Question on branch cuts and branch points The 2019 Stack Overflow Developer Survey Results Are InUpdating Wagon's FindAllCrossings2D[] functionHow can I superimpose a set of points on a ContourPlot?Sqrt — how to get negative branch?How to return the points Xi,Zi generated by ContourPlotInverse functions - how are they calculated?Differential Equation in Complex Plane and Parametric PlotStreamPlot slow in version 10Contour Integration along a contour containing two branch pointsHow to plot real roots of complex polynomials avoiding branch cuts?Branch cuts of sqrtContourPlot problem with Sinc funtion

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Question on branch cuts and branch points



The 2019 Stack Overflow Developer Survey Results Are InUpdating Wagon's FindAllCrossings2D[] functionHow can I superimpose a set of points on a ContourPlot?Sqrt — how to get negative branch?How to return the points Xi,Zi generated by ContourPlotInverse functions - how are they calculated?Differential Equation in Complex Plane and Parametric PlotStreamPlot slow in version 10Contour Integration along a contour containing two branch pointsHow to plot real roots of complex polynomials avoiding branch cuts?Branch cuts of sqrtContourPlot problem with Sinc funtion










9












$begingroup$


Is it possible to determine branch cuts and branch points for complicated functions using mathematica



Iam trying to determine the brnach cuts and branch points of this complicated function



We have 8 branch cuts
enter image description here



And I calculated the branchPoints in exact and numeric form



And I have tried to visualize the branchPoints and branchcuts but I had a problem
enter image description here$$sqrt(tanh(z) -tanh(2z))^2 +(tanh(z)*tanh(2z)+1)^2-1-2tanh(z)^2 tanh(2z)^2$$



And I calculate
I have tried in mathematica but it's not obvious for me where are the branch cuts ?



ContourPlot[Im[Sqrt[(Tanh[x + I*y] - Tanh[2 x + I*2 y])^2 + (Tanh[x + I*y] 
 Tanh[2 x + I*2 y] + 1)^2-1 - 2 ((Tanh[x + I*2 y])^2)((Tanh[x + I*y])^2) ]],
 x, -10, 10, y, -10, 10, AxesLabel -> Automatic,ContourShading -> Automatic, 
 ColorFunction -> "Rainbow", Contours -> 20]


enter image description here



ContourPlot[Re[Sqrt[(Tanh[x + I*y] - Tanh[2 x + I*2 y])^2 + (Tanh[x + I*y]Tanh[2 x + I*2 y] + 1)^2 - 1 - 2 ((Tanh[x + I*2 y])^2) ((Tanh[x + I*y])^2) ]],
 x, -10, 10, y, -10, 10, AxesLabel -> Automatic,
ContourShading -> Automatic, ColorFunction -> "Rainbow", Contours -> 20]


enter image description here



How Can I visualize the banchPoints and the branchCuts ?






plotting functions complex visualization







share|improve this question









New contributor




topspin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$





This question has an open bounty worth +50
reputation from topspin ending ending at 2019-04-15 21:38:48Z">in 6 days.


The current answers do not contain enough detail.















  • $begingroup$
    The first step might be to find all the zeros of the function under the square root. Perhaps this might help.
    $endgroup$
    – Hugh
    Apr 5 at 13:02






  • 1




    $begingroup$
    Please can you put the equation in a form that can be copied to a mathematica notebook? (Edit your post please.) This is helpful for those of us who might try out approaches.
    $endgroup$
    – Hugh
    Apr 5 at 13:04










  • $begingroup$
    Ok, Thank you . I have just edited my post .
    $endgroup$
    – topspin
    Apr 5 at 14:37






  • 1




    $begingroup$
    I think I should find the zeros of the Imaginary part of the function under the square root , and finding when is the real part is non negative if I am talking about the principal branch excluding the negative real axis . I have tried to find all the zeros of the function under the square root using mathematica but the output was not clear to me
    $endgroup$
    – topspin
    Apr 5 at 14:42











  • $begingroup$
    You have one instance of x + I*2 y in your formula. Do you mean for that to be 2x + I*2 y?
    $endgroup$
    – Chip Hurst
    yesterday















9












$begingroup$


Is it possible to determine branch cuts and branch points for complicated functions using mathematica



Iam trying to determine the brnach cuts and branch points of this complicated function



We have 8 branch cuts
enter image description here



And I calculated the branchPoints in exact and numeric form



And I have tried to visualize the branchPoints and branchcuts but I had a problem
enter image description here$$sqrt(tanh(z) -tanh(2z))^2 +(tanh(z)*tanh(2z)+1)^2-1-2tanh(z)^2 tanh(2z)^2$$



And I calculate
I have tried in mathematica but it's not obvious for me where are the branch cuts ?



ContourPlot[Im[Sqrt[(Tanh[x + I*y] - Tanh[2 x + I*2 y])^2 + (Tanh[x + I*y] 
 Tanh[2 x + I*2 y] + 1)^2-1 - 2 ((Tanh[x + I*2 y])^2)((Tanh[x + I*y])^2) ]],
 x, -10, 10, y, -10, 10, AxesLabel -> Automatic,ContourShading -> Automatic, 
 ColorFunction -> "Rainbow", Contours -> 20]


enter image description here



ContourPlot[Re[Sqrt[(Tanh[x + I*y] - Tanh[2 x + I*2 y])^2 + (Tanh[x + I*y]Tanh[2 x + I*2 y] + 1)^2 - 1 - 2 ((Tanh[x + I*2 y])^2) ((Tanh[x + I*y])^2) ]],
 x, -10, 10, y, -10, 10, AxesLabel -> Automatic,
ContourShading -> Automatic, ColorFunction -> "Rainbow", Contours -> 20]


enter image description here



How Can I visualize the banchPoints and the branchCuts ?






plotting functions complex visualization







share|improve this question









New contributor




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







$endgroup$





This question has an open bounty worth +50
reputation from topspin ending ending at 2019-04-15 21:38:48Z">in 6 days.


The current answers do not contain enough detail.















  • $begingroup$
    The first step might be to find all the zeros of the function under the square root. Perhaps this might help.
    $endgroup$
    – Hugh
    Apr 5 at 13:02






  • 1




    $begingroup$
    Please can you put the equation in a form that can be copied to a mathematica notebook? (Edit your post please.) This is helpful for those of us who might try out approaches.
    $endgroup$
    – Hugh
    Apr 5 at 13:04










  • $begingroup$
    Ok, Thank you . I have just edited my post .
    $endgroup$
    – topspin
    Apr 5 at 14:37






  • 1




    $begingroup$
    I think I should find the zeros of the Imaginary part of the function under the square root , and finding when is the real part is non negative if I am talking about the principal branch excluding the negative real axis . I have tried to find all the zeros of the function under the square root using mathematica but the output was not clear to me
    $endgroup$
    – topspin
    Apr 5 at 14:42











  • $begingroup$
    You have one instance of x + I*2 y in your formula. Do you mean for that to be 2x + I*2 y?
    $endgroup$
    – Chip Hurst
    yesterday













9












9








9


3



$begingroup$


Is it possible to determine branch cuts and branch points for complicated functions using mathematica



Iam trying to determine the brnach cuts and branch points of this complicated function



We have 8 branch cuts
enter image description here



And I calculated the branchPoints in exact and numeric form



And I have tried to visualize the branchPoints and branchcuts but I had a problem
enter image description here$$sqrt(tanh(z) -tanh(2z))^2 +(tanh(z)*tanh(2z)+1)^2-1-2tanh(z)^2 tanh(2z)^2$$



And I calculate
I have tried in mathematica but it's not obvious for me where are the branch cuts ?



ContourPlot[Im[Sqrt[(Tanh[x + I*y] - Tanh[2 x + I*2 y])^2 + (Tanh[x + I*y] 
 Tanh[2 x + I*2 y] + 1)^2-1 - 2 ((Tanh[x + I*2 y])^2)((Tanh[x + I*y])^2) ]],
 x, -10, 10, y, -10, 10, AxesLabel -> Automatic,ContourShading -> Automatic, 
 ColorFunction -> "Rainbow", Contours -> 20]


enter image description here



ContourPlot[Re[Sqrt[(Tanh[x + I*y] - Tanh[2 x + I*2 y])^2 + (Tanh[x + I*y]Tanh[2 x + I*2 y] + 1)^2 - 1 - 2 ((Tanh[x + I*2 y])^2) ((Tanh[x + I*y])^2) ]],
 x, -10, 10, y, -10, 10, AxesLabel -> Automatic,
ContourShading -> Automatic, ColorFunction -> "Rainbow", Contours -> 20]


enter image description here



How Can I visualize the banchPoints and the branchCuts ?






plotting functions complex visualization







share|improve this question









New contributor




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







$endgroup$




Is it possible to determine branch cuts and branch points for complicated functions using mathematica



Iam trying to determine the brnach cuts and branch points of this complicated function



We have 8 branch cuts
enter image description here



And I calculated the branchPoints in exact and numeric form



And I have tried to visualize the branchPoints and branchcuts but I had a problem
enter image description here$$sqrt(tanh(z) -tanh(2z))^2 +(tanh(z)*tanh(2z)+1)^2-1-2tanh(z)^2 tanh(2z)^2$$



And I calculate
I have tried in mathematica but it's not obvious for me where are the branch cuts ?



ContourPlot[Im[Sqrt[(Tanh[x + I*y] - Tanh[2 x + I*2 y])^2 + (Tanh[x + I*y] 
 Tanh[2 x + I*2 y] + 1)^2-1 - 2 ((Tanh[x + I*2 y])^2)((Tanh[x + I*y])^2) ]],
 x, -10, 10, y, -10, 10, AxesLabel -> Automatic,ContourShading -> Automatic, 
 ColorFunction -> "Rainbow", Contours -> 20]


enter image description here



ContourPlot[Re[Sqrt[(Tanh[x + I*y] - Tanh[2 x + I*2 y])^2 + (Tanh[x + I*y]Tanh[2 x + I*2 y] + 1)^2 - 1 - 2 ((Tanh[x + I*2 y])^2) ((Tanh[x + I*y])^2) ]],
 x, -10, 10, y, -10, 10, AxesLabel -> Automatic,
ContourShading -> Automatic, ColorFunction -> "Rainbow", Contours -> 20]


enter image description here



How Can I visualize the banchPoints and the branchCuts ?






plotting functions complex visualization




plotting functions complex visualization






share|improve this question









New contributor




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











share|improve this question









New contributor




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









share|improve this question




share|improve this question








edited yesterday







topspin













New contributor




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









asked Apr 5 at 11:56









topspintopspin

964




964




New contributor




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





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






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






This question has an open bounty worth +50
reputation from topspin ending ending at 2019-04-15 21:38:48Z">in 6 days.


The current answers do not contain enough detail.








This question has an open bounty worth +50
reputation from topspin ending ending at 2019-04-15 21:38:48Z">in 6 days.


The current answers do not contain enough detail.













