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Question on branch cuts and branch points
Updating Wagon's FindAllCrossings2D[] functionSqrt — how to get negative branch?Differential Equation in Complex Plane and Parametric PlotContour Integration along a contour containing two branch pointsDoes not evaluate the Integral and a plane plotHow to plot real roots of complex polynomials avoiding branch cuts?Branch cuts of sqrtHow can I find a list of points that satisfy constraints?ContourPlot problem with Sinc funtionplotting functions containing branch cuts and crossingsHow would one go about plotting this parameterized curve using numerical resources (analitically it's too hard)?
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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
$$sqrt(tanh(z) -tanh(2z))^2 +(tanh(z)*tanh(2z)+1)^2-1-2tanh(z)^2 tanh(2z)^2$$
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]
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]
plotting functions complex
asked yesterday
topspintopspin
1363
New contributor
topspin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
$begingroup$
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
$$sqrt(tanh(z) -tanh(2z))^2 +(tanh(z)*tanh(2z)+1)^2-1-2tanh(z)^2 tanh(2z)^2$$
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]
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]
plotting functions complex
asked yesterday
topspintopspin
1363
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.
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The first step might be to find all the zeros of the function under the square root. Perhaps this might help.
$endgroup$
– Hugh
yesterday
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
yesterday
$begingroup$
Ok, Thank you . I have just edited my post .
$endgroup$
– topspin
yesterday
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
yesterday
add a comment |
$begingroup$
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
$$sqrt(tanh(z) -tanh(2z))^2 +(tanh(z)*tanh(2z)+1)^2-1-2tanh(z)^2 tanh(2z)^2$$
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]
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]
plotting functions complex
asked yesterday
topspintopspin
1363
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
$$sqrt(tanh(z) -tanh(2z))^2 +(tanh(z)*tanh(2z)+1)^2-1-2tanh(z)^2 tanh(2z)^2$$
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]
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]
plotting functions complex
plotting functions complex
asked yesterday
topspintopspin
1363
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 yesterday
topspintopspin
1363
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topspin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 17 hours ago
topspin
asked yesterday
topspintopspin
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topspin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked yesterday
topspintopspin
1363
asked yesterday
topspintopspin
1363
1363
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.
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$begingroup$
The first step might be to find all the zeros of the function under the square root. Perhaps this might help.
$endgroup$
– Hugh
yesterday
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
yesterday
$begingroup$
Ok, Thank you . I have just edited my post .
$endgroup$
– topspin
yesterday
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
yesterday
add a comment |
$begingroup$
The first step might be to find all the zeros of the function under the square root. Perhaps this might help.
$endgroup$
– Hugh
yesterday
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
yesterday
$begingroup$
Ok, Thank you . I have just edited my post .
$endgroup$
– topspin
yesterday
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
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
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
yesterday
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
yesterday
$begingroup$
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
yesterday
$begingroup$
Ok, Thank you . I have just edited my post .
$endgroup$
– topspin
yesterday
$begingroup$
Ok, Thank you . I have just edited my post .
$endgroup$
– topspin
yesterday
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
yesterday
$begingroup$
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
yesterday
add a comment |
2 Answers
2
active
oldest
votes
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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])))
answered yesterday
Carl WollCarl Woll
72.9k396188
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Thank you very much .
$endgroup$
– topspin
yesterday
$begingroup$
Is there any way to visualize those branch points and the branch cuts in Mathematica Instead of ContourPlot ?
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– topspin
yesterday
$begingroup$
Or visualizing the branch points and the branch cuts using ContourPlot .
$endgroup$
– topspin
yesterday
$begingroup$
Just to compare: the Maple's result dropbox.com/s/zh7mq932rlvb1uk/branch_cuts.pdf?dl=0 seems to be different.
$endgroup$
– user64494
yesterday
add a comment |
$begingroup$
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:
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]
answered 15 hours ago
DominicDominic
876
$endgroup$
add a comment |
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$begingroup$
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])))
answered yesterday
Carl WollCarl Woll
72.9k396188
$endgroup$
$begingroup$
Thank you very much .
$endgroup$
– topspin
yesterday
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Is there any way to visualize those branch points and the branch cuts in Mathematica Instead of ContourPlot ?
$endgroup$
– topspin
yesterday
$begingroup$
Or visualizing the branch points and the branch cuts using ContourPlot .
