Shifting the branch cut of the complex logarithm
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I have some troubles understanding branch cuts of the complex logarithm, or to be more precise, the shift thereof. So typically, the branch cut is along the negative real axis, connecting the branch points at $0$ and $infty$. We know, that this choice is not unique - I could, for instance, also choose the positive real axis.
Now, this is where I'm puzzled: is it also possible to shift it to any different spot, say, to the negative imaginary axis? This would lead to a single-valued function on all of the real axis except at the origin, where we still have a singularity - right?
logarithms branch-cuts
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I have some troubles understanding branch cuts of the complex logarithm, or to be more precise, the shift thereof. So typically, the branch cut is along the negative real axis, connecting the branch points at $0$ and $infty$. We know, that this choice is not unique - I could, for instance, also choose the positive real axis.
Now, this is where I'm puzzled: is it also possible to shift it to any different spot, say, to the negative imaginary axis? This would lead to a single-valued function on all of the real axis except at the origin, where we still have a singularity - right?
logarithms branch-cuts
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rammelmueller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
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down vote
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up vote
0
down vote
favorite
I have some troubles understanding branch cuts of the complex logarithm, or to be more precise, the shift thereof. So typically, the branch cut is along the negative real axis, connecting the branch points at $0$ and $infty$. We know, that this choice is not unique - I could, for instance, also choose the positive real axis.
Now, this is where I'm puzzled: is it also possible to shift it to any different spot, say, to the negative imaginary axis? This would lead to a single-valued function on all of the real axis except at the origin, where we still have a singularity - right?
logarithms branch-cuts
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rammelmueller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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I have some troubles understanding branch cuts of the complex logarithm, or to be more precise, the shift thereof. So typically, the branch cut is along the negative real axis, connecting the branch points at $0$ and $infty$. We know, that this choice is not unique - I could, for instance, also choose the positive real axis.
Now, this is where I'm puzzled: is it also possible to shift it to any different spot, say, to the negative imaginary axis? This would lead to a single-valued function on all of the real axis except at the origin, where we still have a singularity - right?
logarithms branch-cuts
logarithms branch-cuts
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asked 12 hours ago


rammelmueller
1033
1033
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3 Answers
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Surely you can do that. Define $log , z=log|z|+itheta$ with $-pi /2 <theta < 3pi /2$, where $z=|z|e^{itheta}$, and you will get an analytic branch of logarithm in $mathbb Csetminus {{-it: tgeq 0}}$.
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Essentially it is not possible to define the complex logarithm on a closed curve around the origin. To this end the cutting is usually done from the origin in a straight line (in any direction). However you could in theory also cut along a general curve (from the origin). See also the wikipedia article for information: https://en.wikipedia.org/wiki/Complex_logarithm#Branch_cuts
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$$log z=log |z|+iangle z$$
and you can choose where you want the phase jump of $angle z$ to occur.
You would have for positive $x$
$$log(-x)=log x-ipi.$$
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Surely you can do that. Define $log , z=log|z|+itheta$ with $-pi /2 <theta < 3pi /2$, where $z=|z|e^{itheta}$, and you will get an analytic branch of logarithm in $mathbb Csetminus {{-it: tgeq 0}}$.
add a comment |
up vote
2
down vote
accepted
Surely you can do that. Define $log , z=log|z|+itheta$ with $-pi /2 <theta < 3pi /2$, where $z=|z|e^{itheta}$, and you will get an analytic branch of logarithm in $mathbb Csetminus {{-it: tgeq 0}}$.
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Surely you can do that. Define $log , z=log|z|+itheta$ with $-pi /2 <theta < 3pi /2$, where $z=|z|e^{itheta}$, and you will get an analytic branch of logarithm in $mathbb Csetminus {{-it: tgeq 0}}$.
Surely you can do that. Define $log , z=log|z|+itheta$ with $-pi /2 <theta < 3pi /2$, where $z=|z|e^{itheta}$, and you will get an analytic branch of logarithm in $mathbb Csetminus {{-it: tgeq 0}}$.
answered 12 hours ago


Kavi Rama Murthy
39k31748
39k31748
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add a comment |
up vote
2
down vote
Essentially it is not possible to define the complex logarithm on a closed curve around the origin. To this end the cutting is usually done from the origin in a straight line (in any direction). However you could in theory also cut along a general curve (from the origin). See also the wikipedia article for information: https://en.wikipedia.org/wiki/Complex_logarithm#Branch_cuts
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maxmilgram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
2
down vote
Essentially it is not possible to define the complex logarithm on a closed curve around the origin. To this end the cutting is usually done from the origin in a straight line (in any direction). However you could in theory also cut along a general curve (from the origin). See also the wikipedia article for information: https://en.wikipedia.org/wiki/Complex_logarithm#Branch_cuts
New contributor
maxmilgram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
2
down vote
up vote
2
down vote
Essentially it is not possible to define the complex logarithm on a closed curve around the origin. To this end the cutting is usually done from the origin in a straight line (in any direction). However you could in theory also cut along a general curve (from the origin). See also the wikipedia article for information: https://en.wikipedia.org/wiki/Complex_logarithm#Branch_cuts
New contributor
maxmilgram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Essentially it is not possible to define the complex logarithm on a closed curve around the origin. To this end the cutting is usually done from the origin in a straight line (in any direction). However you could in theory also cut along a general curve (from the origin). See also the wikipedia article for information: https://en.wikipedia.org/wiki/Complex_logarithm#Branch_cuts
New contributor
maxmilgram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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answered 12 hours ago
maxmilgram
3435
3435
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add a comment |
add a comment |
up vote
0
down vote
$$log z=log |z|+iangle z$$
and you can choose where you want the phase jump of $angle z$ to occur.
You would have for positive $x$
$$log(-x)=log x-ipi.$$
add a comment |
up vote
0
down vote
$$log z=log |z|+iangle z$$
and you can choose where you want the phase jump of $angle z$ to occur.
You would have for positive $x$
$$log(-x)=log x-ipi.$$
add a comment |
up vote
0
down vote
up vote
0
down vote
$$log z=log |z|+iangle z$$
and you can choose where you want the phase jump of $angle z$ to occur.
You would have for positive $x$
$$log(-x)=log x-ipi.$$
$$log z=log |z|+iangle z$$
and you can choose where you want the phase jump of $angle z$ to occur.
You would have for positive $x$
$$log(-x)=log x-ipi.$$
answered 11 hours ago
Yves Daoust
121k668217
121k668217
add a comment |
add a comment |
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