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Finding and visualizing of branch cuts and branch points
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Updating 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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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 
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
$$sqrt(tanh(z) -tanh(2z))^2 +(tanh(z)*tanh(2z)+1)^2-1-2tanh(z)^2 tanh(2z)^2$$
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]
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]
How Can I visualize the banchPoints and the branchCuts ?
plotting functions complex visualization
asked Apr 5 at 11:56
topspintopspin
1117
$endgroup$
|
show 1 more 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
We have 8 branch cuts
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$$sqrt(tanh(z) -tanh(2z))^2 +(tanh(z)*tanh(2z)+1)^2-1-2tanh(z)^2 tanh(2z)^2$$
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]
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]
How Can I visualize the banchPoints and the branchCuts ?
plotting functions complex visualization
asked Apr 5 at 11:56
topspintopspin
1117
$endgroup$
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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
Apr 5 at 13:02
1
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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
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Ok, Thank you . I have just edited my post .
$endgroup$
– topspin
Apr 5 at 14:37
1
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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
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– topspin
Apr 5 at 14:42
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You have one instance ofx + I*2 yin your formula. Do you mean for that to be2x + I*2 y?
$endgroup$
– Chip Hurst
Apr 8 at 22:06
|
show 1 more 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
We have 8 branch cuts
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$$sqrt(tanh(z) -tanh(2z))^2 +(tanh(z)*tanh(2z)+1)^2-1-2tanh(z)^2 tanh(2z)^2$$
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]
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]
How Can I visualize the banchPoints and the branchCuts ?
plotting functions complex visualization
asked Apr 5 at 11:56
topspintopspin
1117
$endgroup$
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
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$$sqrt(tanh(z) -tanh(2z))^2 +(tanh(z)*tanh(2z)+1)^2-1-2tanh(z)^2 tanh(2z)^2$$
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]
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]
How Can I visualize the banchPoints and the branchCuts ?
plotting functions complex visualization
plotting functions complex visualization
asked Apr 5 at 11:56
topspintopspin
1117
asked Apr 5 at 11:56
topspintopspin
1117
edited Apr 20 at 8:52
topspin
asked Apr 5 at 11:56
topspintopspin
1117
asked Apr 5 at 11:56
topspintopspin
1117
asked Apr 5 at 11:56
topspintopspin
1117
1117
$begingroup$
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 ofx + I*2 yin your formula. Do you mean for that to be2x + I*2 y?
$endgroup$
– Chip Hurst
Apr 8 at 22:06
|
show 1 more 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
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 ofx + I*2 yin your formula. Do you mean for that to be2x + I*2 y?
$endgroup$
– Chip Hurst
Apr 8 at 22:06
$begingroup$
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
Apr 8 at 22:06
$begingroup$
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
Apr 8 at 22:06
|
show 1 more comment
3 Answers
3
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 Apr 5 at 15:06
Carl WollCarl Woll
75.5k3100197
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Thank you very much .
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– topspin
Apr 5 at 17:26
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Is there any way to visualize those branch points and the branch cuts in Mathematica Instead of ContourPlot ?
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– topspin
Apr 5 at 17:28
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Or visualizing the branch points and the branch cuts using ContourPlot .