  • $begingroup$
    The first step might be to find all the zeros of the function under the square root. Perhaps this might help.
    $endgroup$
    – Hugh
    Apr 5 at 13:02






  • 1




    $begingroup$
    Please can you put the equation in a form that can be copied to a mathematica notebook? (Edit your post please.) This is helpful for those of us who might try out approaches.
    $endgroup$
    – Hugh
    Apr 5 at 13:04










  • $begingroup$
    Ok, Thank you . I have just edited my post .
    $endgroup$
    – topspin
    Apr 5 at 14:37






  • 1




    $begingroup$
    I think I should find the zeros of the Imaginary part of the function under the square root , and finding when is the real part is non negative if I am talking about the principal branch excluding the negative real axis . I have tried to find all the zeros of the function under the square root using mathematica but the output was not clear to me
    $endgroup$
    – topspin
    Apr 5 at 14:42











  • $begingroup$
    You have one instance of x + I*2 y in your formula. Do you mean for that to be 2x + I*2 y?
    $endgroup$
    – Chip Hurst
    yesterday
















  • $begingroup$
    The first step might be to find all the zeros of the function under the square root. Perhaps this might help.
    $endgroup$
    – Hugh
    Apr 5 at 13:02






  • 1




    $begingroup$
    Please can you put the equation in a form that can be copied to a mathematica notebook? (Edit your post please.) This is helpful for those of us who might try out approaches.
    $endgroup$
    – Hugh
    Apr 5 at 13:04










  • $begingroup$
    Ok, Thank you . I have just edited my post .
    $endgroup$
    – topspin
    Apr 5 at 14:37






  • 1




    $begingroup$
    I think I should find the zeros of the Imaginary part of the function under the square root , and finding when is the real part is non negative if I am talking about the principal branch excluding the negative real axis . I have tried to find all the zeros of the function under the square root using mathematica but the output was not clear to me
    $endgroup$
    – topspin
    Apr 5 at 14:42











  • $begingroup$
    You have one instance of x + I*2 y in your formula. Do you mean for that to be 2x + I*2 y?
    $endgroup$
    – Chip Hurst
    yesterday















$begingroup$
The first step might be to find all the zeros of the function under the square root. Perhaps this might help.
$endgroup$
– Hugh
Apr 5 at 13:02




$begingroup$
The first step might be to find all the zeros of the function under the square root. Perhaps this might help.
$endgroup$
– Hugh
Apr 5 at 13:02




1




1




$begingroup$
Please can you put the equation in a form that can be copied to a mathematica notebook? (Edit your post please.) This is helpful for those of us who might try out approaches.
$endgroup$
– Hugh
Apr 5 at 13:04




$begingroup$
Please can you put the equation in a form that can be copied to a mathematica notebook? (Edit your post please.) This is helpful for those of us who might try out approaches.
$endgroup$
– Hugh
Apr 5 at 13:04












$begingroup$
Ok, Thank you . I have just edited my post .
$endgroup$
– topspin
Apr 5 at 14:37




$begingroup$
Ok, Thank you . I have just edited my post .
$endgroup$
– topspin
Apr 5 at 14:37




1




1




$begingroup$
I think I should find the zeros of the Imaginary part of the function under the square root , and finding when is the real part is non negative if I am talking about the principal branch excluding the negative real axis . I have tried to find all the zeros of the function under the square root using mathematica but the output was not clear to me
$endgroup$
– topspin
Apr 5 at 14:42





$begingroup$
I think I should find the zeros of the Imaginary part of the function under the square root , and finding when is the real part is non negative if I am talking about the principal branch excluding the negative real axis . I have tried to find all the zeros of the function under the square root using mathematica but the output was not clear to me
$endgroup$
– topspin
Apr 5 at 14:42













$begingroup$
You have one instance of x + I*2 y in your formula. Do you mean for that to be 2x + I*2 y?
$endgroup$
– Chip Hurst
yesterday




$begingroup$
You have one instance of x + I*2 y in your formula. Do you mean for that to be 2x + I*2 y?
$endgroup$
– Chip Hurst
yesterday










3 Answers
3






active

oldest

votes


















13












$begingroup$

Perhaps you can make use of the internal functions ComplexAnalysis`BranchCuts and ComplexAnalysis`BranchPoints. First, use a complex variable z instead of x + I y:



expr = Sqrt[(Tanh[z]-Tanh[2z])^2+(Tanh[z] Tanh[2z]+1)^2-1-2 Tanh[z]^2Tanh[2z]^2];


Then, for example, the branch points are:



pts = ComplexAnalysis`BranchPoints[expr, z]



ConditionalExpression[-(I/(2 π C[1])), C[1] ∈ Integers],
ConditionalExpression[2 I π C[1], C[1] ∈ Integers],
ConditionalExpression[1/(-((I π)/4) + 2 I π C[1]),
C[1] ∈ Integers],
ConditionalExpression[-((I π)/4) + 2 I π C[1],
C[1] ∈ Integers],
ConditionalExpression[1/((I π)/4 + 2 I π C[1]),
C[1] ∈ Integers],
ConditionalExpression[(I π)/4 + 2 I π C[1],
C[1] ∈ Integers],
ConditionalExpression[1/(-((I π)/2) + 2 I π C[1]),
C[1] ∈ Integers],
ConditionalExpression[-((I π)/2) + 2 I π C[1],
C[1] ∈ Integers],
ConditionalExpression[1/((I π)/2 + 2 I π C[1]),
C[1] ∈ Integers],
ConditionalExpression[(I π)/2 + 2 I π C[1],
C[1] ∈ Integers],
ConditionalExpression[1/(-((3 I π)/4) + 2 I π C[1]),
C[1] ∈ Integers],
ConditionalExpression[-((3 I π)/4) + 2 I π C[1],
C[1] ∈ Integers],
ConditionalExpression[1/((3 I π)/4 + 2 I π C[1]),
C[1] ∈ Integers],
ConditionalExpression[(3 I π)/4 + 2 I π C[1],
C[1] ∈ Integers],
ConditionalExpression[1/(I π + 2 I π C[1]),
C[1] ∈ Integers],
ConditionalExpression[I π + 2 I π C[1], C[1] ∈ Integers],
ConditionalExpression[1/(
2 I π C[1] + Log[(-(1/2) + I/2) - Sqrt[-1 - I/2]]),
C[1] ∈ Integers],
ConditionalExpression[2 I π C[1] + Log[(-(1/2) + I/2) - Sqrt[-1 - I/2]],
C[1] ∈ Integers],
ConditionalExpression[1/(2 I π C[1] + Log[(1/2 - I/2) - Sqrt[-1 - I/2]]),
C[1] ∈ Integers],
ConditionalExpression[2 I π C[1] + Log[(1/2 - I/2) - Sqrt[-1 - I/2]],
C[1] ∈ Integers],
ConditionalExpression[1/(
2 I π C[1] + Log[(-(1/2) + I/2) + Sqrt[-1 - I/2]]),
C[1] ∈ Integers],
ConditionalExpression[2 I π C[1] + Log[(-(1/2) + I/2) + Sqrt[-1 - I/2]],
C[1] ∈ Integers],
ConditionalExpression[1/(2 I π C[1] + Log[(1/2 - I/2) + Sqrt[-1 - I/2]]),
C[1] ∈ Integers],
ConditionalExpression[2 I π C[1] + Log[(1/2 - I/2) + Sqrt[-1 - I/2]],
C[1] ∈ Integers],
ConditionalExpression[1/(
2 I π C[1] + Log[(-(1/2) - I/2) - Sqrt[-1 + I/2]]),
C[1] ∈ Integers],
ConditionalExpression[2 I π C[1] + Log[(-(1/2) - I/2) - Sqrt[-1 + I/2]],
C[1] ∈ Integers],
ConditionalExpression[1/(2 I π C[1] + Log[(1/2 + I/2) - Sqrt[-1 + I/2]]),
C[1] ∈ Integers],
ConditionalExpression[2 I π C[1] + Log[(1/2 + I/2) - Sqrt[-1 + I/2]],
C[1] ∈ Integers],
ConditionalExpression[1/(
2 I π C[1] + Log[(-(1/2) - I/2) + Sqrt[-1 + I/2]]),
C[1] ∈ Integers],
ConditionalExpression[2 I π C[1] + Log[(-(1/2) - I/2) + Sqrt[-1 + I/2]],
C[1] ∈ Integers],
ConditionalExpression[1/(2 I π C[1] + Log[(1/2 + I/2) + Sqrt[-1 + I/2]]),
C[1] ∈ Integers],
ConditionalExpression[2 I π C[1] + Log[(1/2 + I/2) + Sqrt[-1 + I/2]],
C[1] ∈ Integers]




The above can be simplified a bit with:



Simplify[pts, C[1] ∈ Integers]



-(I/(2 π C[1])), 2 I π C[1], (4 I)/(π - 8 π C[1]),
1/4 I π (-1 + 8 C[1]), -((4 I)/(π + 8 π C[1])),
1/4 I (π + 8 π C[1]), (2 I)/(π - 4 π C[1]),
1/2 I π (-1 + 4 C[1]), -((2 I)/(π + 4 π C[1])),
1/2 I (π + 4 π C[1]), (4 I)/(3 π - 8 π C[1]),
1/4 I π (-3 + 8 C[1]), -((4 I)/(3 π + 8 π C[1])),
1/4 I π (3 + 8 C[1]), -(I/(π + 2 π C[1])),
I (π + 2 π C[1]), 1/(
2 I π C[1] + Log[(-(1/2) + I/2) - Sqrt[-1 - I/2]]),
2 I π C[1] + Log[(-(1/2) + I/2) - Sqrt[-1 - I/2]], 1/(
2 I π C[1] + Log[(1/2 - I/2) - Sqrt[-1 - I/2]]),
2 I π C[1] + Log[(1/2 - I/2) - Sqrt[-1 - I/2]], 1/(
2 I π C[1] + Log[(-(1/2) + I/2) + Sqrt[-1 - I/2]]),
2 I π C[1] + Log[(-(1/2) + I/2) + Sqrt[-1 - I/2]], 1/(
2 I π C[1] + Log[(1/2 - I/2) + Sqrt[-1 - I/2]]),
2 I π C[1] + Log[(1/2 - I/2) + Sqrt[-1 - I/2]], 1/(
2 I π C[1] + Log[(-(1/2) - I/2) - Sqrt[-1 + I/2]]),
2 I π C[1] + Log[(-(1/2) - I/2) - Sqrt[-1 + I/2]], 1/(
2 I π C[1] + Log[(1/2 + I/2) - Sqrt[-1 + I/2]]),
2 I π C[1] + Log[(1/2 + I/2) - Sqrt[-1 + I/2]], 1/(
2 I π C[1] + Log[(-(1/2) - I/2) + Sqrt[-1 + I/2]]),
2 I π C[1] + Log[(-(1/2) - I/2) + Sqrt[-1 + I/2]], 1/(
2 I π C[1] + Log[(1/2 + I/2) + Sqrt[-1 + I/2]]),
2 I π C[1] + Log[(1/2 + I/2) + Sqrt[-1 + I/2]]




Similarly, the branch cuts can be found with:



ComplexAnalysis`BranchCuts[expr, z]



