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?










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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?










    share|cite|improve this question







    New contributor




    rammelmueller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      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?










      share|cite|improve this question







      New contributor




      rammelmueller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      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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      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 $mathbb Csetminus {{-it: tgeq 0}}$.






                share|cite|improve this answer

























                  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 $mathbb Csetminus {{-it: tgeq 0}}$.






                  share|cite|improve this answer























                    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 $mathbb Csetminus {{-it: tgeq 0}}$.






                    share|cite|improve this answer












                    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 $mathbb Csetminus {{-it: tgeq 0}}$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 12 hours ago









                    Kavi Rama Murthy

                    39k31748




                    39k31748






















                        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






                        share|cite|improve this answer








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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






                          share|cite|improve this answer








                          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.




















                            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






                            share|cite|improve this answer








                            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







                            share|cite|improve this answer








                            New contributor




                            maxmilgram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                            share|cite|improve this answer



                            share|cite|improve this answer






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                            answered 12 hours ago









                            maxmilgram

                            3435




                            3435




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                                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 $x$



                                $$log(-x)=log x-ipi.$$






                                share|cite|improve this answer

























                                  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 $x$



                                  $$log(-x)=log x-ipi.$$






                                  share|cite|improve this answer























                                    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 $x$



                                    $$log(-x)=log x-ipi.$$






                                    share|cite|improve this answer












                                    $$log z=log |z|+iangle z$$



                                    and you can choose where you want the phase jump of $angle z$ to occur.



                                    You would have for positive $x$



                                    $$log(-x)=log x-ipi.$$







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered 11 hours ago









                                    Yves Daoust

                                    121k668217




                                    121k668217






















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