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– topspin
Apr 5 at 17:51
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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
add a comment |
$begingroup$
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 = Compile[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
]
As of version 12 we can use ComplexPlot to visualize the domain coloring:
exprz = (Tanh[z] - Tanh[2 z])^2 + (Tanh[z] Tanh[2 z] + 1)^2 - 1 - 2 ((Tanh[2 z])^2) ((Tanh[z])^2);
exprxy = exprz /. z -> x + I y;
branchCutRegion[x_, y_, __] = Re[exprxy] <= 0;
bpvals = Union[x, y /. Solve[(expr == 0 || 1/Together[TrigToExp[expr]] == 0) && -10 < x < 10 && -10 < y < 10, x, y]];
domaincoloring = ComplexPlot[exprz, z, -10 - 10 I, 10 + 10 I,
ColorFunction -> "CyclicLogAbsArg", ImageSize -> 800];
cutplot = ContourPlot[Im[exprxy] == 0, x, -10, 10, y, -10, 10,
RegionFunction -> branchCutRegion, PlotPoints -> 100, ContourStyle -> Black];
Show[
domaincoloring,
cutplot,
Epilog -> EdgeForm[Black], GrayLevel[.8], Disk[#, Scaled[.0045]] & /@ bpvals
]
To achieve the same image from my original answer, you can use
domaincoloring = ComplexPlot[exprz, z, -10 - 10 I, 10 + 10 I,
ColorFunction -> Hue[Divide[Mod[#8, 2π], 2π],
Power[1 + 0.3*Log[#7 + 1], -1],
Power[1 + 0.5*Log[#7 + 1], -1]] &, None,
ColorFunctionScaling -> False,
Exclusions -> None,
ImageSize -> 800
];
answered Apr 8 at 22:23
Chip HurstChip Hurst
23.8k15995
$endgroup$
$begingroup$
In accordance with dropbox.com/s/zh7mq932rlvb1uk/branch_cuts.pdf?dl=0
$endgroup$
– user64494
Apr 9 at 7:22
$begingroup$
Thank you very much .
$endgroup$
– topspin
Apr 12 at 8:38
$begingroup$
How to get the output domain coloring as you have got ?
$endgroup$
– topspin
Apr 12 at 8:41
$begingroup$
I have a problem running your second code for domain cloring ,I didn't get the output as you got , Would you please check your second code again ?
$endgroup$
– topspin
Apr 12 at 8:42
$begingroup$
In your first code , the lines in blue are the branchcuts right ? I am tring to visulalize the boolean expression and conditional expression for the 8 branchcuts acording to my results
$endgroup$
– topspin
Apr 12 at 8:53
|
show 5 more comments
$begingroup$
First start with the branch points: these are the values of z where the root is not multiple-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 Apr 6 at 14:43
DominicDominic
1477
$endgroup$
$begingroup$
Thank you very much .
$endgroup$
– topspin
Apr 14 at 21:10
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 Apr 5 at 15:06
Carl WollCarl Woll
75.5k3100197
$endgroup$
$begingroup$
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
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 Apr 5 at 15:06
Carl WollCarl Woll
75.5k3100197
$endgroup$
$begingroup$
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
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 Apr 5 at 15:06
Carl WollCarl Woll
75.5k3100197
$endgroup$
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 Apr 5 at 15:06
Carl WollCarl Woll
75.5k3100197
answered Apr 5 at 15:06
Carl WollCarl Woll
75.5k3100197
answered Apr 5 at 15:06
Carl WollCarl Woll
75.5k3100197
answered Apr 5 at 15:06
Carl WollCarl Woll
75.5k3100197
75.5k3100197
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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 = Compile[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
]
As of version 12 we can use ComplexPlot to visualize the domain coloring:
exprz = (Tanh[z] - Tanh[2 z])^2 + (Tanh[z] Tanh[2 z] + 1)^2 - 1 - 2 ((Tanh[2 z])^2) ((Tanh[z])^2);
exprxy = exprz /. z -> x + I y;
branchCutRegion[x_, y_, __] = Re[exprxy] <= 0;
bpvals = Union[x, y /. Solve[(expr == 0 || 1/Together[TrigToExp[expr]] == 0) && -10 < x < 10 && -10 < y < 10, x, y]];
domaincoloring = ComplexPlot[exprz, z, -10 - 10 I, 10 + 10 I,
ColorFunction -> "CyclicLogAbsArg", ImageSize -> 800];
cutplot = ContourPlot[Im[exprxy] == 0, x, -10, 10, y, -10, 10,
RegionFunction -> branchCutRegion, PlotPoints -> 100, ContourStyle -> Black];
Show[
domaincoloring,
cutplot,
Epilog -> EdgeForm[Black], GrayLevel[.8], Disk[#, Scaled[.0045]] & /@ bpvals
]
To achieve the same image from my original answer, you can use
domaincoloring = ComplexPlot[exprz, z, -10 - 10 I, 10 + 10 I,
ColorFunction -> Hue[Divide[Mod[#8, 2π], 2π],
Power[1 + 0.3*Log[#7 + 1], -1],
Power[1 + 0.5*Log[#7 + 1], -1]] &, None,
ColorFunctionScaling -> False,
Exclusions -> None,
ImageSize -> 800
];
answered Apr 8 at 22:23
Chip HurstChip Hurst
23.8k15995
$endgroup$
$begingroup$
In accordance with dropbox.com/s/zh7mq932rlvb1uk/branch_cuts.pdf?dl=0
$endgroup$
– user64494
Apr 9 at 7:22
$begingroup$
Thank you very much .
