Branch point of log(z)
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I'm currently following a class on complex analysis (for physicists) and I'm having trouble with the concept of branch points. In my textbook, they say that log(z) has a branch point at zero and at infinity. They also state that every other point is not a branch point and i'm having a hard time proving that $z=2$ is actually not a branch point. I've searched a lot without finding anything convincing.
So far I tried the following :
Let's say we look at the function $f(z) = ln(z)$ at $z=2$. To show that this is, or is not a branch point, we can draw a circle of radius 1 around this point. Thus, let us take $z_i = 2+e^icdot 0$ and $z_f=2+e^2pi i$. In that case, $f(z_i)=ln(2+1) = ln(3e^i2pi n)$ and $f(z_f)=ln(2+1) = ln(3e^i 2pi n)$ and thus $ f(z_i)=f(z_f)$, which is what we want. However, if I take a circle of radius 3, the result of this methode is the sames, altough it shouldn't since the circle goes around zero. Indeed, in that case, this method leads to :
$f(z_i)=ln(2+3) = ln(5e^i2pi n)$ and $f(z_f)=ln(2+3) = ln(5e^i2pi n)$...
I guess there is something wrong with what I'm doing, but I don't know what...
I would like to show that $z=2$ is not a branch point using contours, not by other means.
Thanks a lot!
complex-analysis logarithms branch-points
asked Feb 18 '17 at 21:52
Xavier-rp
11
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I'm currently following a class on complex analysis (for physicists) and I'm having trouble with the concept of branch points. In my textbook, they say that log(z) has a branch point at zero and at infinity. They also state that every other point is not a branch point and i'm having a hard time proving that $z=2$ is actually not a branch point. I've searched a lot without finding anything convincing.
So far I tried the following :
Let's say we look at the function $f(z) = ln(z)$ at $z=2$. To show that this is, or is not a branch point, we can draw a circle of radius 1 around this point. Thus, let us take $z_i = 2+e^icdot 0$ and $z_f=2+e^2pi i$. In that case, $f(z_i)=ln(2+1) = ln(3e^i2pi n)$ and $f(z_f)=ln(2+1) = ln(3e^i 2pi n)$ and thus $ f(z_i)=f(z_f)$, which is what we want. However, if I take a circle of radius 3, the result of this methode is the sames, altough it shouldn't since the circle goes around zero. Indeed, in that case, this method leads to :
$f(z_i)=ln(2+3) = ln(5e^i2pi n)$ and $f(z_f)=ln(2+3) = ln(5e^i2pi n)$...
I guess there is something wrong with what I'm doing, but I don't know what...
I would like to show that $z=2$ is not a branch point using contours, not by other means.
Thanks a lot!
complex-analysis logarithms branch-points
asked Feb 18 '17 at 21:52
Xavier-rp
11
If $f(z) = e^z$, then $f(log 2) = 2$, and $f'(log 2) neq 0$, hence $f$ has a local, analytic inverse $g$. It follows that $log$ does not have branch point at $2$.
– copper.hat
Feb 18 '17 at 22:00
Is there a way to find this answer using contours?
– Xavier-rp
Feb 18 '17 at 22:02
I don't understand what you mean by 'using contours'. If $g$ (which is $log$ plus a constant) is analytic at $2$ then the value of $g$ locally is independent of any path.
– copper.hat
Feb 18 '17 at 22:04
Page 4 of that textbook : (math.mit.edu/classes/18.305/Notes/n00Branch_Points_B_Cuts.pdf) states that ''consider again the case of the function log(z), take z0 = 2 and a closed curve around z = 2 that also encloses z = 0. The value of log(z) wil l change as this curve is traced, but this does not make z = 2 a branch point. In fact, for curves close to z = 2 there is no change, hence z = 2 is not a branch point of log(z)!) ''
– Xavier-rp
Feb 18 '17 at 22:09
The inverse function theorem shows that $log$ is analytic in a convex neighbourhood of $2$. So there is no issue with definition. If you take a path around $2$ that also encloses $0$, then the function will need to be extended ('define it continuously' in the book) as one goes around the path and on return to the starting point will have accumulated some multiple of $2 pi$ (depending on the net encirclements). The function $log$ is defined everywhere in a neighbourhood of $2$ so there is no need to extend.
