Branch cut for $sqrt{z^2-a^2}$ in the lower half of the complex plane (instead of $[-a,a]$) - How does this...
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I have given a function $sqrt{z^2-a^2}$ with $a>0$.
At first i have chosen the branch cut on the real axis at $-a<z<a$: $$f(z)=sqrt{z-a}sqrt{z+a}=sqrt{|z-a||z+a|} exp({ifrac{theta_1+theta_2}{2}})$$
My question is: What happens, if i turn the branch cuts by $frac{pi}{2}$ to ${ z=pm a -iy, y>0}$? And how does this change the (value of the) function?
Change of the branch cut(s)
Naively i would assume for the square root function $sqrt{z} rightarrow e^{i pi /4}sqrt{z}=sqrt{iz}$ , which leads to $$f(z) rightarrow g(z)=isqrt{z-a} sqrt{z+a}=sqrt{{a^2-z^2}}$$
$ $
This problem is part of an exercise:
By changing the branch cuts $$R(z)=Aleft[z-sqrt{z^2-a^2}right] rightarrow tilde{R}(z)$$ $$h(z)=frac{1}{z-epsilon-tilde{R}(z)}$$ and the function $h(z)$ has a pole at $z=E_0-i Gamma /2$.
complex-analysis complex-numbers contour-integration complex-integration branch-cuts
asked Nov 15 at 17:44
FFderLuchs
11
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I have given a function $sqrt{z^2-a^2}$ with $a>0$.
At first i have chosen the branch cut on the real axis at $-a<z<a$: $$f(z)=sqrt{z-a}sqrt{z+a}=sqrt{|z-a||z+a|} exp({ifrac{theta_1+theta_2}{2}})$$
My question is: What happens, if i turn the branch cuts by $frac{pi}{2}$ to ${ z=pm a -iy, y>0}$? And how does this change the (value of the) function?
Change of the branch cut(s)
Naively i would assume for the square root function $sqrt{z} rightarrow e^{i pi /4}sqrt{z}=sqrt{iz}$ , which leads to $$f(z) rightarrow g(z)=isqrt{z-a} sqrt{z+a}=sqrt{{a^2-z^2}}$$
$ $
This problem is part of an exercise:
By changing the branch cuts $$R(z)=Aleft[z-sqrt{z^2-a^2}right] rightarrow tilde{R}(z)$$ $$h(z)=frac{1}{z-epsilon-tilde{R}(z)}$$ and the function $h(z)$ has a pole at $z=E_0-i Gamma /2$.
complex-analysis complex-numbers contour-integration complex-integration branch-cuts
asked Nov 15 at 17:44
FFderLuchs
11
IMO changing the branch cut isn't meaningful, the important concept being the analytic continuation of a given $f$ over some curve $gamma$. By definition all the branches are obtained this way.
– reuns
Nov 15 at 19:09
what happens when continuing (a given branch of) $sqrt{z-c}$ analytically over some curve ? So what happens to $prod_j sqrt{z-c_j}$ ? In your image look at the white area where the function is analytic, not at the branch cuts.
– reuns
Nov 15 at 19:26
For a fixed choice of branch cut for $sqrt{cdot}$, $sqrt{x^2 - a^2} neq sqrt{x - a} sqrt{x - a}$.
– Travis
Nov 15 at 19:37
@reuns As long as the curve does not cross the branch cuts, nothing should happen?
– FFderLuchs
Nov 15 at 20:35
Assuming that your redefinition doesn't change the value at $z = i$, the value will not change anywhere on the upper half plane, or on the portions of the lower half plane outside the strips (with $|Re z| > a$). On the strip ($|Re z| < a$ and $Im z < 0$), the value will change by one period.
– Paul Sinclair
Nov 16 at 0:48
add a comment |
up vote
0
down vote
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up vote
0
down vote
favorite
I have given a function $sqrt{z^2-a^2}$ with $a>0$.
