Branch cuts for $(z^2+1)^1/3$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Need help to understand branch cutsProblem identifying branch cuts of a square root functionBranch cuts of $frac1sqrtz^2 + m^2$Branch cuts for $sqrtz^2+1$Branch points and Branch cutsAlgorithmic/procedural way to find branch points and branch cuts?Conceptualizing Branch CutsContour integration on branch cutsArgument range for branch cut of $[-1,1]$?Elementary questions about branch points and branch cuts
Can a new player join a group only when a new campaign starts?
Do I really need to have a message in a novel to appeal to readers?
Maximum summed subsequences with non-adjacent items
How do I find out the mythology and history of my Fortress?
What's the meaning of "fortified infraction restraint"?
Is there a kind of relay only consumes power when switching?
A term for a woman complaining about things/begging in a cute/childish way
Crossing US/Canada Border for less than 24 hours
How could we fake a moon landing now?
Why does the remaining Rebel fleet at the end of Rogue One seem dramatically larger than the one in A New Hope?
How can I reduce the gap between left and right of cdot with a macro?
As a beginner, should I get a Squier Strat with a SSS config or a HSS?
Why should I vote and accept answers?
What is the font for "b" letter?
Did Deadpool rescue all of the X-Force?
NumericArray versus PackedArray in MMA12
Most bit efficient text communication method?
AppleTVs create a chatty alternate WiFi network
Hangman Game with C++
Why is it faster to reheat something than it is to cook it?
How to compare two different files line by line in unix?
How fail-safe is nr as stop bytes?
Should I use a zero-interest credit card for a large one-time purchase?
Why does it sometimes sound good to play a grace note as a lead in to a note in a melody?
Branch cuts for $(z^2+1)^1/3$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Need help to understand branch cutsProblem identifying branch cuts of a square root functionBranch cuts of $frac1sqrtz^2 + m^2$Branch cuts for $sqrtz^2+1$Branch points and Branch cutsAlgorithmic/procedural way to find branch points and branch cuts?Conceptualizing Branch CutsContour integration on branch cutsArgument range for branch cut of $[-1,1]$?Elementary questions about branch points and branch cuts
$begingroup$
I'm just learning about branch cuts so I'm hoping to get some clarification on this.
As in the title, I'm looking at $f(z)=(z^2+1)^1/3$. The obvious way to write this is $f(z)=exp(frac13ln(z^2+1))$. There are two main candidates for my branch cut (of $ln w$) here: $A:[0,-infty), B:[0,infty)$. I know I can choose any ray from the origin for $ln w$, but I think these are the only 2 interesting ones.
If I choose $A$ as my branch cut then that corresponds to branch cuts of $[i,iinfty)cup[-i,-iinfty)$ in $f(z)$. I believe this is the natural choice as it uses the principle branch of the logarithm and also a 'nice' branch for $f$.
On the other hand, if I choose $B$ as the branch cut, this corresponds to $[-i,i]$. This is where I'm slightly confused, from thinking about what happens to specific regions geometrically, it seems there is also a requirement for a branch cut along $(-infty,infty)$.
I was under the impression that branch cuts should be between branch points, but isn't the branch points of this function only $pm i$? In this case why is there also one along the entire real axis? Is this just a demonstration of why the $A$ cut is more natural?
complex-analysis branch-cuts branch-points
asked Apr 1 at 17:49
Oscar SOscar S
112
$endgroup$
add a comment |
$begingroup$
I'm just learning about branch cuts so I'm hoping to get some clarification on this.
As in the title, I'm looking at $f(z)=(z^2+1)^1/3$. The obvious way to write this is $f(z)=exp(frac13ln(z^2+1))$. There are two main candidates for my branch cut (of $ln w$) here: $A:[0,-infty), B:[0,infty)$. I know I can choose any ray from the origin for $ln w$, but I think these are the only 2 interesting ones.