$endgroup$
– topspin
Apr 12 at 8:38
$begingroup$
How to get the output domain coloring as you have got ?
$endgroup$
– topspin
Apr 12 at 8:41
$begingroup$
I have a problem running your second code for domain cloring ,I didn't get the output as you got , Would you please check your second code again ?
$endgroup$
– topspin
Apr 12 at 8:42
$begingroup$
In your first code , the lines in blue are the branchcuts right ? I am tring to visulalize the boolean expression and conditional expression for the 8 branchcuts acording to my results
$endgroup$
– topspin
Apr 12 at 8:53
|
show 5 more comments
$begingroup$
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 = Compile[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
]
As of version 12 we can use ComplexPlot to visualize the domain coloring:
exprz = (Tanh[z] - Tanh[2 z])^2 + (Tanh[z] Tanh[2 z] + 1)^2 - 1 - 2 ((Tanh[2 z])^2) ((Tanh[z])^2);
exprxy = exprz /. z -> x + I y;
branchCutRegion[x_, y_, __] = Re[exprxy] <= 0;
bpvals = Union[x, y /. Solve[(expr == 0 || 1/Together[TrigToExp[expr]] == 0) && -10 < x < 10 && -10 < y < 10, x, y]];
domaincoloring = ComplexPlot[exprz, z, -10 - 10 I, 10 + 10 I,
ColorFunction -> "CyclicLogAbsArg", ImageSize -> 800];
cutplot = ContourPlot[Im[exprxy] == 0, x, -10, 10, y, -10, 10,
RegionFunction -> branchCutRegion, PlotPoints -> 100, ContourStyle -> Black];
Show[
domaincoloring,
cutplot,
Epilog -> EdgeForm[Black], GrayLevel[.8], Disk[#, Scaled[.0045]] & /@ bpvals
]
To achieve the same image from my original answer, you can use
domaincoloring = ComplexPlot[exprz, z, -10 - 10 I, 10 + 10 I,
ColorFunction -> Hue[Divide[Mod[#8, 2π], 2π],
Power[1 + 0.3*Log[#7 + 1], -1],
Power[1 + 0.5*Log[#7 + 1], -1]] &, None,
ColorFunctionScaling -> False,
Exclusions -> None,
ImageSize -> 800
];
answered Apr 8 at 22:23
Chip HurstChip Hurst
23.8k15995
$endgroup$
$begingroup$
In accordance with dropbox.com/s/zh7mq932rlvb1uk/branch_cuts.pdf?dl=0
$endgroup$
– user64494
Apr 9 at 7:22
$begingroup$
Thank you very much .
$endgroup$
– topspin
Apr 12 at 8:38
$begingroup$
How to get the output domain coloring as you have got ?
$endgroup$
– topspin
Apr 12 at 8:41
$begingroup$
I have a problem running your second code for domain cloring ,I didn't get the output as you got , Would you please check your second code again ?
