgroup$
– user40276
May 26 '15 at 16:19
$begingroup$
Thanks! I'll check it out.
$endgroup$
– Braindead
May 26 '15 at 16:19
|
show 1 more comment
$begingroup$
I have been studying complex analysis for a while, but I still cannot "get" how multivalued functions work. Despite having it explained to me many times, my brain cannot process it.
For example,
- I do not know / understand what $f(z) = sqrt{z^2}$, $g(z) = (sqrt{z})^2$ are.
- I don't know how to decide which branch to use when I see expressions like $f(z) = sqrt{z(z-1)(z-lambda)}$
- I don't understand how integrals involving roots work, especially when it seems like the integral is being taken along the branch cuts.
I feel like many books I have seen treats these as being obvious. Is there a reference that really drills on this topic? Kind of like 'multivalued functions for dummies'?
EDIT:
$g(z) = sqrt{z}^2 = z$ (with a removable singularity at $z=0$.)
$f(z) = sqrt{z^2}$ is a multivalued function.
EDIT2:
$f(z) = sqrt{z(z-1)(z-lambda)} = exp(frac12 log(g(z)))$
where
$g(z) = z(z-1)(z-lambda).$ Let $z_0$ some point that is not $0, 1, lambda$. Then, it is possible to define $L$ in a small disk around $z_0$, with
$$L(z) = ln|z| + i theta,$$
If $z_0$ is not a negative real number, $theta$ can be $[-pi,pi]$. If $z_0$ is a negative real number, then $theta$ can be $[0,2pi]$.
Then for $zne 0,1,lambda$, we can define
$$log z = int_{z_0}^z dfrac{g'(z)}{g(z)} dz + L(z_0),$$
which is multivalued.
Then,
$$sqrt{g(z)} =exp(dfrac{1}{2} left(int_{z_0}^z dfrac{g'(z)}{g(z)} dz + L(z_0)right))$$
By the argument principle, this is a single valued function on $D$ as long as $D$ does not allow the path from $z_0$ to $z$ to loop around an odd number of zeros of $g$. This can be prevented by pairing up the zeros with a branch cut. For an odd number of zeroes, the left over one can be connected to the point at infinity.
So any combination of branch cuts that connects two of the zeros, and the third zero to the point at infinity will make $sqrt{g(z)}$ single valued.
Similarly,
$sqrt[n]{g(z)}$ would require the branch cuts to group $n$-zeroes at a time, to prevent any loops around $k$-zeroes for $k<n$. If there is any shortage, $infty$ fire one branch cut off to infinity.
complex-analysis reference-request multivalued-functions
asked May 22 '15 at 16:12
BraindeadBraindead
3,4001739
$endgroup$
I have been studying complex analysis for a while, but I still cannot "get" how multivalued functions work. Despite having it explained to me many times, my brain cannot process it.
For example,
- I do not know / understand what $f(z) = sqrt{z^2}$, $g(z) = (sqrt{z})^2$ are.
- I don't know how to decide which branch to use when I see expressions like $f(z) = sqrt{z(z-1)(z-lambda)}$
- I don't understand how integrals involving roots work, especially when it seems like the integral is being taken along the branch cuts.
I feel like many books I have seen treats these as being obvious. Is there a reference that really drills on this topic? Kind of like 'multivalued functions for dummies'?
EDIT:
$g(z) = sqrt{z}^2 = z$ (with a removable singularity at $z=0$.)
$f(z) = sqrt{z^2}$ is a multivalued function.
EDIT2:
$f(z) = sqrt{z(z-1)(z-lambda)} = exp(frac12 log(g(z)))$
where
$g(z) = z(z-1)(z-lambda).$ Let $z_0$ some point that is not $0, 1, lambda$. Then, it is possible to define $L$ in a small disk around $z_0$, with
$$L(z) = ln|z| + i theta,$$
If $z_0$ is not a negative real number, $theta$ can be $[-pi,pi]$. If $z_0$ is a negative real number, then $theta$ can be $[0,2pi]$.
Then for $zne 0,1,lambda$, we can define
$$log z = int_{z_0}^z dfrac{g'(z)}{g(z)} dz + L(z_0),$$
which is multivalued.
Then,
$$sqrt{g(z)} =exp(dfrac{1}{2} left(int_{z_0}^z dfrac{g'(z)}{g(z)} dz + L(z_0)right))$$
By the argument principle, this is a single valued function on $D$ as long as $D$ does not allow the path from $z_0$ to $z$ to loop around an odd number of zeros of $g$. This can be prevented by pairing up the zeros with a branch cut. For an odd number of zeroes, the left over one can be connected to the point at infinity.
So any combination of branch cuts that connects two of the zeros, and the third zero to the point at infinity will make $sqrt{g(z)}$ single valued.