At first i have chosen the branch cut on the real axis at $-a<z<a$: $$f(z)=sqrt{z-a}sqrt{z+a}=sqrt{|z-a||z+a|} exp({ifrac{theta_1+theta_2}{2}})$$
My question is: What happens, if i turn the branch cuts by $frac{pi}{2}$ to ${ z=pm a -iy, y>0}$? And how does this change the (value of the) function?
Change of the branch cut(s)
Naively i would assume for the square root function $sqrt{z} rightarrow e^{i pi /4}sqrt{z}=sqrt{iz}$ , which leads to $$f(z) rightarrow g(z)=isqrt{z-a} sqrt{z+a}=sqrt{{a^2-z^2}}$$
$ $
This problem is part of an exercise:
By changing the branch cuts $$R(z)=Aleft[z-sqrt{z^2-a^2}right] rightarrow tilde{R}(z)$$ $$h(z)=frac{1}{z-epsilon-tilde{R}(z)}$$ and the function $h(z)$ has a pole at $z=E_0-i Gamma /2$.
complex-analysis complex-numbers contour-integration complex-integration branch-cuts
asked Nov 15 at 17:44
FFderLuchs
11
I have given a function $sqrt{z^2-a^2}$ with $a>0$.
At first i have chosen the branch cut on the real axis at $-a<z<a$: $$f(z)=sqrt{z-a}sqrt{z+a}=sqrt{|z-a||z+a|} exp({ifrac{theta_1+theta_2}{2}})$$
My question is: What happens, if i turn the branch cuts by $frac{pi}{2}$ to ${ z=pm a -iy, y>0}$? And how does this change the (value of the) function?
Change of the branch cut(s)
Naively i would assume for the square root function $sqrt{z} rightarrow e^{i pi /4}sqrt{z}=sqrt{iz}$ , which leads to $$f(z) rightarrow g(z)=isqrt{z-a} sqrt{z+a}=sqrt{{a^2-z^2}}$$
$ $
This problem is part of an exercise:
By changing the branch cuts $$R(z)=Aleft[z-sqrt{z^2-a^2}right] rightarrow tilde{R}(z)$$ $$h(z)=frac{1}{z-epsilon-tilde{R}(z)}$$ and the function $h(z)$ has a pole at $z=E_0-i Gamma /2$.
complex-analysis complex-numbers contour-integration complex-integration branch-cuts
complex-analysis complex-numbers contour-integration complex-integration branch-cuts
asked Nov 15 at 17:44
FFderLuchs
11
asked Nov 15 at 17:44
FFderLuchs
11
edited Nov 15 at 21:46
asked Nov 15 at 17:44
FFderLuchs
11
asked Nov 15 at 17:44
FFderLuchs
11
asked Nov 15 at 17:44
FFderLuchs
11
11
IMO changing the branch cut isn't meaningful, the important concept being the analytic continuation of a given $f$ over some curve $gamma$. By definition all the branches are obtained this way.
– reuns
Nov 15 at 19:09
what happens when continuing (a given branch of) $sqrt{z-c}$ analytically over some curve ? So what happens to $prod_j sqrt{z-c_j}$ ? In your image look at the white area where the function is analytic, not at the branch cuts.
– reuns
Nov 15 at 19:26
For a fixed choice of branch cut for $sqrt{cdot}$, $sqrt{x^2 - a^2} neq sqrt{x - a} sqrt{x - a}$.
– Travis
Nov 15 at 19:37
@reuns As long as the curve does not cross the branch cuts, nothing should happen?
– FFderLuchs
Nov 15 at 20:35
Assuming that your redefinition doesn't change the value at $z = i$, the value will not change anywhere on the upper half plane, or on the portions of the lower half plane outside the strips (with $|Re z| > a$). On the strip ($|Re z| < a$ and $Im z < 0$), the value will change by one period.
– Paul Sinclair
Nov 16 at 0:48
add a comment |
IMO changing the branch cut isn't meaningful, the important concept being the analytic continuation of a given $f$ over some curve $gamma$. By definition all the branches are obtained this way.
– reuns
Nov 15 at 19:09
what happens when continuing (a given branch of) $sqrt{z-c}$ analytically over some curve ? So what happens to $prod_j sqrt{z-c_j}$ ? In your image look at the white area where the function is analytic, not at the branch cuts.