If I choose $A$ as my branch cut then that corresponds to branch cuts of $[i,iinfty)cup[-i,-iinfty)$ in $f(z)$. I believe this is the natural choice as it uses the principle branch of the logarithm and also a 'nice' branch for $f$.
On the other hand, if I choose $B$ as the branch cut, this corresponds to $[-i,i]$. This is where I'm slightly confused, from thinking about what happens to specific regions geometrically, it seems there is also a requirement for a branch cut along $(-infty,infty)$.
I was under the impression that branch cuts should be between branch points, but isn't the branch points of this function only $pm i$? In this case why is there also one along the entire real axis? Is this just a demonstration of why the $A$ cut is more natural?
complex-analysis branch-cuts branch-points
asked Apr 1 at 17:49
Oscar SOscar S
112
$endgroup$
$begingroup$
$f(z) = (z+i)^1/3(z-i)^1/3$. When rotating one time ccw around $z=i$ then $f(z) mapsto e^2ipi /3 f(z)$, when rotating ccw around $z=-i$, $f(z) mapsto e^2ipi /3 f(z)$, and those are the generators of the monodromy group. The branch is any open set $Usubset BbbC-i,-i$ such that every closed curve $gamma subset U$ is trivial in the monodromy group.
$endgroup$
– reuns
Apr 2 at 4:30
$begingroup$
@reuns could you clarify what you mean by monodromy group? I haven't come across this term before and the Wikipedia page is far above my understanding.
$endgroup$
– Oscar S
Apr 2 at 8:55
$begingroup$
It is the group I defined, telling how $f$ transforms when it is continued analytically along a closed curve $subset BbbC-i,-i$. In general it is the homotopy group with basepoint $a$ quotiented by the group of closed curves $gamma: a to a$ such that $f$ around $a$ is equal to its analytic continuation $f_gamma$ around $a$.
$endgroup$
– reuns
Apr 2 at 20:23
add a comment |
$begingroup$
I'm just learning about branch cuts so I'm hoping to get some clarification on this.
As in the title, I'm looking at $f(z)=(z^2+1)^1/3$. The obvious way to write this is $f(z)=exp(frac13ln(z^2+1))$. There are two main candidates for my branch cut (of $ln w$) here: $A:[0,-infty), B:[0,infty)$. I know I can choose any ray from the origin for $ln w$, but I think these are the only 2 interesting ones.
If I choose $A$ as my branch cut then that corresponds to branch cuts of $[i,iinfty)cup[-i,-iinfty)$ in $f(z)$. I believe this is the natural choice as it uses the principle branch of the logarithm and also a 'nice' branch for $f$.
On the other hand, if I choose $B$ as the branch cut, this corresponds to $[-i,i]$. This is where I'm slightly confused, from thinking about what happens to specific regions geometrically, it seems there is also a requirement for a branch cut along $(-infty,infty)$.
I was under the impression that branch cuts should be between branch points, but isn't the branch points of this function only $pm i$? In this case why is there also one along the entire real axis? Is this just a demonstration of why the $A$ cut is more natural?
complex-analysis branch-cuts branch-points
asked Apr 1 at 17:49
Oscar SOscar S
112
$endgroup$
I'm just learning about branch cuts so I'm hoping to get some clarification on this.
As in the title, I'm looking at $f(z)=(z^2+1)^1/3$. The obvious way to write this is $f(z)=exp(frac13ln(z^2+1))$. There are two main candidates for my branch cut (of $ln w$) here: $A:[0,-infty), B:[0,infty)$. I know I can choose any ray from the origin for $ln w$, but I think these are the only 2 interesting ones.
If I choose $A$ as my branch cut then that corresponds to branch cuts of $[i,iinfty)cup[-i,-iinfty)$ in $f(z)$. I believe this is the natural choice as it uses the principle branch of the logarithm and also a 'nice' branch for $f$.
On the other hand, if I choose $B$ as the branch cut, this corresponds to $[-i,i]$. This is where I'm slightly confused, from thinking about what happens to specific regions geometrically, it seems there is also a requirement for a branch cut along $(-infty,infty)$.