C[1] ∈
Integers && ((1/2 Log[Root[1 - 2 #1 - 2 #1^2 - 2 #1^3 + #1^4 &, 1]] <
Re[z] < 0 && (Im[
z] == -ArcTan[Sqrt[(3 + 4 E^(2 Re[z]) + 3 E^(4 Re[z]))/(
1 + E^(4 Re[z]))]] + π C[1] ||
Im[z] == ArcTan[Sqrt[(3 + 4 E^(2 Re[z]) + 3 E^(4 Re[z]))/(
1 + E^(4 Re[z]))]] + π C[1])) || (Re[z] ==
0 && (1/2 (-π + 2 π C[1]) < Im[z] <
1/4 (-π + 4 π C[1]) ||
1/4 (-π + 4 π C[1]) < Im[z] < π C[1] || π C[1] <
Im[z] < 1/4 (π + 4 π C[1]) ||
1/4 (π + 4 π C[1]) < Im[z] <
1/2 (π + 2 π C[1]))) || (0 < Re[z] <
1/2 Log[Root[1 - 2 #1 - 2 #1^2 - 2 #1^3 + #1^4 &, 2]] && (Im[
z] == -ArcTan[Sqrt[(3 + 4 E^(2 Re[z]) + 3 E^(4 Re[z]))/(
1 + E^(4 Re[z]))]] + π C[1] ||
Im[z] == ArcTan[Sqrt[(3 + 4 E^(2 Re[z]) + 3 E^(4 Re[z]))/(
1 + E^(4 Re[z]))]] + π C[1])))







share|improve this answer









$endgroup$












  • $begingroup$
    Thank you very much .
    $endgroup$
    – topspin
    Apr 5 at 17:26










  • $begingroup$
    Is there any way to visualize those branch points and the branch cuts in Mathematica Instead of ContourPlot ?
    $endgroup$
    – topspin
    Apr 5 at 17:28











  • $begingroup$
    Or visualizing the branch points and the branch cuts using ContourPlot .
    $endgroup$
    – topspin
    Apr 5 at 17:51










  • $begingroup$
    Just to compare: the Maple's result dropbox.com/s/zh7mq932rlvb1uk/branch_cuts.pdf?dl=0 seems to be different.
    $endgroup$
    – user64494
    Apr 6 at 6:08


















6












$begingroup$

In this case, the only branch cuts and branch points will come from the square root. The cuts of $sqrtf(z)$ occurs along the half line $textIm(f(z)) = 0 ,wedge, textRe(f(z)) leq 0$. The branch points lie at $f(z) = 0$ or $f(z) = tildeinfty$.



Your example:



With[z = x + I y,
 expr = (Tanh[z] - Tanh[2 z])^2 + (Tanh[z] Tanh[2 z] + 1)^2 - 1 - 2 ((Tanh[2 z])^2) ((Tanh[z])^2);
 branchCutRegion[x_, y_, __] = Re[expr] <= 0;
];

bpvals = Union[x, y /. Solve[(expr == 0 || 1/Together[TrigToExp[expr]] == 0) && -10 < x < 10 && -10 < y < 10, x, y]];


Here we needed to help Solve find the branch points corresponding to $tildeinfty$.



We can visualize the cut by plotting the constraint on the imaginary part, restricted to the region defined by the constraint on the real part. Here I've overlaid the branch points:



ContourPlot[Im[expr] == 0, x, -10, 10, y, -10, 10, 
 RegionFunction -> branchCutRegion, PlotPoints -> 100, 
 Epilog -> Red, Point[bpvals]
]




For fun we can add a plot of the expression under the cuts. Here I'll use domain coloring. Here, the complex argument varies with hue and the absolute value varies with saturation and brightness -- the darker the pixel, the larger the absolute value. I've also binned the absolute value to show some contours.



binnedabs = fastCompile[z, _Complex,
 Module[f, abs, rnd, sgn, val,
 f = (Tanh[z] - Tanh[2 z])^2 + (Tanh[z] Tanh[2 z] + 1)^2 - 1 - 2 Tanh[2 z]^2 Tanh[z]^2;
 abs = Abs[f];
 rnd = Round[abs, .2];
 val = If[rnd == 0, f, rnd Sign[f]];
 
 Divide[Mod[Arg[val], 2π], 2π], 
 Power[1 + 0.3*Log[Abs[val] + 1], -1], 
 Power[1 + 0.5*Log[Abs[val] + 1], -1]
 
 ],
 CompilationTarget -> "C",
 Parallelization -> True,
 RuntimeAttributes -> Listable,
 RuntimeOptions -> "Speed"
];

lattice = Array[List, 2048, 2048, -10., 10., -10., 10.].I, 1;

raster = Raster[binnedabs[lattice], -10, -10, 10, 10, ColorFunction -> Hue];

cutplot = ContourPlot[Im[expr] == 0, x, -10, 10, y, -10, 10, 
 RegionFunction -> branchCutRegion, PlotPoints -> 100, ContourStyle -> Black];

Show[
 cutplot, 
 ImageSize -> 800,
 Prolog -> raster, 
 Epilog -> EdgeForm[Black], GrayLevel[.8], Disk[#, Scaled[.0045]] & /@ bpvals
]


enter image description here






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$endgroup$












  • $begingroup$
    In accordance with dropbox.com/s/zh7mq932rlvb1uk/branch_cuts.pdf?dl=0
    $endgroup$
    – user64494
    14 hours ago


















4












$begingroup$

First start with the branch points: these are the values of z where the root is not single-valued.
First:



myexp = Together[
 TrigToExp[
 FullSimplify[(Tanh[z] - Tanh[2 z])^2 + (Tanh[z] Tanh[2 z] + 1)^2 - 
 1 - 2 Tanh[z]^2 Tanh[2 z]^2]
 ]]


$$fracleft(e^2 z-1right)^2 left(4 e^2 z+10 e^4 z+4 e^6 z+e^8 z+1right)left(e^2 z+1right)^2 left(e^4 z+1right)^2$$



Now solve for the zeros of the denominator and numerator. I'll do the numerator: First obtain a polynomial in e^z and then solve the polynomial in terms of a polynomial in just z:



 Expand[Numerator[
 Together[TrigToExp[
 FullSimplify[(Tanh[z] - Tanh[2 z])^2 + (Tanh[z] Tanh[2 z] + 
 1)^2 - 1 - 2 Tanh[z]^2 Tanh[2 z]^2]
 ]]]]
 mySol = z /. 
 Solve[1 + 2 z^2 + 3 z^4 - 12 z^6 + 3 z^8 + 2 z^10 + z^12 == 0, z];


Now make the substitution Log[z] and keep in mind Log[z]=Log[Abs[z]]+i (Arg(z)+2k pi) so that we have a set of branch points for all integer k. I will do k=0,1,-1 and then plot the results:enter image description here



p1 = Show[
 Graphics[Red, 
 Point @@ Re[#], Im[#] & /@ (N[Log[#]] & /@ mySol)], 
 Axes -> True, PlotRange -> 5];
p2 = Show[
 Graphics[Blue, 
 Point @@ Re[#], Im[#] & /@ (N[(Log[#] + 2 [Pi] I)] & /@ 
 mySol)], Axes -> True, PlotRange -> 15];
p3 = Show[
 Graphics[Green // Darker, 
 Point @@ Re[#], Im[#] & /@ (N[(Log[#] - 2 [Pi] I)] & /@ 
 mySol)], Axes -> True, PlotRange -> 15];
Show[p1, p2, p3, PlotRange -> 15]





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    13












    $begingroup$

    Perhaps you can make use of the internal functions ComplexAnalysis`BranchCuts and ComplexAnalysis`BranchPoints. First, use a complex variable z instead of x + I y:



    expr = Sqrt[(Tanh[z]-Tanh[2z])^2+(Tanh[z] Tanh[2z]+1)^2-1-2 Tanh[z]^2Tanh[2z]^2];


    Then, for example, the branch points are:



    pts = ComplexAnalysis`BranchPoints[expr, z]



    ConditionalExpression[-(I/(2 π C[1])), C[1] ∈ Integers],
    ConditionalExpression[2 I π C[1], C[1] ∈ Integers],
    ConditionalExpression[1/(-((I π)/4) + 2 I π C[1]),
    C[1] ∈ Integers],
    ConditionalExpression[-((I π)/4) + 2 I π C[1],
    C[1] ∈ Integers],
    ConditionalExpression[1/((I π)/4 + 2 I π C[1]),
    C[1] ∈ Integers],
    ConditionalExpression[(I π)/4 + 2 I π C[1],
    C[1] ∈ Integers],
    ConditionalExpression[1/(-((I π)/2) + 2 I π C[1]),
    C[1] ∈ Integers],
    ConditionalExpression[-((I π)/2) + 2 I π C[1],
    C[1] ∈ Integers],
    ConditionalExpression[1/((I π)/2 + 2 I π C[1]),
    C[1] ∈ Integers],
    ConditionalExpression[(I π)/2 + 2 I π C[1],
    C[1] ∈ Integers],
    ConditionalExpression[1/(-((3 I π)/4) + 2 I π C[1]),
    C[1] ∈ Integers],
    ConditionalExpression[-((3 I π)/4) + 2 I π C[1],
    C[1] ∈ Integers],
    ConditionalExpression[1/((3 I π)/4 + 2 I π C[1]),
    C[1] ∈ Integers],
    ConditionalExpression[(3 I π)/4 + 2 I π C[1],
    C[1] ∈ Integers],
    ConditionalExpression[1/(I π + 2 I π C[1]),
    C[1] ∈ Integers],
    ConditionalExpression[I π + 2 I π C[1], C[1] ∈ Integers],
    ConditionalExpression[1/(
    2 I π C[1] + Log[(-(1/2) + I/2) - Sqrt[-1 - I/2]]),
    C[1] ∈ Integers],
    ConditionalExpression[2 I π C[1] + Log[(-(1/2) + I/2) - Sqrt[-1 - I/2]],
    C[1] ∈ Integers],
    ConditionalExpression[1/(2 I π C[1] + Log[(1/2 - I/2) - Sqrt[-1 - I/2]]),
    C[1] ∈ Integers],
    ConditionalExpression[2 I π C[1] + Log[(1/2 - I/2) - Sqrt[-1 - I/2]],
    C[1] ∈ Integers],
    ConditionalExpression[1/(
    2 I π C[1] + Log[(-(1/2) + I/2) + Sqrt[-1 - I/2]]),
    C[1] ∈ Integers],
    ConditionalExpression[2 I π C[1] + Log[(-(1/2) + I/2) + Sqrt[-1 - I/2]],
    C[1] ∈ Integers],
    ConditionalExpression[1/(2 I π C[1] + Log[(1/2 - I/2) + Sqrt[-1 - I/2]]),
    C[1] ∈ Integers],
    ConditionalExpression[2 I π C[1] + Log[(1/2 - I/2) + Sqrt[-1 - I/2]],
    C[1] ∈ Integers],
    ConditionalExpression[1/(
    2 I π C[1] + Log[(-(1/2) - I/2) - Sqrt[-1 + I/2]]),
    C[1] ∈ Integers],
    ConditionalExpression[2 I π C[1] + Log[(-(1/2) - I/2) - Sqrt[-1 + I/2]],
    C[1] ∈ Integers],
    ConditionalExpression[1/(2 I π C[1] + Log[(1/2 + I/2) - Sqrt[-1 + I/2]]),
    C[1] ∈ Integers],
    ConditionalExpression[2 I π C[1] + Log[(1/2 + I/2) - Sqrt[-1 + I/2]],
    C[1] ∈ Integers],
    ConditionalExpression[1/(
    2 I π C[1] + Log[(-(1/2) - I/2) + Sqrt[-1 + I/2]]),
    C[1] ∈ Integers],
    ConditionalExpression[2 I π C[1] + Log[(-(1/2) - I/2) + Sqrt[-1 + I/2]],
    C[1] ∈ Integers],
    ConditionalExpression[1/(2 I π C[1] + Log[(1/2 + I/2) + Sqrt[-1 + I/2]]),
    C[1] ∈ Integers],
    ConditionalExpression[2 I π C[1] + Log[(1/2 + I/2) + Sqrt[-1 + I/2]],
    C[1] ∈ Integers]




    The above can be simplified a bit with:



    Simplify[pts, C[1] ∈ Integers]



    -(I/(2 π C[1])), 2 I π C[1], (4 I)/(π - 8 π C[1]),
    1/4 I π (-1 + 8 C[1]), -((4 I)/(π + 8 π C[1])),
    1/4 I (π + 8 π C[1]), (2 I)/(π - 4 π C[1]),
    1/2 I π (-1 + 4 C[1]), -((2 I)/(π + 4 π C[1])),
    1/2 I (π + 4 π C[1]), (4 I)/(3 π - 8 π C[1]),
    1/4 I π (-3 + 8 C[1]), -((4 I)/(3 π + 8 π C[1])),
    1/4 I π (3 + 8 C[1]), -(I/(π + 2 π C[1])),
    I (π + 2 π C[1]), 1/(
    2 I π C[1] + Log[(-(1/2) + I/2) - Sqrt[-1 - I/2]]),
    2 I π C[1] + Log[(-(1/2) + I/2) - Sqrt[-1 - I/2]], 1/(
    2 I π C[1] + Log[(1/2 - I/2) - Sqrt[-1 - I/2]]),
    2 I π C[1] + Log[(1/2 - I/2) - Sqrt[-1 - I/2]], 1/(
    2 I π C[1] + Log[(-(1/2) + I/2) + Sqrt[-1 - I/2]]),
    2 I π C[1] + Log[(-(1/2) + I/2) + Sqrt[-1 - I/2]], 1/(
    2 I π C[1] + Log[(1/2 - I/2) + Sqrt[-1 - I/2]]),
    2 I π C[1] + Log[(1/2 - I/2) + Sqrt[-1 - I/2]], 1/(
    2 I π C[1] + Log[(-(1/2) - I/2) - Sqrt[-1 + I/2]]),
    2 I π C[1] + Log[(-(1/2) - I/2) - Sqrt[-1 + I/2]], 1/(
    2 I π C[1] + Log[(1/2 + I/2) - Sqrt[-1 + I/2]]),
    2 I π C[1] + Log[(1/2 + I/2) - Sqrt[-1 + I/2]], 1/(
    2 I π C[1] + Log[(-(1/2) - I/2) + Sqrt[-1 + I/2]]),
    2 I π C[1] + Log[(-(1/2) - I/2) + Sqrt[-1 + I/2]], 1/(
    2 I π C[1] + Log[(1/2 + I/2) + Sqrt[-1 + I/2]]),
    2 I π C[1] + Log[(1/2 + I/2) + Sqrt[-1 + I/2]]




    Similarly, the branch cuts can be found with:



    ComplexAnalysis`BranchCuts[expr, z]



    C[1] ∈
    Integers && ((1/2 Log[Root[1 - 2 #1 - 2 #1^2 - 2 #1^3 + #1^4 &, 1]] <
    Re[z] < 0 && (Im[
    z] == -ArcTan[Sqrt[(3 + 4 E^(2 Re[z]) + 3 E^(4 Re[z]))/(
    1 + E^(4 Re[z]))]] + π C[1] ||
    Im[z] == ArcTan[Sqrt[(3 + 4 E^(2 Re[z]) + 3 E^(4 Re[z]))/(
    1 + E^(4 Re[z]))]] + π C[1])) || (Re[z] ==
    0 && (1/2 (-π + 2 π C[1]) < Im[z] <
    1/4 (-π + 4 π C[1]) ||
    1/4 (-π + 4 π C[1]) < Im[z] < π C[1] || π C[1] <
    Im[z] < 1/4 (π + 4 π C[1]) ||
    1/4 (π + 4 π C[1]) < Im[z] <
    1/2 (π + 2 π C[1]))) || (0 < Re[z] <
    1/2 Log[Root[1 - 2 #1 - 2 #1^2 - 2 #1^3 + #1^4 &, 2]] && (Im[
    z] == -ArcTan[Sqrt[(3 + 4 E^(2 Re[z]) + 3 E^(4 Re[z]))/(
    1 + E^(4 Re[z]))]] + π C[1] ||
    Im[z] == ArcTan[Sqrt[(3 + 4 E^(2 Re[z]) + 3 E^(4 Re[z]))/(
    1 + E^(4 Re[z]))]] + π C[1])))







    share|improve this answer









    $endgroup$












    • $begingroup$
      Thank you very much .
      $endgroup$
      – topspin
      Apr 5 at 17:26










    • $begingroup$
      Is there any way to visualize those branch points and the branch cuts in Mathematica Instead of ContourPlot ?
      $endgroup$
      – topspin
      Apr 5 at 17:28











    • $begingroup$
      Or visualizing the branch points and the branch cuts using ContourPlot .
      $endgroup$
      – topspin
      Apr 5 at 17:51










    • $begingroup$
      Just to compare: the Maple's result dropbox.com/s/zh7mq932rlvb1uk/branch_cuts.pdf?dl=0 seems to be different.
      $endgroup$
      – user64494
      Apr 6 at 6:08















    13












    $begingroup$

    Perhaps you can make use of the internal functions ComplexAnalysis`BranchCuts and ComplexAnalysis`BranchPoints. First, use a complex variable z instead of x + I y:



    expr = Sqrt[(Tanh[z]-Tanh[2z])^2+(Tanh[z] Tanh[2z]+1)^2-1-2 Tanh[z]^2Tanh[2z]^2];


    Then, for example, the branch points are:



    pts = ComplexAnalysis`BranchPoints[expr, z]



    ConditionalExpression[-(I/(2 π C[1])), C[1] ∈ Integers],
    ConditionalExpression[2 I π C[1], C[1] ∈ Integers],
    ConditionalExpression[1/(-((I π)/4) + 2 I π C[1]),
    C[1] ∈ Integers],
    ConditionalExpression[-((I π)/4) + 2 I π C[1],
    C[1] ∈ Integers],
    ConditionalExpression[1/((I π)/4 + 2 I π C[1]),
    C[1] ∈ Integers],
    ConditionalExpression[(I π)/4 + 2 I π C[1],
    C[1] ∈ Integers],
    ConditionalExpression[1/(-((I π)/2) + 2 I π C[1]),
    C[1] ∈ Integers],
    ConditionalExpression[-((I π)/2) + 2 I π C[1],
    C[1] ∈ Integers],
    ConditionalExpression[1/((I π)/2 + 2 I π C[1]),
    C[1] ∈ Integers],
    ConditionalExpression[(I π)/2 + 2 I π C[1],
    C[1] ∈ Integers],
    ConditionalExpression[1/(-((3 I π)/4) + 2 I π C[1]),
    C[1] ∈ Integers],
    ConditionalExpression[-((3 I π)/4) + 2 I π C[1],
    C[1] ∈ Integers],
    ConditionalExpression[1/((3 I π)/4 + 2 I π C[1]),
    C[1] ∈ Integers],
    ConditionalExpression[(3 I π)/4 + 2 I π C[1],
    C[1] ∈ Integers],
    ConditionalExpression[1/(I π + 2 I π C[1]),
    C[1] ∈ Integers],
    ConditionalExpression[I π + 2 I π C[1], C[1] ∈ Integers],
    ConditionalExpression[1/(
    2 I π C[1] + Log[(-(1/2) + I/2) - Sqrt[-1 - I/2]]),
    C[1] ∈ Integers],
    ConditionalExpression[2 I π C[1] + Log[(-(1/2) + I/2) - Sqrt[-1 - I/2]],
    C[1] ∈ Integers],
    ConditionalExpression[1/(2 I π C[1] + Log[(1/2 - I/2) - Sqrt[-1 - I/2]]),
    C[1] ∈ Integers],
    ConditionalExpression[2 I π C[1] + Log[(1/2 - I/2) - Sqrt[-1 - I/2]],
    C[1] ∈ Integers],
    ConditionalExpression[1/(
    2 I π C[1] + Log[(-(1/2) + I/2) + Sqrt[-1 - I/2]]),
    C[1] ∈ Integers],
    ConditionalExpression[2 I π C[1] + Log[(-(1/2) + I/2) + Sqrt[-1 - I/2]],
    C[1] ∈ Integers],
    ConditionalExpression[1/(2 I π C[1] + Log[(1/2 - I/2) + Sqrt[-1 - I/2]]),
    C[1] ∈ Integers],
    ConditionalExpression[2 I π C[1] + Log[(1/2 - I/2) + Sqrt[-1 - I/2]],
    C[1] ∈ Integers],
    ConditionalExpression[1/(
    2 I π C[1] + Log[(-(1/2) - I/2) - Sqrt[-1 + I/2]]),
    C[1] ∈ Integers],
    ConditionalExpression[2 I π C[1] + Log[(-(1/2) - I/2) - Sqrt[-1 + I/2]],
    C[1] ∈ Integers],
    ConditionalExpression[1/(2 I π C[1] + Log[(1/2 + I/2) - Sqrt[-1 + I/2]]),
    C[1] ∈ Integers],
    ConditionalExpression[2 I π C[1] + Log[(1/2 + I/2) - Sqrt[-1 + I/2]],
    C[1] ∈ Integers],
    ConditionalExpression[1/(
    2 I π C[1] + Log[(-(1/2) - I/2) + Sqrt[-1 + I/2]]),
    C[1] ∈ Integers],
    ConditionalExpression[2 I π C[1] + Log[(-(1/2) - I/2) + Sqrt[-1 + I/2]],
    C[1] ∈ Integers],
    ConditionalExpression[1/(2 I π C[1] + Log[(1/2 + I/2) + Sqrt[-1 + I/2]]),
    C[1] ∈ Integers],
    ConditionalExpression[2 I π C[1] + Log[(1/2 + I/2) + Sqrt[-1 + I/2]],
    C[1] ∈ Integers]




    The above can be simplified a bit with:



    Simplify[pts, C[1] ∈ Integers]



    -(I/(2 π C[1])), 2 I π C[1], (4 I)/(π - 8 π C[1]),
    1/4 I π (-1 + 8 C[1]), -((4 I)/(π + 8 π C[1])),
    1/4 I (π + 8 π C[1]), (2 I)/(π - 4 π C[1]),
    1/2 I π (-1 + 4 C[1]), -((2 I)/(π + 4 π C[1])),
    1/2 I (π + 4 π C[1]), (4 I)/(3 π - 8 π C[1]),
    1/4 I π (-3 + 8 C[1]), -((4 I)/(3 π + 8 π C[1])),
    1/4 I π (3 + 8 C[1]), -(I/(π + 2 π C[1])),
    I (π + 2 π C[1]), 1/(
    2 I π C[1] + Log[(-(1/2) + I/2) - Sqrt[-1 - I/2]]),
    2 I π C[1] + Log[(-(1/2) + I/2) - Sqrt[-1 - I/2]], 1/(
    2 I π C[1] + Log[(1/2 - I/2) - Sqrt[-1 - I/2]]),
    2 I π C[1] + Log[(1/2 - I/2) - Sqrt[-1 - I/2]], 1/(
    2 I π C[1] + Log[(-(1/2) + I/2) + Sqrt[-1 - I/2]]),
    2 I π C[1] + Log[(-(1/2) + I/2) + Sqrt[-1 - I/2]], 1/(
    2 I π C[1] + Log[(1/2 - I/2) + Sqrt[-1 - I/2]]),
    2 I π C[1] + Log[(1/2 - I/2) + Sqrt[-1 - I/2]], 1/(
    2 I π C[1] + Log[(-(1/2) - I/2) - Sqrt[-1 + I/2]]),
    2 I π C[1] + Log[(-(1/2) - I/2) - Sqrt[-1 + I/2]], 1/(
    2 I π C[1] + Log[(1/2 + I/2) - Sqrt[-1 + I/2]]),
    2 I π C[1] + Log[(1/2 + I/2) - Sqrt[-1 + I/2]], 1/(
    2 I π C[1] + Log[(-(1/2) - I/2) + Sqrt[-1 + I/2]]),
    2 I π C[1] + Log[(-(1/2) - I/2) + Sqrt[-1 + I/2]], 1/(
    2 I π C[1] + Log[(1/2 + I/2) + Sqrt[-1 + I/2]]),
    2 I π C[1] + Log[(1/2 + I/2) + Sqrt[-1 + I/2]]