– copper.hat
Feb 18 '17 at 22:24
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm currently following a class on complex analysis (for physicists) and I'm having trouble with the concept of branch points. In my textbook, they say that log(z) has a branch point at zero and at infinity. They also state that every other point is not a branch point and i'm having a hard time proving that $z=2$ is actually not a branch point. I've searched a lot without finding anything convincing.
So far I tried the following :
Let's say we look at the function $f(z) = ln(z)$ at $z=2$. To show that this is, or is not a branch point, we can draw a circle of radius 1 around this point. Thus, let us take $z_i = 2+e^icdot 0$ and $z_f=2+e^2pi i$. In that case, $f(z_i)=ln(2+1) = ln(3e^i2pi n)$ and $f(z_f)=ln(2+1) = ln(3e^i 2pi n)$ and thus $ f(z_i)=f(z_f)$, which is what we want. However, if I take a circle of radius 3, the result of this methode is the sames, altough it shouldn't since the circle goes around zero. Indeed, in that case, this method leads to :
$f(z_i)=ln(2+3) = ln(5e^i2pi n)$ and $f(z_f)=ln(2+3) = ln(5e^i2pi n)$...
I guess there is something wrong with what I'm doing, but I don't know what...
I would like to show that $z=2$ is not a branch point using contours, not by other means.
Thanks a lot!
complex-analysis logarithms branch-points
asked Feb 18 '17 at 21:52
Xavier-rp
11
I'm currently following a class on complex analysis (for physicists) and I'm having trouble with the concept of branch points. In my textbook, they say that log(z) has a branch point at zero and at infinity. They also state that every other point is not a branch point and i'm having a hard time proving that $z=2$ is actually not a branch point. I've searched a lot without finding anything convincing.
So far I tried the following :
Let's say we look at the function $f(z) = ln(z)$ at $z=2$. To show that this is, or is not a branch point, we can draw a circle of radius 1 around this point. Thus, let us take $z_i = 2+e^icdot 0$ and $z_f=2+e^2pi i$. In that case, $f(z_i)=ln(2+1) = ln(3e^i2pi n)$ and $f(z_f)=ln(2+1) = ln(3e^i 2pi n)$ and thus $ f(z_i)=f(z_f)$, which is what we want. However, if I take a circle of radius 3, the result of this methode is the sames, altough it shouldn't since the circle goes around zero. Indeed, in that case, this method leads to :
$f(z_i)=ln(2+3) = ln(5e^i2pi n)$ and $f(z_f)=ln(2+3) = ln(5e^i2pi n)$...
I guess there is something wrong with what I'm doing, but I don't know what...
I would like to show that $z=2$ is not a branch point using contours, not by other means.
Thanks a lot!
complex-analysis logarithms branch-points
asked Feb 18 '17 at 21:52
Xavier-rp
11
edited Feb 18 '17 at 22:03
asked Feb 18 '17 at 21:52
Xavier-rp
11
asked Feb 18 '17 at 21:52
Xavier-rp
11
asked Feb 18 '17 at 21:52
Xavier-rp
11
11
If $f(z) = e^z$, then $f(log 2) = 2$, and $f'(log 2) neq 0$, hence $f$ has a local, analytic inverse $g$. It follows that $log$ does not have branch point at $2$.
– copper.hat
Feb 18 '17 at 22:00
Is there a way to find this answer using contours?
– Xavier-rp
Feb 18 '17 at 22:02
I don't understand what you mean by 'using contours'. If $g$ (which is $log$ plus a constant) is analytic at $2$ then the value of $g$ locally is independent of any path.