$endgroup$
– topspin
Apr 12 at 8:42
$begingroup$
In your first code , the lines in blue are the branchcuts right ? I am tring to visulalize the boolean expression and conditional expression for the 8 branchcuts acording to my results
$endgroup$
– topspin
Apr 12 at 8:53
|
show 5 more comments
$begingroup$
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 = Compile[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
]
As of version 12 we can use ComplexPlot to visualize the domain coloring:
exprz = (Tanh[z] - Tanh[2 z])^2 + (Tanh[z] Tanh[2 z] + 1)^2 - 1 - 2 ((Tanh[2 z])^2) ((Tanh[z])^2);
exprxy = exprz /. z -> x + I y;
branchCutRegion[x_, y_, __] = Re[exprxy] <= 0;
bpvals = Union[x, y /. Solve[(expr == 0 || 1/Together[TrigToExp[expr]] == 0) && -10 < x < 10 && -10 < y < 10, x, y]];
domaincoloring = ComplexPlot[exprz, z, -10 - 10 I, 10 + 10 I,
ColorFunction -> "CyclicLogAbsArg", ImageSize -> 800];
cutplot = ContourPlot[Im[exprxy] == 0, x, -10, 10, y, -10, 10,
RegionFunction -> branchCutRegion, PlotPoints -> 100, ContourStyle -> Black];
Show[
domaincoloring,
cutplot,
Epilog -> EdgeForm[Black], GrayLevel[.8], Disk[#, Scaled[.0045]] & /@ bpvals
]
To achieve the same image from my original answer, you can use
domaincoloring = ComplexPlot[exprz, z, -10 - 10 I, 10 + 10 I,
ColorFunction -> Hue[Divide[Mod[#8, 2π], 2π],
Power[1 + 0.3*Log[#7 + 1], -1],
Power[1 + 0.5*Log[#7 + 1], -1]] &, None,
ColorFunctionScaling -> False,
Exclusions -> None,
ImageSize -> 800
];
answered Apr 8 at 22:23
Chip HurstChip Hurst
23.8k15995
$endgroup$
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 = Compile[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
]
As of version 12 we can use ComplexPlot to visualize the domain coloring:
exprz = (Tanh[z] - Tanh[2 z])^2 + (Tanh[z] Tanh[2 z] + 1)^2 - 1 - 2 ((Tanh[2 z])^2) ((Tanh[z])^2);
exprxy = exprz /. z -> x + I y;
branchCutRegion[x_, y_, __] = Re[exprxy] <= 0;
bpvals = Union[x, y /. Solve[(expr == 0 || 1/Together[TrigToExp[expr]] == 0) && -10 < x < 10 && -10 < y < 10, x, y]];
domaincoloring = ComplexPlot[exprz, z, -10 - 10 I, 10 + 10 I,
ColorFunction -> "CyclicLogAbsArg", ImageSize -> 800];
cutplot = ContourPlot[Im[exprxy] == 0, x, -10, 10, y, -10, 10,
RegionFunction -> branchCutRegion, PlotPoints -> 100, ContourStyle -> Black];
Show[
domaincoloring,
cutplot,
Epilog -> EdgeForm[Black], GrayLevel[.8], Disk[#, Scaled[.0045]] & /@ bpvals
]
To achieve the same image from my original answer, you can use
domaincoloring = ComplexPlot[exprz, z, -10 - 10 I, 10 + 10 I,
ColorFunction -> Hue[Divide[Mod[#8, 2π], 2π],
Power[1 + 0.3*Log[#7 + 1], -1],
Power[1 + 0.5*Log[#7 + 1], -1]] &, None,
ColorFunctionScaling -> False,
Exclusions -> None,
ImageSize -> 800
];
answered Apr 8 at 22:23
Chip HurstChip Hurst
23.8k15995
edited Apr 15 at 13:46
answered Apr 8 at 22:23
Chip HurstChip Hurst
23.8k15995
answered Apr 8 at 22:23
Chip HurstChip Hurst
23.8k15995
answered Apr 8 at 22:23
Chip HurstChip Hurst
23.8k15995
23.8k15995
$begingroup$
In accordance with dropbox.com/s/zh7mq932rlvb1uk/branch_cuts.pdf?dl=0
$endgroup$
– user64494
Apr 9 at 7:22
$begingroup$
Thank you very much .
$endgroup$
– topspin
Apr 12 at 8:38
$begingroup$
How to get the output domain coloring as you have got ?
$endgroup$
– topspin
Apr 12 at 8:41
$begingroup$
I have a problem running your second code for domain cloring ,I didn't get the output as you got , Would you please check your second code again ?
