Choosing multiple branch cuts
$begingroup$
We have $z^{frac{1}{3}}$. I need to find three branch cuts of this function. I know branch cuts are made such that the function becomes single valued. However, I'm really uncertain on how to find the branch cuts for this function. I've read up on the topic and discovered that they can be chosen at random, depending on the problem you're dealing with. This doesn't seem helpful.
I have determined that the branch point for this function is rather obvious: it's $0$. How do I use this fact to produce three branch cuts? If I needed only one, then I would be able to determine one. How does it work for three? Thanks in advance.
complex-analysis branch-cuts branch-points
asked Nov 26 '18 at 17:20
user569579user569579
735
$endgroup$
add a comment |
$begingroup$
We have $z^{frac{1}{3}}$. I need to find three branch cuts of this function. I know branch cuts are made such that the function becomes single valued. However, I'm really uncertain on how to find the branch cuts for this function. I've read up on the topic and discovered that they can be chosen at random, depending on the problem you're dealing with. This doesn't seem helpful.
I have determined that the branch point for this function is rather obvious: it's $0$. How do I use this fact to produce three branch cuts? If I needed only one, then I would be able to determine one. How does it work for three? Thanks in advance.
complex-analysis branch-cuts branch-points
asked Nov 26 '18 at 17:20
user569579user569579
735
$endgroup$
$begingroup$
You are correct that $z^{1/3}$ is single valued if the plane is cut from $0$ to the point at $infty$. The path that connects $0$ and $infty$ is arbitrary. That is to say, the branch cut need not be a straight line. Once chosen, for each $z$, one Riemann sheet of a branch has $thetain (theta_0(z),theta_0(z)+2pi]$. For that same $z$, the next Riemann sheet has $theta in (theta_0(z)+2pi, theta_0(z)+4pi]$.
$endgroup$
– Mark Viola
Nov 26 '18 at 18:00
$begingroup$
Does this mean that the second and third branch would be given by specifying different range of values for $arg(z)$, rather than by specifying a second and third branch cut? Maybe that is where the confusion lies. I deduce from your comment that branches aren't the same as branch cuts. Is that correct?
$endgroup$
– user569579
Nov 26 '18 at 18:05
$begingroup$
You need only choose one branch cut. The branches are the Riemann sheets.
$endgroup$
– Mark Viola
Nov 26 '18 at 18:12
$begingroup$
I am not yet familiar with that terminology (Riemann sheets), but I'll look it up. Thank you very much.
$endgroup$
– user569579
Nov 26 '18 at 18:16
$begingroup$
You're welcome. My pleasure.
$endgroup$
– Mark Viola
Nov 26 '18 at 18:20
add a comment |
$begingroup$
We have $z^{frac{1}{3}}$. I need to find three branch cuts of this function. I know branch cuts are made such that the function becomes single valued. However, I'm really uncertain on how to find the branch cuts for this function. I've read up on the topic and discovered that they can be chosen at random, depending on the problem you're dealing with. This doesn't seem helpful.
I have determined that the branch point for this function is rather obvious: it's $0$. How do I use this fact to produce three branch cuts? If I needed only one, then I would be able to determine one. How does it work for three? Thanks in advance.
complex-analysis branch-cuts branch-points
asked Nov 26 '18 at 17:20
user569579user569579
735
$endgroup$
We have $z^{frac{1}{3}}$. I need to find three branch cuts of this function. I know branch cuts are made such that the function becomes single valued. However, I'm really uncertain on how to find the branch cuts for this function. I've read up on the topic and discovered that they can be chosen at random, depending on the problem you're dealing with. This doesn't seem helpful.
I have determined that the branch point for this function is rather obvious: it's $0$. How do I use this fact to produce three branch cuts? If I needed only one, then I would be able to determine one. How does it work for three? Thanks in advance.
complex-analysis branch-cuts branch-points
complex-analysis branch-cuts branch-points
asked Nov 26 '18 at 17:20
user569579user569579
735
asked Nov 26 '18 at 17:20
user569579user569579
735
asked Nov 26 '18 at 17:20
user569579user569579
735
asked Nov 26 '18 at 17:20
user569579user569579
735
asked Nov 26 '18 at 17:20
user569579user569579
735
735
$begingroup$
You are correct that $z^{1/3}$ is single valued if the plane is cut from $0$ to the point at $infty$. The path that connects $0$ and $infty$ is arbitrary. That is to say, the branch cut need not be a straight line. Once chosen, for each $z$, one Riemann sheet of a branch has $thetain (theta_0(z),theta_0(z)+2pi]$. For that same $z$, the next Riemann sheet has $theta in (theta_0(z)+2pi, theta_0(z)+4pi]$.