I was under the impression that branch cuts should be between branch points, but isn't the branch points of this function only $pm i$? In this case why is there also one along the entire real axis? Is this just a demonstration of why the $A$ cut is more natural?
complex-analysis branch-cuts branch-points
complex-analysis branch-cuts branch-points
asked Apr 1 at 17:49
Oscar SOscar S
112
asked Apr 1 at 17:49
Oscar SOscar S
112
asked Apr 1 at 17:49
Oscar SOscar S
112
asked Apr 1 at 17:49
Oscar SOscar S
112
asked Apr 1 at 17:49
Oscar SOscar S
112
112
$begingroup$
$f(z) = (z+i)^1/3(z-i)^1/3$. When rotating one time ccw around $z=i$ then $f(z) mapsto e^2ipi /3 f(z)$, when rotating ccw around $z=-i$, $f(z) mapsto e^2ipi /3 f(z)$, and those are the generators of the monodromy group. The branch is any open set $Usubset BbbC-i,-i$ such that every closed curve $gamma subset U$ is trivial in the monodromy group.
$endgroup$
– reuns
Apr 2 at 4:30
$begingroup$
@reuns could you clarify what you mean by monodromy group? I haven't come across this term before and the Wikipedia page is far above my understanding.
$endgroup$
– Oscar S
Apr 2 at 8:55
$begingroup$
It is the group I defined, telling how $f$ transforms when it is continued analytically along a closed curve $subset BbbC-i,-i$. In general it is the homotopy group with basepoint $a$ quotiented by the group of closed curves $gamma: a to a$ such that $f$ around $a$ is equal to its analytic continuation $f_gamma$ around $a$.
$endgroup$
– reuns
Apr 2 at 20:23
add a comment |
$begingroup$
$f(z) = (z+i)^1/3(z-i)^1/3$. When rotating one time ccw around $z=i$ then $f(z) mapsto e^2ipi /3 f(z)$, when rotating ccw around $z=-i$, $f(z) mapsto e^2ipi /3 f(z)$, and those are the generators of the monodromy group. The branch is any open set $Usubset BbbC-i,-i$ such that every closed curve $gamma subset U$ is trivial in the monodromy group.
$endgroup$
– reuns
Apr 2 at 4:30
$begingroup$
@reuns could you clarify what you mean by monodromy group? I haven't come across this term before and the Wikipedia page is far above my understanding.
$endgroup$
– Oscar S
Apr 2 at 8:55
$begingroup$
It is the group I defined, telling how $f$ transforms when it is continued analytically along a closed curve $subset BbbC-i,-i$. In general it is the homotopy group with basepoint $a$ quotiented by the group of closed curves $gamma: a to a$ such that $f$ around $a$ is equal to its analytic continuation $f_gamma$ around $a$.
$endgroup$
– reuns
Apr 2 at 20:23
$begingroup$
$f(z) = (z+i)^1/3(z-i)^1/3$. When rotating one time ccw around $z=i$ then $f(z) mapsto e^2ipi /3 f(z)$, when rotating ccw around $z=-i$, $f(z) mapsto e^2ipi /3 f(z)$, and those are the generators of the monodromy group. The branch is any open set $Usubset BbbC-i,-i$ such that every closed curve $gamma subset U$ is trivial in the monodromy group.
$endgroup$
– reuns
Apr 2 at 4:30
$begingroup$
$f(z) = (z+i)^1/3(z-i)^1/3$. When rotating one time ccw around $z=i$ then $f(z) mapsto e^2ipi /3 f(z)$, when rotating ccw around $z=-i$, $f(z) mapsto e^2ipi /3 f(z)$, and those are the generators of the monodromy group. The branch is any open set $Usubset BbbC-i,-i$ such that every closed curve $gamma subset U$ is trivial in the monodromy group.