Similarly,
$sqrt[n]{g(z)}$ would require the branch cuts to group $n$-zeroes at a time, to prevent any loops around $k$-zeroes for $k<n$. If there is any shortage, $infty$ fire one branch cut off to infinity.
complex-analysis reference-request multivalued-functions
complex-analysis reference-request multivalued-functions
asked May 22 '15 at 16:12
BraindeadBraindead
3,4001739
asked May 22 '15 at 16:12
BraindeadBraindead
3,4001739
edited May 26 '15 at 15:52
Braindead
asked May 22 '15 at 16:12
BraindeadBraindead
3,4001739
asked May 22 '15 at 16:12
BraindeadBraindead
3,4001739
asked May 22 '15 at 16:12
BraindeadBraindead
3,4001739
3,4001739
1
$begingroup$
You're looking for algebraic functions. For each algebraic function, you can associate a Riemann surface such that the algebraic function becomes a holomorphic function. See Markushevich's book "Theory of Functions of a Complex Variable" for a practical procedure using triangulations. If you want something more abstract, there is Foster's "Lectures on Riemann Surfaces".
$endgroup$
– user40276
May 26 '15 at 16:14
$begingroup$
Thanks. Are there multiple volumes of Markushevich?
$endgroup$
– Braindead
May 26 '15 at 16:16
$begingroup$
And this is definitely not a trivial or obvious topic, since all the theory of elliptic curves and abelian varieties come from this (actually motives can be realized in this way!).
$endgroup$
– user40276
May 26 '15 at 16:17
$begingroup$
I think in the new edition all volumes were merged. For the old edition, I believe it's somewhere in the third volume.
$endgroup$
– user40276
May 26 '15 at 16:19
$begingroup$
Thanks! I'll check it out.
$endgroup$
– Braindead
May 26 '15 at 16:19
|
show 1 more comment
1
$begingroup$
You're looking for algebraic functions. For each algebraic function, you can associate a Riemann surface such that the algebraic function becomes a holomorphic function. See Markushevich's book "Theory of Functions of a Complex Variable" for a practical procedure using triangulations. If you want something more abstract, there is Foster's "Lectures on Riemann Surfaces".
$endgroup$
– user40276
May 26 '15 at 16:14
$begingroup$
Thanks. Are there multiple volumes of Markushevich?
$endgroup$
– Braindead
May 26 '15 at 16:16
$begingroup$
And this is definitely not a trivial or obvious topic, since all the theory of elliptic curves and abelian varieties come from this (actually motives can be realized in this way!).
$endgroup$
– user40276
May 26 '15 at 16:17
$begingroup$
I think in the new edition all volumes were merged. For the old edition, I believe it's somewhere in the third volume.
$endgroup$
– user40276
May 26 '15 at 16:19
$begingroup$
Thanks! I'll check it out.
$endgroup$
– Braindead
May 26 '15 at 16:19
1
1
$begingroup$
You're looking for algebraic functions. For each algebraic function, you can associate a Riemann surface such that the algebraic function becomes a holomorphic function. See Markushevich's book "Theory of Functions of a Complex Variable" for a practical procedure using triangulations. If you want something more abstract, there is Foster's "Lectures on Riemann Surfaces".
$endgroup$
– user40276
May 26 '15 at 16:14
$begingroup$
You're looking for algebraic functions. For each algebraic function, you can associate a Riemann surface such that the algebraic function becomes a holomorphic function. See Markushevich's book "Theory of Functions of a Complex Variable" for a practical procedure using triangulations. If you want something more abstract, there is Foster's "Lectures on Riemann Surfaces".
$endgroup$
– user40276
May 26 '15 at 16:14
$begingroup$
Thanks. Are there multiple volumes of Markushevich?
$endgroup$
– Braindead
May 26 '15 at 16:16
$begingroup$
Thanks. Are there multiple volumes of Markushevich?
$endgroup$
– Braindead
May 26 '15 at 16:16
$begingroup$
And this is definitely not a trivial or obvious topic, since all the theory of elliptic curves and abelian varieties come from this (actually motives can be realized in this way!).
$endgroup$
– user40276
May 26 '15 at 16:17
$begingroup$
And this is definitely not a trivial or obvious topic, since all the theory of elliptic curves and abelian varieties come from this (actually motives can be realized in this way!).
$endgroup$
– user40276
May 26 '15 at 16:17
$begingroup$
I think in the new edition all volumes were merged. For the old edition, I believe it's somewhere in the third volume.
$endgroup$
– user40276
May 26 '15 at 16:19
$begingroup$
I think in the new edition all volumes were merged. For the old edition, I believe it's somewhere in the third volume.
$endgroup$
– user40276
May 26 '15 at 16:19
$begingroup$
Thanks! I'll check it out.
$endgroup$
– Braindead
May 26 '15 at 16:19
$begingroup$
Thanks! I'll check it out.
$endgroup$
– Braindead
May 26 '15 at 16:19
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
It is actually very simple. Draw some disjoint curves connecting singularities,
or starting at the singularities and going to infinity, so that these curves separate the plane into simply connected regions. Then instead of a "multivalued function" you have several ordinary (single-valued) functions in each region.
These ordinary functions are called "branches" of the multivalued function.