$endgroup$
– topspin
Apr 12 at 8:42
$begingroup$
In your first code , the lines in blue are the branchcuts right ? I am tring to visulalize the boolean expression and conditional expression for the 8 branchcuts acording to my results
$endgroup$
– topspin
Apr 12 at 8:53
|
show 5 more comments
$begingroup$
In accordance with dropbox.com/s/zh7mq932rlvb1uk/branch_cuts.pdf?dl=0
$endgroup$
– user64494
Apr 9 at 7:22
$begingroup$
Thank you very much .
$endgroup$
– topspin
Apr 12 at 8:38
$begingroup$
How to get the output domain coloring as you have got ?
$endgroup$
– topspin
Apr 12 at 8:41
$begingroup$
I have a problem running your second code for domain cloring ,I didn't get the output as you got , Would you please check your second code again ?
$endgroup$
– topspin
Apr 12 at 8:42
$begingroup$
In your first code , the lines in blue are the branchcuts right ? I am tring to visulalize the boolean expression and conditional expression for the 8 branchcuts acording to my results
$endgroup$
– topspin
Apr 12 at 8:53
$begingroup$
In accordance with dropbox.com/s/zh7mq932rlvb1uk/branch_cuts.pdf?dl=0
$endgroup$
– user64494
Apr 9 at 7:22
$begingroup$
In accordance with dropbox.com/s/zh7mq932rlvb1uk/branch_cuts.pdf?dl=0
$endgroup$
– user64494
Apr 9 at 7:22
$begingroup$
Thank you very much .
$endgroup$
– topspin
Apr 12 at 8:38
$begingroup$
Thank you very much .
$endgroup$
– topspin
Apr 12 at 8:38
$begingroup$
How to get the output domain coloring as you have got ?
$endgroup$
– topspin
Apr 12 at 8:41
$begingroup$
How to get the output domain coloring as you have got ?
$endgroup$
– topspin
Apr 12 at 8:41
$begingroup$
I have a problem running your second code for domain cloring ,I didn't get the output as you got , Would you please check your second code again ?
$endgroup$
– topspin
Apr 12 at 8:42
$begingroup$
I have a problem running your second code for domain cloring ,I didn't get the output as you got , Would you please check your second code again ?
$endgroup$
– topspin
Apr 12 at 8:42
$begingroup$
In your first code , the lines in blue are the branchcuts right ? I am tring to visulalize the boolean expression and conditional expression for the 8 branchcuts acording to my results
$endgroup$
– topspin
Apr 12 at 8:53
$begingroup$
In your first code , the lines in blue are the branchcuts right ? I am tring to visulalize the boolean expression and conditional expression for the 8 branchcuts acording to my results
$endgroup$
– topspin
Apr 12 at 8:53
|
show 5 more comments
$begingroup$
First start with the branch points: these are the values of z where the root is not multiple-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 Apr 6 at 14:43
DominicDominic
1477
$endgroup$
$begingroup$
Thank you very much .
$endgroup$
– topspin
Apr 14 at 21:10
add a comment |
$begingroup$
First start with the branch points: these are the values of z where the root is not multiple-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 Apr 6 at 14:43
DominicDominic
1477
$endgroup$
$begingroup$
Thank you very much .
$endgroup$
– topspin
Apr 14 at 21:10
add a comment |
$begingroup$
First start with the branch points: these are the values of z where the root is not multiple-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 Apr 6 at 14:43
DominicDominic
1477
$endgroup$
First start with the branch points: these are the values of z where the root is not multiple-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 Apr 6 at 14:43
DominicDominic
1477
edited Apr 10 at 23:15
answered Apr 6 at 14:43
DominicDominic
1477
answered Apr 6 at 14:43
DominicDominic
1477
answered Apr 6 at 14:43
DominicDominic
1477
1477
$begingroup$
Thank you very much .
$endgroup$
– topspin
Apr 14 at 21:10
add a comment |
$begingroup$
Thank you very much .
$endgroup$
– topspin
Apr 14 at 21:10
$begingroup$
Thank you very much .
$endgroup$
– topspin
Apr 14 at 21:10
$begingroup$
Thank you very much .
$endgroup$
– topspin
Apr 14 at 21:10
add a comment |
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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
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 yin your formula. Do you mean for that to be2x + I*2 y?$endgroup$
– Chip Hurst
Apr 8 at 22:06