– copper.hat
Feb 18 '17 at 22:04
Page 4 of that textbook : (math.mit.edu/classes/18.305/Notes/n00Branch_Points_B_Cuts.pdf) states that ''consider again the case of the function log(z), take z0 = 2 and a closed curve around z = 2 that also encloses z = 0. The value of log(z) wil l change as this curve is traced, but this does not make z = 2 a branch point. In fact, for curves close to z = 2 there is no change, hence z = 2 is not a branch point of log(z)!) ''
– Xavier-rp
Feb 18 '17 at 22:09
The inverse function theorem shows that $log$ is analytic in a convex neighbourhood of $2$. So there is no issue with definition. If you take a path around $2$ that also encloses $0$, then the function will need to be extended ('define it continuously' in the book) as one goes around the path and on return to the starting point will have accumulated some multiple of $2 pi$ (depending on the net encirclements). The function $log$ is defined everywhere in a neighbourhood of $2$ so there is no need to extend.
– copper.hat
Feb 18 '17 at 22:24
add a comment |Â
If $f(z) = e^z$, then $f(log 2) = 2$, and $f'(log 2) neq 0$, hence $f$ has a local, analytic inverse $g$. It follows that $log$ does not have branch point at $2$.
– copper.hat
Feb 18 '17 at 22:00
Is there a way to find this answer using contours?
– Xavier-rp
Feb 18 '17 at 22:02
I don't understand what you mean by 'using contours'. If $g$ (which is $log$ plus a constant) is analytic at $2$ then the value of $g$ locally is independent of any path.
– copper.hat
Feb 18 '17 at 22:04
Page 4 of that textbook : (math.mit.edu/classes/18.305/Notes/n00Branch_Points_B_Cuts.pdf) states that ''consider again the case of the function log(z), take z0 = 2 and a closed curve around z = 2 that also encloses z = 0. The value of log(z) wil l change as this curve is traced, but this does not make z = 2 a branch point. In fact, for curves close to z = 2 there is no change, hence z = 2 is not a branch point of log(z)!) ''
– Xavier-rp
Feb 18 '17 at 22:09
The inverse function theorem shows that $log$ is analytic in a convex neighbourhood of $2$. So there is no issue with definition. If you take a path around $2$ that also encloses $0$, then the function will need to be extended ('define it continuously' in the book) as one goes around the path and on return to the starting point will have accumulated some multiple of $2 pi$ (depending on the net encirclements). The function $log$ is defined everywhere in a neighbourhood of $2$ so there is no need to extend.
– copper.hat
Feb 18 '17 at 22:24
If $f(z) = e^z$, then $f(log 2) = 2$, and $f'(log 2) neq 0$, hence $f$ has a local, analytic inverse $g$. It follows that $log$ does not have branch point at $2$.
– copper.hat
Feb 18 '17 at 22:00
If $f(z) = e^z$, then $f(log 2) = 2$, and $f'(log 2) neq 0$, hence $f$ has a local, analytic inverse $g$. It follows that $log$ does not have branch point at $2$.
– copper.hat
Feb 18 '17 at 22:00
Is there a way to find this answer using contours?
– Xavier-rp
Feb 18 '17 at 22:02
Is there a way to find this answer using contours?
– Xavier-rp
Feb 18 '17 at 22:02
I don't understand what you mean by 'using contours'. If $g$ (which is $log$ plus a constant) is analytic at $2$ then the value of $g$ locally is independent of any path.
– copper.hat
Feb 18 '17 at 22:04
I don't understand what you mean by 'using contours'. If $g$ (which is $log$ plus a constant) is analytic at $2$ then the value of $g$ locally is independent of any path.
– copper.hat
Feb 18 '17 at 22:04
Page 4 of that textbook : (math.mit.edu/classes/18.305/Notes/n00Branch_Points_B_Cuts.pdf) states that ''consider again the case of the function log(z), take z0 = 2 and a closed curve around z = 2 that also encloses z = 0. The value of log(z) wil l change as this curve is traced, but this does not make z = 2 a branch point. In fact, for curves close to z = 2 there is no change, hence z = 2 is not a branch point of log(z)!) ''
– Xavier-rp
Feb 18 '17 at 22:09
Page 4 of that textbook : (math.mit.edu/classes/18.305/Notes/n00Branch_Points_B_Cuts.pdf) states that ''consider again the case of the function log(z), take z0 = 2 and a closed curve around z = 2 that also encloses z = 0. The value of log(z) wil l change as this curve is traced, but this does not make z = 2 a branch point. In fact, for curves close to z = 2 there is no change, hence z = 2 is not a branch point of log(z)!) ''
– Xavier-rp
Feb 18 '17 at 22:09
The inverse function theorem shows that $log$ is analytic in a convex neighbourhood of $2$. So there is no issue with definition. If you take a path around $2$ that also encloses $0$, then the function will need to be extended ('define it continuously' in the book) as one goes around the path and on return to the starting point will have accumulated some multiple of $2 pi$ (depending on the net encirclements). The function $log$ is defined everywhere in a neighbourhood of $2$ so there is no need to extend.