$endgroup$
– topspin
yesterday
$begingroup$
Just to compare: the Maple's result dropbox.com/s/zh7mq932rlvb1uk/branch_cuts.pdf?dl=0 seems to be different.
$endgroup$
– user64494
yesterday
add a comment |
$begingroup$
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])))
answered yesterday
Carl WollCarl Woll
72.9k396188
$endgroup$
$begingroup$
Thank you very much .
$endgroup$
– topspin
yesterday
$begingroup$
Is there any way to visualize those branch points and the branch cuts in Mathematica Instead of ContourPlot ?
$endgroup$
– topspin
yesterday
$begingroup$
Or visualizing the branch points and the branch cuts using ContourPlot .
$endgroup$
– topspin
yesterday
$begingroup$
Just to compare: the Maple's result dropbox.com/s/zh7mq932rlvb1uk/branch_cuts.pdf?dl=0 seems to be different.
$endgroup$
– user64494
yesterday
add a comment |
$begingroup$
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])))
answered yesterday
Carl WollCarl Woll
72.9k396188
$endgroup$
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])))
answered yesterday
Carl WollCarl Woll
72.9k396188
answered yesterday
Carl WollCarl Woll
72.9k396188
answered yesterday
Carl WollCarl Woll
72.9k396188
answered yesterday
Carl WollCarl Woll
72.9k396188
72.9k396188
$begingroup$
Thank you very much .
$endgroup$
– topspin
yesterday
$begingroup$
Is there any way to visualize those branch points and the branch cuts in Mathematica Instead of ContourPlot ?
$endgroup$
– topspin
yesterday
$begingroup$
Or visualizing the branch points and the branch cuts using ContourPlot .
$endgroup$
– topspin
yesterday
$begingroup$
Just to compare: the Maple's result dropbox.com/s/zh7mq932rlvb1uk/branch_cuts.pdf?dl=0 seems to be different.
$endgroup$
– user64494
yesterday
add a comment |
$begingroup$
Thank you very much .
$endgroup$
– topspin
yesterday
$begingroup$
Is there any way to visualize those branch points and the branch cuts in Mathematica Instead of ContourPlot ?
$endgroup$
– topspin
yesterday
$begingroup$
Or visualizing the branch points and the branch cuts using ContourPlot .
$endgroup$
– topspin
yesterday
$begingroup$
Just to compare: the Maple's result dropbox.com/s/zh7mq932rlvb1uk/branch_cuts.pdf?dl=0 seems to be different.
$endgroup$
– user64494
yesterday
$begingroup$
Thank you very much .
$endgroup$
– topspin
yesterday
$begingroup$
Thank you very much .
$endgroup$
– topspin
yesterday
$begingroup$
Is there any way to visualize those branch points and the branch cuts in Mathematica Instead of ContourPlot ?
$endgroup$
– topspin
yesterday
$begingroup$
Is there any way to visualize those branch points and the branch cuts in Mathematica Instead of ContourPlot ?
$endgroup$
– topspin
yesterday
$begingroup$
Or visualizing the branch points and the branch cuts using ContourPlot .
$endgroup$
– topspin
yesterday
$begingroup$
Or visualizing the branch points and the branch cuts using ContourPlot .
$endgroup$
– topspin
yesterday
$begingroup$
Just to compare: the Maple's result dropbox.com/s/zh7mq932rlvb1uk/branch_cuts.pdf?dl=0 seems to be different.
$endgroup$
– user64494
yesterday
$begingroup$
Just to compare: the Maple's result dropbox.com/s/zh7mq932rlvb1uk/branch_cuts.pdf?dl=0 seems to be different.
$endgroup$
– user64494
yesterday
add a comment |
$begingroup$
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:
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]
answered 15 hours ago
DominicDominic
876
$endgroup$
add a comment |
$begingroup$
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:
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]
answered 15 hours ago
DominicDominic
876
$endgroup$
add a comment |
$begingroup$
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:
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]
answered 15 hours ago
DominicDominic
876
$endgroup$
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:
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]
answered 15 hours ago
DominicDominic
876
answered 15 hours ago
DominicDominic
876
answered 15 hours ago
DominicDominic
876
answered 15 hours ago
DominicDominic
876
876
add a comment |
add a comment |
topspin is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
The first step might be to find all the zeros of the function under the square root. Perhaps this might help.
$endgroup$
– Hugh
yesterday
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
yesterday
$begingroup$
Ok, Thank you . I have just edited my post .
$endgroup$
– topspin
yesterday
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
yesterday