    Similarly, the branch cuts can be found with:



    ComplexAnalysis`BranchCuts[expr, z]



    C[1] ∈
    Integers && ((1/2 Log[Root[1 - 2 #1 - 2 #1^2 - 2 #1^3 + #1^4 &, 1]] <
    Re[z] < 0 && (Im[
    z] == -ArcTan[Sqrt[(3 + 4 E^(2 Re[z]) + 3 E^(4 Re[z]))/(
    1 + E^(4 Re[z]))]] + π C[1] ||
    Im[z] == ArcTan[Sqrt[(3 + 4 E^(2 Re[z]) + 3 E^(4 Re[z]))/(
    1 + E^(4 Re[z]))]] + π C[1])) || (Re[z] ==
    0 && (1/2 (-π + 2 π C[1]) < Im[z] <
    1/4 (-π + 4 π C[1]) ||
    1/4 (-π + 4 π C[1]) < Im[z] < π C[1] || π C[1] <
    Im[z] < 1/4 (π + 4 π C[1]) ||
    1/4 (π + 4 π C[1]) < Im[z] <
    1/2 (π + 2 π C[1]))) || (0 < Re[z] <
    1/2 Log[Root[1 - 2 #1 - 2 #1^2 - 2 #1^3 + #1^4 &, 2]] && (Im[
    z] == -ArcTan[Sqrt[(3 + 4 E^(2 Re[z]) + 3 E^(4 Re[z]))/(
    1 + E^(4 Re[z]))]] + π C[1] ||
    Im[z] == ArcTan[Sqrt[(3 + 4 E^(2 Re[z]) + 3 E^(4 Re[z]))/(
    1 + E^(4 Re[z]))]] + π C[1])))







    share|improve this answer









    $endgroup$












    • $begingroup$
      Thank you very much .
      $endgroup$
      – topspin
      Apr 5 at 17:26










    • $begingroup$
      Is there any way to visualize those branch points and the branch cuts in Mathematica Instead of ContourPlot ?
      $endgroup$
      – topspin
      Apr 5 at 17:28











    • $begingroup$
      Or visualizing the branch points and the branch cuts using ContourPlot .
      $endgroup$
      – topspin
      Apr 5 at 17:51










    • $begingroup$
      Just to compare: the Maple's result dropbox.com/s/zh7mq932rlvb1uk/branch_cuts.pdf?dl=0 seems to be different.
      $endgroup$
      – user64494
      Apr 6 at 6:08













    13












    13








    13





    $begingroup$

    Perhaps you can make use of the internal functions ComplexAnalysis`BranchCuts and ComplexAnalysis`BranchPoints. First, use a complex variable z instead of x + I y:



    expr = Sqrt[(Tanh[z]-Tanh[2z])^2+(Tanh[z] Tanh[2z]+1)^2-1-2 Tanh[z]^2Tanh[2z]^2];


    Then, for example, the branch points are:



    pts = ComplexAnalysis`BranchPoints[expr, z]



    ConditionalExpression[-(I/(2 π C[1])), C[1] ∈ Integers],
    ConditionalExpression[2 I π C[1], C[1] ∈ Integers],
    ConditionalExpression[1/(-((I π)/4) + 2 I π C[1]),
    C[1] ∈ Integers],
    ConditionalExpression[-((I π)/4) + 2 I π C[1],
    C[1] ∈ Integers],
    ConditionalExpression[1/((I π)/4 + 2 I π C[1]),
    C[1] ∈ Integers],
    ConditionalExpression[(I π)/4 + 2 I π C[1],
    C[1] ∈ Integers],
    ConditionalExpression[1/(-((I π)/2) + 2 I π C[1]),
    C[1] ∈ Integers],
    ConditionalExpression[-((I π)/2) + 2 I π C[1],
    C[1] ∈ Integers],
    ConditionalExpression[1/((I π)/2 + 2 I π C[1]),
    C[1] ∈ Integers],
    ConditionalExpression[(I π)/2 + 2 I π C[1],
    C[1] ∈ Integers],
    ConditionalExpression[1/(-((3 I π)/4) + 2 I π C[1]),
    C[1] ∈ Integers],
    ConditionalExpression[-((3 I π)/4) + 2 I π C[1],
    C[1] ∈ Integers],
    ConditionalExpression[1/((3 I π)/4 + 2 I π C[1]),
    C[1] ∈ Integers],
    ConditionalExpression[(3 I π)/4 + 2 I π C[1],
    C[1] ∈ Integers],
    ConditionalExpression[1/(I π + 2 I π C[1]),
    C[1] ∈ Integers],
    ConditionalExpression[I π + 2 I π C[1], C[1] ∈ Integers],
    ConditionalExpression[1/(
    2 I π C[1] + Log[(-(1/2) + I/2) - Sqrt[-1 - I/2]]),
    C[1] ∈ Integers],
    ConditionalExpression[2 I π C[1] + Log[(-(1/2) + I/2) - Sqrt[-1 - I/2]],
    C[1] ∈ Integers],
    ConditionalExpression[1/(2 I π C[1] + Log[(1/2 - I/2) - Sqrt[-1 - I/2]]),
    C[1] ∈ Integers],
    ConditionalExpression[2 I π C[1] + Log[(1/2 - I/2) - Sqrt[-1 - I/2]],
    C[1] ∈ Integers],
    ConditionalExpression[1/(
    2 I π C[1] + Log[(-(1/2) + I/2) + Sqrt[-1 - I/2]]),
    C[1] ∈ Integers],
    ConditionalExpression[2 I π C[1] + Log[(-(1/2) + I/2) + Sqrt[-1 - I/2]],
    C[1] ∈ Integers],
    ConditionalExpression[1/(2 I π C[1] + Log[(1/2 - I/2) + Sqrt[-1 - I/2]]),
    C[1] ∈ Integers],
    ConditionalExpression[2 I π C[1] + Log[(1/2 - I/2) + Sqrt[-1 - I/2]],
    C[1] ∈ Integers],
    ConditionalExpression[1/(
    2 I π C[1] + Log[(-(1/2) - I/2) - Sqrt[-1 + I/2]]),
    C[1] ∈ Integers],
    ConditionalExpression[2 I π C[1] + Log[(-(1/2) - I/2) - Sqrt[-1 + I/2]],
    C[1] ∈ Integers],
    ConditionalExpression[1/(2 I π C[1] + Log[(1/2 + I/2) - Sqrt[-1 + I/2]]),
    C[1] ∈ Integers],
    ConditionalExpression[2 I π C[1] + Log[(1/2 + I/2) - Sqrt[-1 + I/2]],
    C[1] ∈ Integers],
    ConditionalExpression[1/(
    2 I π C[1] + Log[(-(1/2) - I/2) + Sqrt[-1 + I/2]]),
    C[1] ∈ Integers],
    ConditionalExpression[2 I π C[1] + Log[(-(1/2) - I/2) + Sqrt[-1 + I/2]],
    C[1] ∈ Integers],
    ConditionalExpression[1/(2 I π C[1] + Log[(1/2 + I/2) + Sqrt[-1 + I/2]]),
    C[1] ∈ Integers],
    ConditionalExpression[2 I π C[1] + Log[(1/2 + I/2) + Sqrt[-1 + I/2]],
    C[1] ∈ Integers]




    The above can be simplified a bit with:



    Simplify[pts, C[1] ∈ Integers]



    -(I/(2 π C[1])), 2 I π C[1], (4 I)/(π - 8 π C[1]),
    1/4 I π (-1 + 8 C[1]), -((4 I)/(π + 8 π C[1])),
    1/4 I (π + 8 π C[1]), (2 I)/(π - 4 π C[1]),
    1/2 I π (-1 + 4 C[1]), -((2 I)/(π + 4 π C[1])),
    1/2 I (π + 4 π C[1]), (4 I)/(3 π - 8 π C[1]),
    1/4 I π (-3 + 8 C[1]), -((4 I)/(3 π + 8 π C[1])),
    1/4 I π (3 + 8 C[1]), -(I/(π + 2 π C[1])),
    I (π + 2 π C[1]), 1/(
    2 I π C[1] + Log[(-(1/2) + I/2) - Sqrt[-1 - I/2]]),
    2 I π C[1] + Log[(-(1/2) + I/2) - Sqrt[-1 - I/2]], 1/(
    2 I π C[1] + Log[(1/2 - I/2) - Sqrt[-1 - I/2]]),
    2 I π C[1] + Log[(1/2 - I/2) - Sqrt[-1 - I/2]], 1/(
    2 I π C[1] + Log[(-(1/2) + I/2) + Sqrt[-1 - I/2]]),
    2 I π C[1] + Log[(-(1/2) + I/2) + Sqrt[-1 - I/2]], 1/(
    2 I π C[1] + Log[(1/2 - I/2) + Sqrt[-1 - I/2]]),
    2 I π C[1] + Log[(1/2 - I/2) + Sqrt[-1 - I/2]], 1/(
    2 I π C[1] + Log[(-(1/2) - I/2) - Sqrt[-1 + I/2]]),
    2 I π C[1] + Log[(-(1/2) - I/2) - Sqrt[-1 + I/2]], 1/(
    2 I π C[1] + Log[(1/2 + I/2) - Sqrt[-1 + I/2]]),
    2 I π C[1] + Log[(1/2 + I/2) - Sqrt[-1 + I/2]], 1/(
    2 I π C[1] + Log[(-(1/2) - I/2) + Sqrt[-1 + I/2]]),
    2 I π C[1] + Log[(-(1/2) - I/2) + Sqrt[-1 + I/2]], 1/(
    2 I π C[1] + Log[(1/2 + I/2) + Sqrt[-1 + I/2]]),
    2 I π C[1] + Log[(1/2 + I/2) + Sqrt[-1 + I/2]]




    Similarly, the branch cuts can be found with:



    ComplexAnalysis`BranchCuts[expr, z]



