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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 enter image description here

And I calculated the branchPoints in exact and numeric form

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

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

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

enter image description here

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

enter image description here

How Can I visualize the banchPoints and the branchCuts ?

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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. $\endgroup$ Commented Apr 5, 2019 at 13:02
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    $\begingroup$ 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. $\endgroup$ Commented Apr 5, 2019 at 13:04
  • $\begingroup$ Ok, Thank you . I have just edited my post . $\endgroup$ Commented Apr 5, 2019 at 14:37
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    $\begingroup$ 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 $\endgroup$ Commented Apr 5, 2019 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? $\endgroup$ Commented Apr 8, 2019 at 22:06

3 Answers 3

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In this case, the only branch cuts and branch points will come from the square root. The cuts of $\sqrt{f(z)}$ occurs along the half line $\text{Im}(f(z)) = 0 \,\wedge\, \text{Re}(f(z)) \leq 0$. The branch points lie at $f(z) = 0$ or $f(z) = \tilde\infty$.

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 $\tilde\infty$.

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} ]

enter image description here

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} ]

enter image description here

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 ];
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  • $\begingroup$ In accordance with dropbox.com/s/zh7mq932rlvb1uk/branch_cuts.pdf?dl=0 $\endgroup$ Commented Apr 9, 2019 at 7:22
  • $\begingroup$ Thank you very much . $\endgroup$ Commented Apr 12, 2019 at 8:38
  • $\begingroup$ How to get the output domain coloring as you have got ? $\endgroup$ Commented Apr 12, 2019 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 ? $\endgroup$ Commented Apr 12, 2019 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 $\endgroup$ Commented Apr 12, 2019 at 8:53
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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])))

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  • $\begingroup$ Thank you very much . $\endgroup$ Commented Apr 5, 2019 at 17:26
  • $\begingroup$ Is there any way to visualize those branch points and the branch cuts in Mathematica Instead of ContourPlot ? $\endgroup$ Commented Apr 5, 2019 at 17:28
  • $\begingroup$ Or visualizing the branch points and the branch cuts using ContourPlot . $\endgroup$ Commented Apr 5, 2019 at 17:51
  • $\begingroup$ Just to compare: the Maple's result dropbox.com/s/zh7mq932rlvb1uk/branch_cuts.pdf?dl=0 seems to be different. $\endgroup$ Commented Apr 6, 2019 at 6:08
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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] ]]

$$\frac{\left(e^{2 z}-1\right)^2 \left(4 e^{2 z}+10 e^{4 z}+4 e^{6 z}+e^{8 z}+1\right)}{\left(e^{2 z}+1\right)^2 \left(e^{4 z}+1\right)^2}$$

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

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

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

p1 = Show[ Graphics[{Red, Point @@ {{Re[#], Im[#]} & /@ (N[Log[#]] & /@ mySol)}}], Axes -> True, PlotRange -> 5]; p2 = Show[ Graphics[{Blue, Point @@ {{Re[#], Im[#]} & /@ (N[(Log[#] + 2 \[Pi] I)] & /@ mySol)}}], Axes -> True, PlotRange -> 15]; p3 = Show[ Graphics[{Green // Darker, Point @@ {{Re[#], Im[#]} & /@ (N[(Log[#] - 2 \[Pi] I)] & /@ mySol)}}], Axes -> True, PlotRange -> 15]; Show[{p1, p2, p3}, PlotRange -> 15]
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  • $\begingroup$ Thank you very much . $\endgroup$ Commented Apr 14, 2019 at 21:10

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