– reuns
Nov 15 at 19:26
For a fixed choice of branch cut for $sqrt{cdot}$, $sqrt{x^2 - a^2} neq sqrt{x - a} sqrt{x - a}$.
– Travis
Nov 15 at 19:37
@reuns As long as the curve does not cross the branch cuts, nothing should happen?
– FFderLuchs
Nov 15 at 20:35
Assuming that your redefinition doesn't change the value at $z = i$, the value will not change anywhere on the upper half plane, or on the portions of the lower half plane outside the strips (with $|Re z| > a$). On the strip ($|Re z| < a$ and $Im z < 0$), the value will change by one period.
– Paul Sinclair
Nov 16 at 0:48
IMO changing the branch cut isn't meaningful, the important concept being the analytic continuation of a given $f$ over some curve $gamma$. By definition all the branches are obtained this way.
– reuns
Nov 15 at 19:09
IMO changing the branch cut isn't meaningful, the important concept being the analytic continuation of a given $f$ over some curve $gamma$. By definition all the branches are obtained this way.
– reuns
Nov 15 at 19:09
what happens when continuing (a given branch of) $sqrt{z-c}$ analytically over some curve ? So what happens to $prod_j sqrt{z-c_j}$ ? In your image look at the white area where the function is analytic, not at the branch cuts.
– reuns
Nov 15 at 19:26
what happens when continuing (a given branch of) $sqrt{z-c}$ analytically over some curve ? So what happens to $prod_j sqrt{z-c_j}$ ? In your image look at the white area where the function is analytic, not at the branch cuts.
– reuns
Nov 15 at 19:26
For a fixed choice of branch cut for $sqrt{cdot}$, $sqrt{x^2 - a^2} neq sqrt{x - a} sqrt{x - a}$.
– Travis
Nov 15 at 19:37
For a fixed choice of branch cut for $sqrt{cdot}$, $sqrt{x^2 - a^2} neq sqrt{x - a} sqrt{x - a}$.
– Travis
Nov 15 at 19:37
@reuns As long as the curve does not cross the branch cuts, nothing should happen?
– FFderLuchs
Nov 15 at 20:35
@reuns As long as the curve does not cross the branch cuts, nothing should happen?
– FFderLuchs
Nov 15 at 20:35
Assuming that your redefinition doesn't change the value at $z = i$, the value will not change anywhere on the upper half plane, or on the portions of the lower half plane outside the strips (with $|Re z| > a$). On the strip ($|Re z| < a$ and $Im z < 0$), the value will change by one period.
– Paul Sinclair
Nov 16 at 0:48
Assuming that your redefinition doesn't change the value at $z = i$, the value will not change anywhere on the upper half plane, or on the portions of the lower half plane outside the strips (with $|Re z| > a$). On the strip ($|Re z| < a$ and $Im z < 0$), the value will change by one period.
– Paul Sinclair
Nov 16 at 0:48
add a comment |
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IMO changing the branch cut isn't meaningful, the important concept being the analytic continuation of a given $f$ over some curve $gamma$. By definition all the branches are obtained this way.
– reuns
Nov 15 at 19:09
what happens when continuing (a given branch of) $sqrt{z-c}$ analytically over some curve ? So what happens to $prod_j sqrt{z-c_j}$ ? In your image look at the white area where the function is analytic, not at the branch cuts.
– reuns
Nov 15 at 19:26
For a fixed choice of branch cut for $sqrt{cdot}$, $sqrt{x^2 - a^2} neq sqrt{x - a} sqrt{x - a}$.
– Travis
Nov 15 at 19:37
@reuns As long as the curve does not cross the branch cuts, nothing should happen?
– FFderLuchs
Nov 15 at 20:35
Assuming that your redefinition doesn't change the value at $z = i$, the value will not change anywhere on the upper half plane, or on the portions of the lower half plane outside the strips (with $|Re z| > a$). On the strip ($|Re z| < a$ and $Im z < 0$), the value will change by one period.
– Paul Sinclair
Nov 16 at 0:48