    C[1] ∈
    Integers && ((1/2 Log[Root[1 - 2 #1 - 2 #1^2 - 2 #1^3 + #1^4 &, 1]] <
    Re[z] < 0 && (Im[
    z] == -ArcTan[Sqrt[(3 + 4 E^(2 Re[z]) + 3 E^(4 Re[z]))/(
    1 + E^(4 Re[z]))]] + π C[1] ||
    Im[z] == ArcTan[Sqrt[(3 + 4 E^(2 Re[z]) + 3 E^(4 Re[z]))/(
    1 + E^(4 Re[z]))]] + π C[1])) || (Re[z] ==
    0 && (1/2 (-π + 2 π C[1]) < Im[z] <
    1/4 (-π + 4 π C[1]) ||
    1/4 (-π + 4 π C[1]) < Im[z] < π C[1] || π C[1] <
    Im[z] < 1/4 (π + 4 π C[1]) ||
    1/4 (π + 4 π C[1]) < Im[z] <
    1/2 (π + 2 π C[1]))) || (0 < Re[z] <
    1/2 Log[Root[1 - 2 #1 - 2 #1^2 - 2 #1^3 + #1^4 &, 2]] && (Im[
    z] == -ArcTan[Sqrt[(3 + 4 E^(2 Re[z]) + 3 E^(4 Re[z]))/(
    1 + E^(4 Re[z]))]] + π C[1] ||
    Im[z] == ArcTan[Sqrt[(3 + 4 E^(2 Re[z]) + 3 E^(4 Re[z]))/(
    1 + E^(4 Re[z]))]] + π C[1])))







    share|improve this answer









    $endgroup$



    Perhaps you can make use of the internal functions ComplexAnalysis`BranchCuts and ComplexAnalysis`BranchPoints. First, use a complex variable z instead of x + I y:



    expr = Sqrt[(Tanh[z]-Tanh[2z])^2+(Tanh[z] Tanh[2z]+1)^2-1-2 Tanh[z]^2Tanh[2z]^2];


    Then, for example, the branch points are:



    pts = ComplexAnalysis`BranchPoints[expr, z]



    ConditionalExpression[-(I/(2 π C[1])), C[1] ∈ Integers],
    ConditionalExpression[2 I π C[1], C[1] ∈ Integers],
    ConditionalExpression[1/(-((I π)/4) + 2 I π C[1]),
    C[1] ∈ Integers],
    ConditionalExpression[-((I π)/4) + 2 I π C[1],
    C[1] ∈ Integers],
    ConditionalExpression[1/((I π)/4 + 2 I π C[1]),
    C[1] ∈ Integers],
    ConditionalExpression[(I π)/4 + 2 I π C[1],
    C[1] ∈ Integers],
    ConditionalExpression[1/(-((I π)/2) + 2 I π C[1]),
    C[1] ∈ Integers],
    ConditionalExpression[-((I π)/2) + 2 I π C[1],
    C[1] ∈ Integers],
    ConditionalExpression[1/((I π)/2 + 2 I π C[1]),
    C[1] ∈ Integers],
    ConditionalExpression[(I π)/2 + 2 I π C[1],
    C[1] ∈ Integers],
    ConditionalExpression[1/(-((3 I π)/4) + 2 I π C[1]),
    C[1] ∈ Integers],
    ConditionalExpression[-((3 I π)/4) + 2 I π C[1],
    C[1] ∈ Integers],
    ConditionalExpression[1/((3 I π)/4 + 2 I π C[1]),
    C[1] ∈ Integers],
    ConditionalExpression[(3 I π)/4 + 2 I π C[1],
    C[1] ∈ Integers],
    ConditionalExpression[1/(I π + 2 I π C[1]),
    C[1] ∈ Integers],
    ConditionalExpression[I π + 2 I π C[1], C[1] ∈ Integers],
    ConditionalExpression[1/(
    2 I π C[1] + Log[(-(1/2) + I/2) - Sqrt[-1 - I/2]]),
    C[1] ∈ Integers],
    ConditionalExpression[2 I π C[1] + Log[(-(1/2) + I/2) - Sqrt[-1 - I/2]],
    C[1] ∈ Integers],
    ConditionalExpression[1/(2 I π C[1] + Log[(1/2 - I/2) - Sqrt[-1 - I/2]]),
    C[1] ∈ Integers],
    ConditionalExpression[2 I π C[1] + Log[(1/2 - I/2) - Sqrt[-1 - I/2]],
    C[1] ∈ Integers],
    ConditionalExpression[1/(
    2 I π C[1] + Log[(-(1/2) + I/2) + Sqrt[-1 - I/2]]),
    C[1] ∈ Integers],
    ConditionalExpression[2 I π C[1] + Log[(-(1/2) + I/2) + Sqrt[-1 - I/2]],
    C[1] ∈ Integers],
    ConditionalExpression[1/(2 I π C[1] + Log[(1/2 - I/2) + Sqrt[-1 - I/2]]),
    C[1] ∈ Integers],
    ConditionalExpression[2 I π C[1] + Log[(1/2 - I/2) + Sqrt[-1 - I/2]],
    C[1] ∈ Integers],
    ConditionalExpression[1/(
    2 I π C[1] + Log[(-(1/2) - I/2) - Sqrt[-1 + I/2]]),
    C[1] ∈ Integers],
    ConditionalExpression[2 I π C[1] + Log[(-(1/2) - I/2) - Sqrt[-1 + I/2]],
    C[1] ∈ Integers],
    ConditionalExpression[1/(2 I π C[1] + Log[(1/2 + I/2) - Sqrt[-1 + I/2]]),
    C[1] ∈ Integers],
    ConditionalExpression[2 I π C[1] + Log[(1/2 + I/2) - Sqrt[-1 + I/2]],
    C[1] ∈ Integers],
    ConditionalExpression[1/(
    2 I π C[1] + Log[(-(1/2) - I/2) + Sqrt[-1 + I/2]]),
    C[1] ∈ Integers],
    ConditionalExpression[2 I π C[1] + Log[(-(1/2) - I/2) + Sqrt[-1 + I/2]],
    C[1] ∈ Integers],
    ConditionalExpression[1/(2 I π C[1] + Log[(1/2 + I/2) + Sqrt[-1 + I/2]]),
    C[1] ∈ Integers],
    ConditionalExpression[2 I π C[1] + Log[(1/2 + I/2) + Sqrt[-1 + I/2]],
    C[1] ∈ Integers]




    The above can be simplified a bit with:



    Simplify[pts, C[1] ∈ Integers]



    -(I/(2 π C[1])), 2 I π C[1], (4 I)/(π - 8 π C[1]),
    1/4 I π (-1 + 8 C[1]), -((4 I)/(π + 8 π C[1])),
    1/4 I (π + 8 π C[1]), (2 I)/(π - 4 π C[1]),
    1/2 I π (-1 + 4 C[1]), -((2 I)/(π + 4 π C[1])),
    1/2 I (π + 4 π C[1]), (4 I)/(3 π - 8 π C[1]),
    1/4 I π (-3 + 8 C[1]), -((4 I)/(3 π + 8 π C[1])),
    1/4 I π (3 + 8 C[1]), -(I/(π + 2 π C[1])),
    I (π + 2 π C[1]), 1/(
    2 I π C[1] + Log[(-(1/2) + I/2) - Sqrt[-1 - I/2]]),
    2 I π C[1] + Log[(-(1/2) + I/2) - Sqrt[-1 - I/2]], 1/(
    2 I π C[1] + Log[(1/2 - I/2) - Sqrt[-1 - I/2]]),
    2 I π C[1] + Log[(1/2 - I/2) - Sqrt[-1 - I/2]], 1/(
    2 I π C[1] + Log[(-(1/2) + I/2) + Sqrt[-1 - I/2]]),
    2 I π C[1] + Log[(-(1/2) + I/2) + Sqrt[-1 - I/2]], 1/(
    2 I π C[1] + Log[(1/2 - I/2) + Sqrt[-1 - I/2]]),
    2 I π C[1] + Log[(1/2 - I/2) + Sqrt[-1 - I/2]], 1/(
    2 I π C[1] + Log[(-(1/2) - I/2) - Sqrt[-1 + I/2]]),
    2 I π C[1] + Log[(-(1/2) - I/2) - Sqrt[-1 + I/2]], 1/(
    2 I π C[1] + Log[(1/2 + I/2) - Sqrt[-1 + I/2]]),
    2 I π C[1] + Log[(1/2 + I/2) - Sqrt[-1 + I/2]], 1/(
    2 I π C[1] + Log[(-(1/2) - I/2) + Sqrt[-1 + I/2]]),
    2 I π C[1] + Log[(-(1/2) - I/2) + Sqrt[-1 + I/2]], 1/(
    2 I π C[1] + Log[(1/2 + I/2) + Sqrt[-1 + I/2]]),
    2 I π C[1] + Log[(1/2 + I/2) + Sqrt[-1 + I/2]]




    Similarly, the branch cuts can be found with:



    ComplexAnalysis`BranchCuts[expr, z]



    C[1] ∈
    Integers && ((1/2 Log[Root[1 - 2 #1 - 2 #1^2 - 2 #1^3 + #1^4 &, 1]] <
    Re[z] < 0 && (Im[
    z] == -ArcTan[Sqrt[(3 + 4 E^(2 Re[z]) + 3 E^(4 Re[z]))/(
    1 + E^(4 Re[z]))]] + π C[1] ||
    Im[z] == ArcTan[Sqrt[(3 + 4 E^(2 Re[z]) + 3 E^(4 Re[z]))/(
    1 + E^(4 Re[z]))]] + π C[1])) || (Re[z] ==
    0 && (1/2 (-π + 2 π C[1]) < Im[z] <
    1/4 (-π + 4 π C[1]) ||
    1/4 (-π + 4 π C[1]) < Im[z] < π C[1] || π C[1] <
    Im[z] < 1/4 (π + 4 π C[1]) ||
    1/4 (π + 4 π C[1]) < Im[z] <
    1/2 (π + 2 π C[1]))) || (0 < Re[z] <
    1/2 Log[Root[1 - 2 #1 - 2 #1^2 - 2 #1^3 + #1^4 &, 2]] && (Im[
    z] == -ArcTan[Sqrt[(3 + 4 E^(2 Re[z]) + 3 E^(4 Re[z]))/(
    1 + E^(4 Re[z]))]] + π C[1] ||
    Im[z] == ArcTan[Sqrt[(3 + 4 E^(2 Re[z]) + 3 E^(4 Re[z]))/(
    1 + E^(4 Re[z]))]] + π C[1])))








    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Apr 5 at 15:06









    Carl WollCarl Woll

    73.2k397191




    73.2k397191











    • $begingroup$
      Thank you very much .
      $endgroup$
      – topspin
      Apr 5 at 17:26










    • $begingroup$
      Is there any way to visualize those branch points and the branch cuts in Mathematica Instead of ContourPlot ?
      $endgroup$
      – topspin
      Apr 5 at 17:28











    • $begingroup$
      Or visualizing the branch points and the branch cuts using ContourPlot .
      $endgroup$
      – topspin
      Apr 5 at 17:51










    • $begingroup$
      Just to compare: the Maple's result dropbox.com/s/zh7mq932rlvb1uk/branch_cuts.pdf?dl=0 seems to be different.
      $endgroup$
      – user64494
      Apr 6 at 6:08
















    • $begingroup$
      Thank you very much .
      $endgroup$
      – topspin
      Apr 5 at 17:26










    • $begingroup$
      Is there any way to visualize those branch points and the branch cuts in Mathematica Instead of ContourPlot ?
      $endgroup$
      – topspin
      Apr 5 at 17:28











    • $begingroup$
      Or visualizing the branch points and the branch cuts using ContourPlot .
      $endgroup$
      – topspin
      Apr 5 at 17:51










    • $begingroup$
      Just to compare: the Maple's result dropbox.com/s/zh7mq932rlvb1uk/branch_cuts.pdf?dl=0 seems to be different.
      $endgroup$
      – user64494
      Apr 6 at 6:08















    $begingroup$
    Thank you very much .
    $endgroup$
    – topspin
    Apr 5 at 17:26




    $begingroup$
    Thank you very much .
    $endgroup$
    – topspin
    Apr 5 at 17:26












    $begingroup$
    Is there any way to visualize those branch points and the branch cuts in Mathematica Instead of ContourPlot ?
    $endgroup$
    – topspin
    Apr 5 at 17:28





    $begingroup$
    Is there any way to visualize those branch points and the branch cuts in Mathematica Instead of ContourPlot ?
    $endgroup$
    – topspin
    Apr 5 at 17:28













    $begingroup$
    Or visualizing the branch points and the branch cuts using ContourPlot .
    $endgroup$
    – topspin
    Apr 5 at 17:51




    $begingroup$
    Or visualizing the branch points and the branch cuts using ContourPlot .
    $endgroup$
    – topspin
    Apr 5 at 17:51












    $begingroup$
    Just to compare: the Maple's result dropbox.com/s/zh7mq932rlvb1uk/branch_cuts.pdf?dl=0 seems to be different.
    $endgroup$
    – user64494
    Apr 6 at 6:08




    $begingroup$
    Just to compare: the Maple's result dropbox.com/s/zh7mq932rlvb1uk/branch_cuts.pdf?dl=0 seems to be different.
    $endgroup$
    – user64494
    Apr 6 at 6:08











    6












    $begingroup$

    In this case, the only branch cuts and branch points will come from the square root. The cuts of $sqrtf(z)$ occurs along the half line $textIm(f(z)) = 0 ,wedge, textRe(f(z)) leq 0$. The branch points lie at $f(z) = 0$ or $f(z) = tildeinfty$.