$endgroup$
– reuns
Apr 2 at 4:30
$begingroup$
@reuns could you clarify what you mean by monodromy group? I haven't come across this term before and the Wikipedia page is far above my understanding.
$endgroup$
– Oscar S
Apr 2 at 8:55
$begingroup$
@reuns could you clarify what you mean by monodromy group? I haven't come across this term before and the Wikipedia page is far above my understanding.
$endgroup$
– Oscar S
Apr 2 at 8:55
$begingroup$
It is the group I defined, telling how $f$ transforms when it is continued analytically along a closed curve $subset BbbC-i,-i$. In general it is the homotopy group with basepoint $a$ quotiented by the group of closed curves $gamma: a to a$ such that $f$ around $a$ is equal to its analytic continuation $f_gamma$ around $a$.
$endgroup$
– reuns
Apr 2 at 20:23
$begingroup$
It is the group I defined, telling how $f$ transforms when it is continued analytically along a closed curve $subset BbbC-i,-i$. In general it is the homotopy group with basepoint $a$ quotiented by the group of closed curves $gamma: a to a$ such that $f$ around $a$ is equal to its analytic continuation $f_gamma$ around $a$.
$endgroup$
– reuns
Apr 2 at 20:23
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Having thought about this a lot more I think I have an explanation.
If one considers the behaviour of $lim_toinfty f(z) sim z^frac23$. Here we clearly have a branch point at $infty$ as well as the ones previously found in the question.
In branch cut $A$, our choice of cut going through infinity is important as this then includes our new branch point. Originally it was thought that the path through infinity was similar to in the case of $g(z)=(1+z^2)^frac12$ where the principle branch cut 'happens' to go through infinity but isn't required to . It turns out this is not the case1.
When we then move to branch cut $B$, we no longer have this branch point at infinity on our branch cut. This is why it is necessary to include the $(-infty, infty)$ branch cut. Presumably any additional branch cut through infinity would also suffice.
Hopefully this answer clears up this question if anyone else has it. If I am wrong or unclear about anything then please let me know and I'll have another look and hopefully understand it better afterwards.
1. This can be seen by noticing that the asymptotic behaviour of $g$ is clearly different $lim_toinfty g(z) sim z$.
answered Apr 1 at 23:17
Oscar SOscar S
112
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3170915%2fbranch-cuts-for-z211-3%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Having thought about this a lot more I think I have an explanation.
If one considers the behaviour of $lim_toinfty f(z) sim z^frac23$. Here we clearly have a branch point at $infty$ as well as the ones previously found in the question.
In branch cut $A$, our choice of cut going through infinity is important as this then includes our new branch point. Originally it was thought that the path through infinity was similar to in the case of $g(z)=(1+z^2)^frac12$ where the principle branch cut 'happens' to go through infinity but isn't required to . It turns out this is not the case1.
When we then move to branch cut $B$, we no longer have this branch point at infinity on our branch cut. This is why it is necessary to include the $(-infty, infty)$ branch cut. Presumably any additional branch cut through infinity would also suffice.
Hopefully this answer clears up this question if anyone else has it. If I am wrong or unclear about anything then please let me know and I'll have another look and hopefully understand it better afterwards.
1. This can be seen by noticing that the asymptotic behaviour of $g$ is clearly different $lim_toinfty g(z) sim z$.
answered Apr 1 at 23:17
Oscar SOscar S
112
$endgroup$
add a comment |
$begingroup$
Having thought about this a lot more I think I have an explanation.
If one considers the behaviour of $lim_toinfty f(z) sim z^frac23$. Here we clearly have a branch point at $infty$ as well as the ones previously found in the question.
In branch cut $A$, our choice of cut going through infinity is important as this then includes our new branch point. Originally it was thought that the path through infinity was similar to in the case of $g(z)=(1+z^2)^frac12$ where the principle branch cut 'happens' to go through infinity but isn't required to . It turns out this is not the case1.
