Determining branch points and closed paths











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My question is: how do we determine, computationally, when a function changes it’s value after we complete a closed path around a potential branch point?



To be specific: take the function $f(z)=log(z^{2}-1)=log(z-1)+log(z+1)$. The branch points are $z=1, z=-1$, and $z=infty$. Now, when we show that $z=1$ is a branch point, we want to show that as we travel around the point $z=1$, on a closed path, $log(z-1)$ changes by a multiple of $2pi i$ but $log(z+1)$ returns to its original value. How do we, computationally, show that a closed path around $z=1$ changes $log(z-1)$ by a multiple of $2pi i$?










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  • The function will have a discontinuity along some path. For instance, $log z$ computed as $logsqrt{x^2+y^2}+iarctan_2(y, x)$ will has a jump of $i2pi$ when crossing the negative axis.
    – Yves Daoust
    Nov 20 at 22:47

















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My question is: how do we determine, computationally, when a function changes it’s value after we complete a closed path around a potential branch point?



To be specific: take the function $f(z)=log(z^{2}-1)=log(z-1)+log(z+1)$. The branch points are $z=1, z=-1$, and $z=infty$. Now, when we show that $z=1$ is a branch point, we want to show that as we travel around the point $z=1$, on a closed path, $log(z-1)$ changes by a multiple of $2pi i$ but $log(z+1)$ returns to its original value. How do we, computationally, show that a closed path around $z=1$ changes $log(z-1)$ by a multiple of $2pi i$?










share|cite|improve this question






















  • The function will have a discontinuity along some path. For instance, $log z$ computed as $logsqrt{x^2+y^2}+iarctan_2(y, x)$ will has a jump of $i2pi$ when crossing the negative axis.
    – Yves Daoust
    Nov 20 at 22:47















up vote
0
down vote

favorite









up vote
0
down vote

favorite











My question is: how do we determine, computationally, when a function changes it’s value after we complete a closed path around a potential branch point?



To be specific: take the function $f(z)=log(z^{2}-1)=log(z-1)+log(z+1)$. The branch points are $z=1, z=-1$, and $z=infty$. Now, when we show that $z=1$ is a branch point, we want to show that as we travel around the point $z=1$, on a closed path, $log(z-1)$ changes by a multiple of $2pi i$ but $log(z+1)$ returns to its original value. How do we, computationally, show that a closed path around $z=1$ changes $log(z-1)$ by a multiple of $2pi i$?










share|cite|improve this question













My question is: how do we determine, computationally, when a function changes it’s value after we complete a closed path around a potential branch point?



To be specific: take the function $f(z)=log(z^{2}-1)=log(z-1)+log(z+1)$. The branch points are $z=1, z=-1$, and $z=infty$. Now, when we show that $z=1$ is a branch point, we want to show that as we travel around the point $z=1$, on a closed path, $log(z-1)$ changes by a multiple of $2pi i$ but $log(z+1)$ returns to its original value. How do we, computationally, show that a closed path around $z=1$ changes $log(z-1)$ by a multiple of $2pi i$?







complex-analysis complex-numbers






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asked Nov 20 at 22:27









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  • The function will have a discontinuity along some path. For instance, $log z$ computed as $logsqrt{x^2+y^2}+iarctan_2(y, x)$ will has a jump of $i2pi$ when crossing the negative axis.
    – Yves Daoust
    Nov 20 at 22:47




















  • The function will have a discontinuity along some path. For instance, $log z$ computed as $logsqrt{x^2+y^2}+iarctan_2(y, x)$ will has a jump of $i2pi$ when crossing the negative axis.
    – Yves Daoust
    Nov 20 at 22:47


















The function will have a discontinuity along some path. For instance, $log z$ computed as $logsqrt{x^2+y^2}+iarctan_2(y, x)$ will has a jump of $i2pi$ when crossing the negative axis.
– Yves Daoust
Nov 20 at 22:47






The function will have a discontinuity along some path. For instance, $log z$ computed as $logsqrt{x^2+y^2}+iarctan_2(y, x)$ will has a jump of $i2pi$ when crossing the negative axis.
– Yves Daoust
Nov 20 at 22:47

















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