    Your example:



    With[z = x + I y,
     expr = (Tanh[z] - Tanh[2 z])^2 + (Tanh[z] Tanh[2 z] + 1)^2 - 1 - 2 ((Tanh[2 z])^2) ((Tanh[z])^2);
     branchCutRegion[x_, y_, __] = Re[expr] <= 0;
    ];

    bpvals = Union[x, y /. Solve[(expr == 0 || 1/Together[TrigToExp[expr]] == 0) && -10 < x < 10 && -10 < y < 10, x, y]];


    Here we needed to help Solve find the branch points corresponding to $tildeinfty$.



    We can visualize the cut by plotting the constraint on the imaginary part, restricted to the region defined by the constraint on the real part. Here I've overlaid the branch points:



    ContourPlot[Im[expr] == 0, x, -10, 10, y, -10, 10, 
     RegionFunction -> branchCutRegion, PlotPoints -> 100, 
     Epilog -> Red, Point[bpvals]
    ]




    For fun we can add a plot of the expression under the cuts. Here I'll use domain coloring. Here, the complex argument varies with hue and the absolute value varies with saturation and brightness -- the darker the pixel, the larger the absolute value. I've also binned the absolute value to show some contours.



    binnedabs = fastCompile[z, _Complex,
     Module[f, abs, rnd, sgn, val,
     f = (Tanh[z] - Tanh[2 z])^2 + (Tanh[z] Tanh[2 z] + 1)^2 - 1 - 2 Tanh[2 z]^2 Tanh[z]^2;
     abs = Abs[f];
     rnd = Round[abs, .2];
     val = If[rnd == 0, f, rnd Sign[f]];
     
     Divide[Mod[Arg[val], 2π], 2π], 
     Power[1 + 0.3*Log[Abs[val] + 1], -1], 
     Power[1 + 0.5*Log[Abs[val] + 1], -1]
     
     ],
     CompilationTarget -> "C",
     Parallelization -> True,
     RuntimeAttributes -> Listable,
     RuntimeOptions -> "Speed"
    ];

    lattice = Array[List, 2048, 2048, -10., 10., -10., 10.].I, 1;

    raster = Raster[binnedabs[lattice], -10, -10, 10, 10, ColorFunction -> Hue];

    cutplot = ContourPlot[Im[expr] == 0, x, -10, 10, y, -10, 10, 
     RegionFunction -> branchCutRegion, PlotPoints -> 100, ContourStyle -> Black];

    Show[
     cutplot, 
     ImageSize -> 800,
     Prolog -> raster, 
     Epilog -> EdgeForm[Black], GrayLevel[.8], Disk[#, Scaled[.0045]] & /@ bpvals
    ]


    enter image description here






    share|improve this answer











    $endgroup$












    • $begingroup$
      In accordance with dropbox.com/s/zh7mq932rlvb1uk/branch_cuts.pdf?dl=0
      $endgroup$
      – user64494
      14 hours ago















    6












    $begingroup$

    In this case, the only branch cuts and branch points will come from the square root. The cuts of $sqrtf(z)$ occurs along the half line $textIm(f(z)) = 0 ,wedge, textRe(f(z)) leq 0$. The branch points lie at $f(z) = 0$ or $f(z) = tildeinfty$.



    Your example:



    With[z = x + I y,
     expr = (Tanh[z] - Tanh[2 z])^2 + (Tanh[z] Tanh[2 z] + 1)^2 - 1 - 2 ((Tanh[2 z])^2) ((Tanh[z])^2);
     branchCutRegion[x_, y_, __] = Re[expr] <= 0;
    ];

    bpvals = Union[x, y /. Solve[(expr == 0 || 1/Together[TrigToExp[expr]] == 0) && -10 < x < 10 && -10 < y < 10, x, y]];


    Here we needed to help Solve find the branch points corresponding to $tildeinfty$.



    We can visualize the cut by plotting the constraint on the imaginary part, restricted to the region defined by the constraint on the real part. Here I've overlaid the branch points:



    ContourPlot[Im[expr] == 0, x, -10, 10, y, -10, 10, 
     RegionFunction -> branchCutRegion, PlotPoints -> 100, 
     Epilog -> Red, Point[bpvals]
    ]




    For fun we can add a plot of the expression under the cuts. Here I'll use domain coloring. Here, the complex argument varies with hue and the absolute value varies with saturation and brightness -- the darker the pixel, the larger the absolute value. I've also binned the absolute value to show some contours.



    binnedabs = fastCompile[z, _Complex,
     Module[f, abs, rnd, sgn, val,
     f = (Tanh[z] - Tanh[2 z])^2 + (Tanh[z] Tanh[2 z] + 1)^2 - 1 - 2 Tanh[2 z]^2 Tanh[z]^2;
     abs = Abs[f];
     rnd = Round[abs, .2];
     val = If[rnd == 0, f, rnd Sign[f]];
     
     Divide[Mod[Arg[val], 2π], 2π], 
     Power[1 + 0.3*Log[Abs[val] + 1], -1], 
     Power[1 + 0.5*Log[Abs[val] + 1], -1]
     
     ],
     CompilationTarget -> "C",
     Parallelization -> True,
     RuntimeAttributes -> Listable,
     RuntimeOptions -> "Speed"
    ];

    lattice = Array[List, 2048, 2048, -10., 10., -10., 10.].I, 1;

    raster = Raster[binnedabs[lattice], -10, -10, 10, 10, ColorFunction -> Hue];

    cutplot = ContourPlot[Im[expr] == 0, x, -10, 10, y, -10, 10, 
     RegionFunction -> branchCutRegion, PlotPoints -> 100, ContourStyle -> Black];

    Show[
     cutplot, 
     ImageSize -> 800,
     Prolog -> raster, 
     Epilog -> EdgeForm[Black], GrayLevel[.8], Disk[#, Scaled[.0045]] & /@ bpvals
    ]


    enter image description here






    share|improve this answer











    $endgroup$












    • $begingroup$
      In accordance with dropbox.com/s/zh7mq932rlvb1uk/branch_cuts.pdf?dl=0
      $endgroup$
      – user64494
      14 hours ago













    6












    6








    6





    $begingroup$

    In this case, the only branch cuts and branch points will come from the square root. The cuts of $sqrtf(z)$ occurs along the half line $textIm(f(z)) = 0 ,wedge, textRe(f(z)) leq 0$. The branch points lie at $f(z) = 0$ or $f(z) = tildeinfty$.



    Your example:



    With[z = x + I y,
     expr = (Tanh[z] - Tanh[2 z])^2 + (Tanh[z] Tanh[2 z] + 1)^2 - 1 - 2 ((Tanh[2 z])^2) ((Tanh[z])^2);
     branchCutRegion[x_, y_, __] = Re[expr] <= 0;
    ];

    bpvals = Union[x, y /. Solve[(expr == 0 || 1/Together[TrigToExp[expr]] == 0) && -10 < x < 10 && -10 < y < 10, x, y]];


    Here we needed to help Solve find the branch points corresponding to $tildeinfty$.



    We can visualize the cut by plotting the constraint on the imaginary part, restricted to the region defined by the constraint on the real part. Here I've overlaid the branch points:



    ContourPlot[Im[expr] == 0, x, -10, 10, y, -10, 10, 
     RegionFunction -> branchCutRegion, PlotPoints -> 100, 
     Epilog -> Red, Point[bpvals]
    ]




    For fun we can add a plot of the expression under the cuts. Here I'll use domain coloring. Here, the complex argument varies with hue and the absolute value varies with saturation and brightness -- the darker the pixel, the larger the absolute value. I've also binned the absolute value to show some contours.



    binnedabs = fastCompile[z, _Complex,
     Module[f, abs, rnd, sgn, val,
     f = (Tanh[z] - Tanh[2 z])^2 + (Tanh[z] Tanh[2 z] + 1)^2 - 1 - 2 Tanh[2 z]^2 Tanh[z]^2;
     abs = Abs[f];
     rnd = Round[abs, .2];
     val = If[rnd == 0, f, rnd Sign[f]];
     
     Divide[Mod[Arg[val], 2π], 2π], 
     Power[1 + 0.3*Log[Abs[val] + 1], -1], 
     Power[1 + 0.5*Log[Abs[val] + 1], -1]
     
     ],
     CompilationTarget -> "C",
     Parallelization -> True,
     RuntimeAttributes -> Listable,
     RuntimeOptions -> "Speed"
    ];

    lattice = Array[List, 2048, 2048, -10., 10., -10., 10.].I, 1;

    raster = Raster[binnedabs[lattice], -10, -10, 10, 10, ColorFunction -> Hue];

    cutplot = ContourPlot[Im[expr] == 0, x, -10, 10, y, -10, 10, 
     RegionFunction -> branchCutRegion, PlotPoints -> 100, ContourStyle -> Black];

    Show[
     cutplot, 
     ImageSize -> 800,
     Prolog -> raster, 
     Epilog -> EdgeForm[Black], GrayLevel[.8], Disk[#, Scaled[.0045]] & /@ bpvals
    ]


    enter image description here






    share|improve this answer











    $endgroup$



    In this case, the only branch cuts and branch points will come from the square root. The cuts of $sqrtf(z)$ occurs along the half line $textIm(f(z)) = 0 ,wedge, textRe(f(z)) leq 0$. The branch points lie at $f(z) = 0$ or $f(z) = tildeinfty$.



    Your example:



    With[z = x + I y,
     expr = (Tanh[z] - Tanh[2 z])^2 + (Tanh[z] Tanh[2 z] + 1)^2 - 1 - 2 ((Tanh[2 z])^2) ((Tanh[z])^2);
     branchCutRegion[x_, y_, __] = Re[expr] <= 0;
    ];

    bpvals = Union[x, y /. Solve[(expr == 0 || 1/Together[TrigToExp[expr]] == 0) && -10 < x < 10 && -10 < y < 10, x, y]];


    Here we needed to help Solve find the branch points corresponding to $tildeinfty$.