– copper.hat
Feb 18 '17 at 22:24
The inverse function theorem shows that $log$ is analytic in a convex neighbourhood of $2$. So there is no issue with definition. If you take a path around $2$ that also encloses $0$, then the function will need to be extended ('define it continuously' in the book) as one goes around the path and on return to the starting point will have accumulated some multiple of $2 pi$ (depending on the net encirclements). The function $log$ is defined everywhere in a neighbourhood of $2$ so there is no need to extend.
– copper.hat
Feb 18 '17 at 22:24
add a comment |Â
1 Answer
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You need that the function can be defined like singlevalue function and this definition must be at least continuos inside the set joined to their interior. This assumption doesn't satisfy in the second contour.
answered Aug 26 at 7:15
Tomás
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1 Answer
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1 Answer
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oldest
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active
oldest
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active
oldest
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up vote
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You need that the function can be defined like singlevalue function and this definition must be at least continuos inside the set joined to their interior. This assumption doesn't satisfy in the second contour.
answered Aug 26 at 7:15
Tomás
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add a comment |Â
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You need that the function can be defined like singlevalue function and this definition must be at least continuos inside the set joined to their interior. This assumption doesn't satisfy in the second contour.
answered Aug 26 at 7:15
Tomás
1
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You need that the function can be defined like singlevalue function and this definition must be at least continuos inside the set joined to their interior. This assumption doesn't satisfy in the second contour.
answered Aug 26 at 7:15
Tomás
1
You need that the function can be defined like singlevalue function and this definition must be at least continuos inside the set joined to their interior. This assumption doesn't satisfy in the second contour.
answered Aug 26 at 7:15
Tomás
1
answered Aug 26 at 7:15
Tomás
1
answered Aug 26 at 7:15
Tomás
1
answered Aug 26 at 7:15
Tomás
1
1
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add a comment |Â
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If $f(z) = e^z$, then $f(log 2) = 2$, and $f'(log 2) neq 0$, hence $f$ has a local, analytic inverse $g$. It follows that $log$ does not have branch point at $2$.
– copper.hat
Feb 18 '17 at 22:00
Is there a way to find this answer using contours?
– Xavier-rp
Feb 18 '17 at 22:02
I don't understand what you mean by 'using contours'. If $g$ (which is $log$ plus a constant) is analytic at $2$ then the value of $g$ locally is independent of any path.
– copper.hat
Feb 18 '17 at 22:04
Page 4 of that textbook : (math.mit.edu/classes/18.305/Notes/n00Branch_Points_B_Cuts.pdf) states that ''consider again the case of the function log(z), take z0 = 2 and a closed curve around z = 2 that also encloses z = 0. The value of log(z) wil l change as this curve is traced, but this does not make z = 2 a branch point. In fact, for curves close to z = 2 there is no change, hence z = 2 is not a branch point of log(z)!) ''
– Xavier-rp
Feb 18 '17 at 22:09
The inverse function theorem shows that $log$ is analytic in a convex neighbourhood of $2$. So there is no issue with definition. If you take a path around $2$ that also encloses $0$, then the function will need to be extended ('define it continuously' in the book) as one goes around the path and on return to the starting point will have accumulated some multiple of $2 pi$ (depending on the net encirclements). The function $log$ is defined everywhere in a neighbourhood of $2$ so there is no need to extend.
– copper.hat
Feb 18 '17 at 22:24