$endgroup$
– Mark Viola
Nov 26 '18 at 18:00
$begingroup$
Does this mean that the second and third branch would be given by specifying different range of values for $arg(z)$, rather than by specifying a second and third branch cut? Maybe that is where the confusion lies. I deduce from your comment that branches aren't the same as branch cuts. Is that correct?
$endgroup$
– user569579
Nov 26 '18 at 18:05
$begingroup$
You need only choose one branch cut. The branches are the Riemann sheets.
$endgroup$
– Mark Viola
Nov 26 '18 at 18:12
$begingroup$
I am not yet familiar with that terminology (Riemann sheets), but I'll look it up. Thank you very much.
$endgroup$
– user569579
Nov 26 '18 at 18:16
$begingroup$
You're welcome. My pleasure.
$endgroup$
– Mark Viola
Nov 26 '18 at 18:20
add a comment |
$begingroup$
You are correct that $z^{1/3}$ is single valued if the plane is cut from $0$ to the point at $infty$. The path that connects $0$ and $infty$ is arbitrary. That is to say, the branch cut need not be a straight line. Once chosen, for each $z$, one Riemann sheet of a branch has $thetain (theta_0(z),theta_0(z)+2pi]$. For that same $z$, the next Riemann sheet has $theta in (theta_0(z)+2pi, theta_0(z)+4pi]$.
$endgroup$
– Mark Viola
Nov 26 '18 at 18:00
$begingroup$
Does this mean that the second and third branch would be given by specifying different range of values for $arg(z)$, rather than by specifying a second and third branch cut? Maybe that is where the confusion lies. I deduce from your comment that branches aren't the same as branch cuts. Is that correct?
$endgroup$
– user569579
Nov 26 '18 at 18:05
$begingroup$
You need only choose one branch cut. The branches are the Riemann sheets.
$endgroup$
– Mark Viola
Nov 26 '18 at 18:12
$begingroup$
I am not yet familiar with that terminology (Riemann sheets), but I'll look it up. Thank you very much.
$endgroup$
– user569579
Nov 26 '18 at 18:16
$begingroup$
You're welcome. My pleasure.
$endgroup$
– Mark Viola
Nov 26 '18 at 18:20
$begingroup$
You are correct that $z^{1/3}$ is single valued if the plane is cut from $0$ to the point at $infty$. The path that connects $0$ and $infty$ is arbitrary. That is to say, the branch cut need not be a straight line. Once chosen, for each $z$, one Riemann sheet of a branch has $thetain (theta_0(z),theta_0(z)+2pi]$. For that same $z$, the next Riemann sheet has $theta in (theta_0(z)+2pi, theta_0(z)+4pi]$.
$endgroup$
– Mark Viola
Nov 26 '18 at 18:00
$begingroup$
You are correct that $z^{1/3}$ is single valued if the plane is cut from $0$ to the point at $infty$. The path that connects $0$ and $infty$ is arbitrary. That is to say, the branch cut need not be a straight line. Once chosen, for each $z$, one Riemann sheet of a branch has $thetain (theta_0(z),theta_0(z)+2pi]$. For that same $z$, the next Riemann sheet has $theta in (theta_0(z)+2pi, theta_0(z)+4pi]$.
$endgroup$
– Mark Viola
Nov 26 '18 at 18:00
$begingroup$
Does this mean that the second and third branch would be given by specifying different range of values for $arg(z)$, rather than by specifying a second and third branch cut? Maybe that is where the confusion lies. I deduce from your comment that branches aren't the same as branch cuts. Is that correct?
$endgroup$
– user569579
Nov 26 '18 at 18:05
$begingroup$
Does this mean that the second and third branch would be given by specifying different range of values for $arg(z)$, rather than by specifying a second and third branch cut? Maybe that is where the confusion lies. I deduce from your comment that branches aren't the same as branch cuts. Is that correct?
$endgroup$
– user569579
Nov 26 '18 at 18:05
$begingroup$
You need only choose one branch cut. The branches are the Riemann sheets.
$endgroup$
– Mark Viola
Nov 26 '18 at 18:12
$begingroup$
You need only choose one branch cut. The branches are the Riemann sheets.
$endgroup$
– Mark Viola
Nov 26 '18 at 18:12
$begingroup$
I am not yet familiar with that terminology (Riemann sheets), but I'll look it up. Thank you very much.
$endgroup$
– user569579
Nov 26 '18 at 18:16
$begingroup$
I am not yet familiar with that terminology (Riemann sheets), but I'll look it up. Thank you very much.
$endgroup$
– user569579
Nov 26 '18 at 18:16
$begingroup$
You're welcome. My pleasure.
$endgroup$
– Mark Viola
Nov 26 '18 at 18:20
$begingroup$
You're welcome. My pleasure.