    We can visualize the cut by plotting the constraint on the imaginary part, restricted to the region defined by the constraint on the real part. Here I've overlaid the branch points:



    ContourPlot[Im[expr] == 0, x, -10, 10, y, -10, 10, 
     RegionFunction -> branchCutRegion, PlotPoints -> 100, 
     Epilog -> Red, Point[bpvals]
    ]




    For fun we can add a plot of the expression under the cuts. Here I'll use domain coloring. Here, the complex argument varies with hue and the absolute value varies with saturation and brightness -- the darker the pixel, the larger the absolute value. I've also binned the absolute value to show some contours.



    binnedabs = fastCompile[z, _Complex,
     Module[f, abs, rnd, sgn, val,
     f = (Tanh[z] - Tanh[2 z])^2 + (Tanh[z] Tanh[2 z] + 1)^2 - 1 - 2 Tanh[2 z]^2 Tanh[z]^2;
     abs = Abs[f];
     rnd = Round[abs, .2];
     val = If[rnd == 0, f, rnd Sign[f]];
     
     Divide[Mod[Arg[val], 2π], 2π], 
     Power[1 + 0.3*Log[Abs[val] + 1], -1], 
     Power[1 + 0.5*Log[Abs[val] + 1], -1]
     
     ],
     CompilationTarget -> "C",
     Parallelization -> True,
     RuntimeAttributes -> Listable,
     RuntimeOptions -> "Speed"
    ];

    lattice = Array[List, 2048, 2048, -10., 10., -10., 10.].I, 1;

    raster = Raster[binnedabs[lattice], -10, -10, 10, 10, ColorFunction -> Hue];

    cutplot = ContourPlot[Im[expr] == 0, x, -10, 10, y, -10, 10, 
     RegionFunction -> branchCutRegion, PlotPoints -> 100, ContourStyle -> Black];

    Show[
     cutplot, 
     ImageSize -> 800,
     Prolog -> raster, 
     Epilog -> EdgeForm[Black], GrayLevel[.8], Disk[#, Scaled[.0045]] & /@ bpvals
    ]


    enter image description here







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 5 hours ago

























    answered 23 hours ago









    Chip HurstChip Hurst

    23.1k15893




    23.1k15893











    • $begingroup$
      In accordance with dropbox.com/s/zh7mq932rlvb1uk/branch_cuts.pdf?dl=0
      $endgroup$
      – user64494
      14 hours ago
















    • $begingroup$
      In accordance with dropbox.com/s/zh7mq932rlvb1uk/branch_cuts.pdf?dl=0
      $endgroup$
      – user64494
      14 hours ago















    $begingroup$
    In accordance with dropbox.com/s/zh7mq932rlvb1uk/branch_cuts.pdf?dl=0
    $endgroup$
    – user64494
    14 hours ago




    $begingroup$
    In accordance with dropbox.com/s/zh7mq932rlvb1uk/branch_cuts.pdf?dl=0
    $endgroup$
    – user64494
    14 hours ago











    4












    $begingroup$

    First start with the branch points: these are the values of z where the root is not single-valued.
    First:



    myexp = Together[
     TrigToExp[
     FullSimplify[(Tanh[z] - Tanh[2 z])^2 + (Tanh[z] Tanh[2 z] + 1)^2 - 
     1 - 2 Tanh[z]^2 Tanh[2 z]^2]
     ]]


    $$fracleft(e^2 z-1right)^2 left(4 e^2 z+10 e^4 z+4 e^6 z+e^8 z+1right)left(e^2 z+1right)^2 left(e^4 z+1right)^2$$



    Now solve for the zeros of the denominator and numerator. I'll do the numerator: First obtain a polynomial in e^z and then solve the polynomial in terms of a polynomial in just z:



     Expand[Numerator[
     Together[TrigToExp[
     FullSimplify[(Tanh[z] - Tanh[2 z])^2 + (Tanh[z] Tanh[2 z] + 
     1)^2 - 1 - 2 Tanh[z]^2 Tanh[2 z]^2]
     ]]]]
     mySol = z /. 
     Solve[1 + 2 z^2 + 3 z^4 - 12 z^6 + 3 z^8 + 2 z^10 + z^12 == 0, z];


    Now make the substitution Log[z] and keep in mind Log[z]=Log[Abs[z]]+i (Arg(z)+2k pi) so that we have a set of branch points for all integer k. I will do k=0,1,-1 and then plot the results:enter image description here



    p1 = Show[
     Graphics[Red, 
     Point @@ Re[#], Im[#] & /@ (N[Log[#]] & /@ mySol)], 
     Axes -> True, PlotRange -> 5];
    p2 = Show[
     Graphics[Blue, 
     Point @@ Re[#], Im[#] & /@ (N[(Log[#] + 2 [Pi] I)] & /@ 
     mySol)], Axes -> True, PlotRange -> 15];
    p3 = Show[
     Graphics[Green // Darker, 
     Point @@ Re[#], Im[#] & /@ (N[(Log[#] - 2 [Pi] I)] & /@ 
     mySol)], Axes -> True, PlotRange -> 15];
    Show[p1, p2, p3, PlotRange -> 15]





    share|improve this answer









    $endgroup$

















      4












      $begingroup$

      First start with the branch points: these are the values of z where the root is not single-valued.
      First:



      myexp = Together[
       TrigToExp[
       FullSimplify[(Tanh[z] - Tanh[2 z])^2 + (Tanh[z] Tanh[2 z] + 1)^2 - 
       1 - 2 Tanh[z]^2 Tanh[2 z]^2]
       ]]


      $$fracleft(e^2 z-1right)^2 left(4 e^2 z+10 e^4 z+4 e^6 z+e^8 z+1right)left(e^2 z+1right)^2 left(e^4 z+1right)^2$$



      Now solve for the zeros of the denominator and numerator. I'll do the numerator: First obtain a polynomial in e^z and then solve the polynomial in terms of a polynomial in just z:



       Expand[Numerator[
       Together[TrigToExp[
       FullSimplify[(Tanh[z] - Tanh[2 z])^2 + (Tanh[z] Tanh[2 z] + 
       1)^2 - 1 - 2 Tanh[z]^2 Tanh[2 z]^2]
       ]]]]
       mySol = z /. 
       Solve[1 + 2 z^2 + 3 z^4 - 12 z^6 + 3 z^8 + 2 z^10 + z^12 == 0, z];


      Now make the substitution Log[z] and keep in mind Log[z]=Log[Abs[z]]+i (Arg(z)+2k pi) so that we have a set of branch points for all integer k. I will do k=0,1,-1 and then plot the results:enter image description here



      p1 = Show[
       Graphics[Red, 
       Point @@ Re[#], Im[#] & /@ (N[Log[#]] & /@ mySol)], 
       Axes -> True, PlotRange -> 5];
      p2 = Show[
       Graphics[Blue, 
       Point @@ Re[#], Im[#] & /@ (N[(Log[#] + 2 [Pi] I)] & /@ 
       mySol)], Axes -> True, PlotRange -> 15];
      p3 = Show[
       Graphics[Green // Darker, 
       Point @@ Re[#], Im[#] & /@ (N[(Log[#] - 2 [Pi] I)] & /@ 
       mySol)], Axes -> True, PlotRange -> 15];
      Show[p1, p2, p3, PlotRange -> 15]





      share|improve this answer









      $endgroup$















        4












        4








        4





        $begingroup$

        First start with the branch points: these are the values of z where the root is not single-valued.
        First:



        myexp = Together[
         TrigToExp[
         FullSimplify[(Tanh[z] - Tanh[2 z])^2 + (Tanh[z] Tanh[2 z] + 1)^2 - 
         1 - 2 Tanh[z]^2 Tanh[2 z]^2]
         ]]


        $$fracleft(e^2 z-1right)^2 left(4 e^2 z+10 e^4 z+4 e^6 z+e^8 z+1right)left(e^2 z+1right)^2 left(e^4 z+1right)^2$$



        Now solve for the zeros of the denominator and numerator. I'll do the numerator: First obtain a polynomial in e^z and then solve the polynomial in terms of a polynomial in just z:



         Expand[Numerator[
         Together[TrigToExp[
         FullSimplify[(Tanh[z] - Tanh[2 z])^2 + (Tanh[z] Tanh[2 z] + 
         1)^2 - 1 - 2 Tanh[z]^2 Tanh[2 z]^2]
         ]]]]
         mySol = z /. 
         Solve[1 + 2 z^2 + 3 z^4 - 12 z^6 + 3 z^8 + 2 z^10 + z^12 == 0, z];


        Now make the substitution Log[z] and keep in mind Log[z]=Log[Abs[z]]+i (Arg(z)+2k pi) so that we have a set of branch points for all integer k. I will do k=0,1,-1 and then plot the results:enter image description here



        p1 = Show[
         Graphics[Red, 
         Point @@ Re[#], Im[#] & /@ (N[Log[#]] & /@ mySol)], 
         Axes -> True, PlotRange -> 5];
        p2 = Show[
         Graphics[Blue, 
         Point @@ Re[#], Im[#] & /@ (N[(Log[#] + 2 [Pi] I)] & /@ 
         mySol)], Axes -> True, PlotRange -> 15];
        p3 = Show[
         Graphics[Green // Darker, 
         Point @@ Re[#], Im[#] & /@ (N[(Log[#] - 2 [Pi] I)] & /@ 
         mySol)], Axes -> True, PlotRange -> 15];
        Show[p1, p2, p3, PlotRange -> 15]





        share|improve this answer









        $endgroup$



        First start with the branch points: these are the values of z where the root is not single-valued.
        First:



        myexp = Together[
         TrigToExp[
         FullSimplify[(Tanh[z] - Tanh[2 z])^2 + (Tanh[z] Tanh[2 z] + 1)^2 - 
         1 - 2 Tanh[z]^2 Tanh[2 z]^2]
         ]]


        $$fracleft(e^2 z-1right)^2 left(4 e^2 z+10 e^4 z+4 e^6 z+e^8 z+1right)left(e^2 z+1right)^2 left(e^4 z+1right)^2$$



        Now solve for the zeros of the denominator and numerator. I'll do the numerator: First obtain a polynomial in e^z and then solve the polynomial in terms of a polynomial in just z:



         Expand[Numerator[
         Together[TrigToExp[
         FullSimplify[(Tanh[z] - Tanh[2 z])^2 + (Tanh[z] Tanh[2 z] + 
         1)^2 - 1 - 2 Tanh[z]^2 Tanh[2 z]^2]
         ]]]]
         mySol = z /. 
         Solve[1 + 2 z^2 + 3 z^4 - 12 z^6 + 3 z^8 + 2 z^10 + z^12 == 0, z];


        Now make the substitution Log[z] and keep in mind Log[z]=Log[Abs[z]]+i (Arg(z)+2k pi) so that we have a set of branch points for all integer k. I will do k=0,1,-1 and then plot the results:enter image description here



        p1 = Show[
         Graphics[Red, 
         Point @@ Re[#], Im[#] & /@ (N[Log[#]] & /@ mySol)], 
         Axes -> True, PlotRange -> 5];
        p2 = Show[
         Graphics[Blue, 
         Point @@ Re[#], Im[#] & /@ (N[(Log[#] + 2 [Pi] I)] & /@ 
         mySol)], Axes -> True, PlotRange -> 15];
        p3 = Show[
         Graphics[Green // Darker, 
         Point @@ Re[#], Im[#] & /@ (N[(Log[#] - 2 [Pi] I)] & /@ 
         mySol)], Axes -> True, PlotRange -> 15];
        Show[p1, p2, p3, PlotRange -> 15]






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Apr 6 at 14:43









        DominicDominic

        1276




        1276




















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