When you choose a branch and try to cross one of your curves into another region
your branch must match one of the branches in that other region.
For example, $sqrt{z}$ has one singularity in the plane, namely $0$ (another one is $infty$). Draw ray from $0$ to $infty$. You obtain one region. In this region you have two ordinary functions, branches of $sqrt{z}$. When you cross the cut, they are interchanged.
answered Nov 29 '17 at 21:04
Alexandre EremenkoAlexandre Eremenko
2,415921
$endgroup$
add a comment |
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$begingroup$
It is actually very simple. Draw some disjoint curves connecting singularities,
or starting at the singularities and going to infinity, so that these curves separate the plane into simply connected regions. Then instead of a "multivalued function" you have several ordinary (single-valued) functions in each region.
These ordinary functions are called "branches" of the multivalued function.
When you choose a branch and try to cross one of your curves into another region
your branch must match one of the branches in that other region.
For example, $sqrt{z}$ has one singularity in the plane, namely $0$ (another one is $infty$). Draw ray from $0$ to $infty$. You obtain one region. In this region you have two ordinary functions, branches of $sqrt{z}$. When you cross the cut, they are interchanged.
answered Nov 29 '17 at 21:04
Alexandre EremenkoAlexandre Eremenko
2,415921
$endgroup$
add a comment |
$begingroup$
It is actually very simple. Draw some disjoint curves connecting singularities,
or starting at the singularities and going to infinity, so that these curves separate the plane into simply connected regions. Then instead of a "multivalued function" you have several ordinary (single-valued) functions in each region.
These ordinary functions are called "branches" of the multivalued function.
When you choose a branch and try to cross one of your curves into another region
your branch must match one of the branches in that other region.
For example, $sqrt{z}$ has one singularity in the plane, namely $0$ (another one is $infty$). Draw ray from $0$ to $infty$. You obtain one region. In this region you have two ordinary functions, branches of $sqrt{z}$. When you cross the cut, they are interchanged.
answered Nov 29 '17 at 21:04
Alexandre EremenkoAlexandre Eremenko
2,415921
$endgroup$
add a comment |
$begingroup$
It is actually very simple. Draw some disjoint curves connecting singularities,
or starting at the singularities and going to infinity, so that these curves separate the plane into simply connected regions. Then instead of a "multivalued function" you have several ordinary (single-valued) functions in each region.
These ordinary functions are called "branches" of the multivalued function.
When you choose a branch and try to cross one of your curves into another region
your branch must match one of the branches in that other region.
For example, $sqrt{z}$ has one singularity in the plane, namely $0$ (another one is $infty$). Draw ray from $0$ to $infty$. You obtain one region. In this region you have two ordinary functions, branches of $sqrt{z}$. When you cross the cut, they are interchanged.
answered Nov 29 '17 at 21:04
Alexandre EremenkoAlexandre Eremenko
2,415921
$endgroup$
It is actually very simple. Draw some disjoint curves connecting singularities,
or starting at the singularities and going to infinity, so that these curves separate the plane into simply connected regions. Then instead of a "multivalued function" you have several ordinary (single-valued) functions in each region.
These ordinary functions are called "branches" of the multivalued function.
When you choose a branch and try to cross one of your curves into another region
your branch must match one of the branches in that other region.
For example, $sqrt{z}$ has one singularity in the plane, namely $0$ (another one is $infty$). Draw ray from $0$ to $infty$. You obtain one region. In this region you have two ordinary functions, branches of $sqrt{z}$. When you cross the cut, they are interchanged.
answered Nov 29 '17 at 21:04
Alexandre EremenkoAlexandre Eremenko
2,415921
answered Nov 29 '17 at 21:04
Alexandre EremenkoAlexandre Eremenko
2,415921
answered Nov 29 '17 at 21:04
Alexandre EremenkoAlexandre Eremenko
2,415921
answered Nov 29 '17 at 21:04
Alexandre EremenkoAlexandre Eremenko
2,415921
2,415921
add a comment |
add a comment |
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$begingroup$
You're looking for algebraic functions. For each algebraic function, you can associate a Riemann surface such that the algebraic function becomes a holomorphic function. See Markushevich's book "Theory of Functions of a Complex Variable" for a practical procedure using triangulations. If you want something more abstract, there is Foster's "Lectures on Riemann Surfaces".
$endgroup$
– user40276
May 26 '15 at 16:14
$begingroup$
Thanks. Are there multiple volumes of Markushevich?
$endgroup$
– Braindead
May 26 '15 at 16:16
$begingroup$
And this is definitely not a trivial or obvious topic, since all the theory of elliptic curves and abelian varieties come from this (actually motives can be realized in this way!).
$endgroup$
– user40276
May 26 '15 at 16:17
$begingroup$
I think in the new edition all volumes were merged. For the old edition, I believe it's somewhere in the third volume.
$endgroup$
– user40276
May 26 '15 at 16:19
$begingroup$
Thanks! I'll check it out.
$endgroup$
– Braindead
May 26 '15 at 16:19