$endgroup$
– Mark Viola
Nov 26 '18 at 18:20
add a comment |
1 Answer
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$begingroup$
I think you are confusing branches and branch points. The only branch points are $0,infty$. To find three branches of the function, cut along $mathbb R^-$. Then you have a standard branch $zmapsto |z|e^{i theta/3}$ where $theta=arg zin(-pi,pi)$. The other two are obtained by multiplying by a cube root of unity.
answered Nov 26 '18 at 17:30
Richard MartinRichard Martin
1,61118
$endgroup$
$begingroup$
Is there no connections between the branch cuts and the branch points? I assumed there would be, that's why I started by finding the branch point $0$.
$endgroup$
– user569579
Nov 26 '18 at 17:33
$begingroup$
Of course there is a connection; but that doesn't mean to say that three branches equates to three branch cuts. Unless you put the cuts on the Riemann surface, I guess, in which case there are indeed three (and all collapse down to $mathbb R^-$ when you project on to $mathbb C$).
$endgroup$
– Richard Martin
Nov 26 '18 at 17:37
$begingroup$
I think I was clear on the distinction between the two, in that case, but at any rate, you helped out a great deal, so thank you very much. If I did confuse the two, then I'm afraid I still don't see where that confusion lies.
$endgroup$
– user569579
Nov 26 '18 at 17:38
add a comment |
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1 Answer
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1 Answer
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$begingroup$
I think you are confusing branches and branch points. The only branch points are $0,infty$. To find three branches of the function, cut along $mathbb R^-$. Then you have a standard branch $zmapsto |z|e^{i theta/3}$ where $theta=arg zin(-pi,pi)$. The other two are obtained by multiplying by a cube root of unity.
answered Nov 26 '18 at 17:30
Richard MartinRichard Martin
1,61118
$endgroup$
$begingroup$
Is there no connections between the branch cuts and the branch points? I assumed there would be, that's why I started by finding the branch point $0$.
$endgroup$
– user569579
Nov 26 '18 at 17:33
$begingroup$
Of course there is a connection; but that doesn't mean to say that three branches equates to three branch cuts. Unless you put the cuts on the Riemann surface, I guess, in which case there are indeed three (and all collapse down to $mathbb R^-$ when you project on to $mathbb C$).
$endgroup$
– Richard Martin
Nov 26 '18 at 17:37
$begingroup$
I think I was clear on the distinction between the two, in that case, but at any rate, you helped out a great deal, so thank you very much. If I did confuse the two, then I'm afraid I still don't see where that confusion lies.
$endgroup$
– user569579
Nov 26 '18 at 17:38
add a comment |
$begingroup$
I think you are confusing branches and branch points. The only branch points are $0,infty$. To find three branches of the function, cut along $mathbb R^-$. Then you have a standard branch $zmapsto |z|e^{i theta/3}$ where $theta=arg zin(-pi,pi)$. The other two are obtained by multiplying by a cube root of unity.
answered Nov 26 '18 at 17:30
Richard MartinRichard Martin
1,61118
$endgroup$
$begingroup$
Is there no connections between the branch cuts and the branch points? I assumed there would be, that's why I started by finding the branch point $0$.
$endgroup$
– user569579
Nov 26 '18 at 17:33
$begingroup$
Of course there is a connection; but that doesn't mean to say that three branches equates to three branch cuts. Unless you put the cuts on the Riemann surface, I guess, in which case there are indeed three (and all collapse down to $mathbb R^-$ when you project on to $mathbb C$).
$endgroup$
– Richard Martin
Nov 26 '18 at 17:37
$begingroup$
I think I was clear on the distinction between the two, in that case, but at any rate, you helped out a great deal, so thank you very much. If I did confuse the two, then I'm afraid I still don't see where that confusion lies.
$endgroup$
– user569579
Nov 26 '18 at 17:38
add a comment |
$begingroup$
I think you are confusing branches and branch points. The only branch points are $0,infty$. To find three branches of the function, cut along $mathbb R^-$. Then you have a standard branch $zmapsto |z|e^{i theta/3}$ where $theta=arg zin(-pi,pi)$. The other two are obtained by multiplying by a cube root of unity.
answered Nov 26 '18 at 17:30
Richard MartinRichard Martin
1,61118
$endgroup$
I think you are confusing branches and branch points. The only branch points are $0,infty$. To find three branches of the function, cut along $mathbb R^-$. Then you have a standard branch $zmapsto |z|e^{i theta/3}$ where $theta=arg zin(-pi,pi)$. The other two are obtained by multiplying by a cube root of unity.
answered Nov 26 '18 at 17:30
Richard MartinRichard Martin
1,61118
answered Nov 26 '18 at 17:30
Richard MartinRichard Martin
1,61118
answered Nov 26 '18 at 17:30
Richard MartinRichard Martin
1,61118
answered Nov 26 '18 at 17:30
Richard MartinRichard Martin
1,61118
1,61118
$begingroup$
Is there no connections between the branch cuts and the branch points? I assumed there would be, that's why I started by finding the branch point $0$.