When we then move to branch cut $B$, we no longer have this branch point at infinity on our branch cut. This is why it is necessary to include the $(-infty, infty)$ branch cut. Presumably any additional branch cut through infinity would also suffice.
Hopefully this answer clears up this question if anyone else has it. If I am wrong or unclear about anything then please let me know and I'll have another look and hopefully understand it better afterwards.
1. This can be seen by noticing that the asymptotic behaviour of $g$ is clearly different $lim_toinfty g(z) sim z$.
answered Apr 1 at 23:17
Oscar SOscar S
112
$endgroup$
add a comment |
$begingroup$
Having thought about this a lot more I think I have an explanation.
If one considers the behaviour of $lim_toinfty f(z) sim z^frac23$. Here we clearly have a branch point at $infty$ as well as the ones previously found in the question.
In branch cut $A$, our choice of cut going through infinity is important as this then includes our new branch point. Originally it was thought that the path through infinity was similar to in the case of $g(z)=(1+z^2)^frac12$ where the principle branch cut 'happens' to go through infinity but isn't required to . It turns out this is not the case1.
When we then move to branch cut $B$, we no longer have this branch point at infinity on our branch cut. This is why it is necessary to include the $(-infty, infty)$ branch cut. Presumably any additional branch cut through infinity would also suffice.
Hopefully this answer clears up this question if anyone else has it. If I am wrong or unclear about anything then please let me know and I'll have another look and hopefully understand it better afterwards.
1. This can be seen by noticing that the asymptotic behaviour of $g$ is clearly different $lim_toinfty g(z) sim z$.
answered Apr 1 at 23:17
Oscar SOscar S
112
$endgroup$
Having thought about this a lot more I think I have an explanation.
If one considers the behaviour of $lim_toinfty f(z) sim z^frac23$. Here we clearly have a branch point at $infty$ as well as the ones previously found in the question.
In branch cut $A$, our choice of cut going through infinity is important as this then includes our new branch point. Originally it was thought that the path through infinity was similar to in the case of $g(z)=(1+z^2)^frac12$ where the principle branch cut 'happens' to go through infinity but isn't required to . It turns out this is not the case1.
When we then move to branch cut $B$, we no longer have this branch point at infinity on our branch cut. This is why it is necessary to include the $(-infty, infty)$ branch cut. Presumably any additional branch cut through infinity would also suffice.
Hopefully this answer clears up this question if anyone else has it. If I am wrong or unclear about anything then please let me know and I'll have another look and hopefully understand it better afterwards.
1. This can be seen by noticing that the asymptotic behaviour of $g$ is clearly different $lim_toinfty g(z) sim z$.
answered Apr 1 at 23:17
Oscar SOscar S
112
answered Apr 1 at 23:17
Oscar SOscar S
112
answered Apr 1 at 23:17
Oscar SOscar S
112
answered Apr 1 at 23:17
Oscar SOscar S
112
112
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3170915%2fbranch-cuts-for-z211-3%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
$f(z) = (z+i)^1/3(z-i)^1/3$. When rotating one time ccw around $z=i$ then $f(z) mapsto e^2ipi /3 f(z)$, when rotating ccw around $z=-i$, $f(z) mapsto e^2ipi /3 f(z)$, and those are the generators of the monodromy group. The branch is any open set $Usubset BbbC-i,-i$ such that every closed curve $gamma subset U$ is trivial in the monodromy group.
$endgroup$
– reuns
Apr 2 at 4:30
$begingroup$
@reuns could you clarify what you mean by monodromy group? I haven't come across this term before and the Wikipedia page is far above my understanding.
$endgroup$
– Oscar S
Apr 2 at 8:55
$begingroup$
It is the group I defined, telling how $f$ transforms when it is continued analytically along a closed curve $subset BbbC-i,-i$. In general it is the homotopy group with basepoint $a$ quotiented by the group of closed curves $gamma: a to a$ such that $f$ around $a$ is equal to its analytic continuation $f_gamma$ around $a$.
$endgroup$
– reuns
Apr 2 at 20:23