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
New contributor
add a comment |
up vote
0
down vote
favorite
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
New contributor
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
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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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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
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
New contributor
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
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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New contributor
answered 12 hours ago
maxmilgram
3435
3435
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up vote
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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 $x$
$$log(-x)=log x-ipi.$$
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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