$endgroup$
– user569579
Nov 26 '18 at 17:33
$begingroup$
Of course there is a connection; but that doesn't mean to say that three branches equates to three branch cuts. Unless you put the cuts on the Riemann surface, I guess, in which case there are indeed three (and all collapse down to $mathbb R^-$ when you project on to $mathbb C$).
$endgroup$
– Richard Martin
Nov 26 '18 at 17:37
$begingroup$
I think I was clear on the distinction between the two, in that case, but at any rate, you helped out a great deal, so thank you very much. If I did confuse the two, then I'm afraid I still don't see where that confusion lies.
$endgroup$
– user569579
Nov 26 '18 at 17:38
add a comment |
$begingroup$
Is there no connections between the branch cuts and the branch points? I assumed there would be, that's why I started by finding the branch point $0$.
$endgroup$
– user569579
Nov 26 '18 at 17:33
$begingroup$
Of course there is a connection; but that doesn't mean to say that three branches equates to three branch cuts. Unless you put the cuts on the Riemann surface, I guess, in which case there are indeed three (and all collapse down to $mathbb R^-$ when you project on to $mathbb C$).
$endgroup$
– Richard Martin
Nov 26 '18 at 17:37
$begingroup$
I think I was clear on the distinction between the two, in that case, but at any rate, you helped out a great deal, so thank you very much. If I did confuse the two, then I'm afraid I still don't see where that confusion lies.
$endgroup$
– user569579
Nov 26 '18 at 17:38
$begingroup$
Is there no connections between the branch cuts and the branch points? I assumed there would be, that's why I started by finding the branch point $0$.
$endgroup$
– user569579
Nov 26 '18 at 17:33
$begingroup$
Is there no connections between the branch cuts and the branch points? I assumed there would be, that's why I started by finding the branch point $0$.
$endgroup$
– user569579
Nov 26 '18 at 17:33
$begingroup$
Of course there is a connection; but that doesn't mean to say that three branches equates to three branch cuts. Unless you put the cuts on the Riemann surface, I guess, in which case there are indeed three (and all collapse down to $mathbb R^-$ when you project on to $mathbb C$).
$endgroup$
– Richard Martin
Nov 26 '18 at 17:37
$begingroup$
Of course there is a connection; but that doesn't mean to say that three branches equates to three branch cuts. Unless you put the cuts on the Riemann surface, I guess, in which case there are indeed three (and all collapse down to $mathbb R^-$ when you project on to $mathbb C$).
$endgroup$
– Richard Martin
Nov 26 '18 at 17:37
$begingroup$
I think I was clear on the distinction between the two, in that case, but at any rate, you helped out a great deal, so thank you very much. If I did confuse the two, then I'm afraid I still don't see where that confusion lies.
$endgroup$
– user569579
Nov 26 '18 at 17:38
$begingroup$
I think I was clear on the distinction between the two, in that case, but at any rate, you helped out a great deal, so thank you very much. If I did confuse the two, then I'm afraid I still don't see where that confusion lies.
$endgroup$
– user569579
Nov 26 '18 at 17:38
add a comment |
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$begingroup$
You are correct that $z^{1/3}$ is single valued if the plane is cut from $0$ to the point at $infty$. The path that connects $0$ and $infty$ is arbitrary. That is to say, the branch cut need not be a straight line. Once chosen, for each $z$, one Riemann sheet of a branch has $thetain (theta_0(z),theta_0(z)+2pi]$. For that same $z$, the next Riemann sheet has $theta in (theta_0(z)+2pi, theta_0(z)+4pi]$.
$endgroup$
– Mark Viola
Nov 26 '18 at 18:00
$begingroup$
Does this mean that the second and third branch would be given by specifying different range of values for $arg(z)$, rather than by specifying a second and third branch cut? Maybe that is where the confusion lies. I deduce from your comment that branches aren't the same as branch cuts. Is that correct?
$endgroup$
– user569579
Nov 26 '18 at 18:05
$begingroup$
You need only choose one branch cut. The branches are the Riemann sheets.
$endgroup$
– Mark Viola
Nov 26 '18 at 18:12
$begingroup$
I am not yet familiar with that terminology (Riemann sheets), but I'll look it up. Thank you very much.
$endgroup$
– user569579
Nov 26 '18 at 18:16
$begingroup$
You're welcome. My pleasure.
$endgroup$
– Mark Viola
Nov 26 '18 at 18:20