# Finding and visualization of branch cuts and branch points

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 $$\sqrt{(\tanh(z) -\tanh(2z))^2 +(\tanh(z)*\tanh(2z)+1)^2-1-2\tanh(z)^2 \tanh(2z)^2}$$

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] 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 ?

• The first step might be to find all the zeros of the function under the square root. Perhaps this might help. – Hugh Apr 5 '19 at 13:02
• 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. – Hugh Apr 5 '19 at 13:04
• Ok, Thank you . I have just edited my post . – Mahmoud Hassan Apr 5 '19 at 14:37
• 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 – Mahmoud Hassan Apr 5 '19 at 14:42
• You have one instance of x + I*2 y in your formula. Do you mean for that to be 2x + I*2 y? – Chip Hurst Apr 8 '19 at 22:06

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 $$f(z) = \tilde\infty$$.

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}
] 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
];

• In accordance with dropbox.com/s/zh7mq932rlvb1uk/branch_cuts.pdf?dl=0 – user64494 Apr 9 '19 at 7:22
• Thank you very much . – Mahmoud Hassan Apr 12 '19 at 8:38
• How to get the output domain coloring as you have got ? – Mahmoud Hassan Apr 12 '19 at 8:41
• 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 ? – Mahmoud Hassan Apr 12 '19 at 8:42
• 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 – Mahmoud Hassan Apr 12 '19 at 8:53

Perhaps you can make use of the internal functions ComplexAnalysisBranchCuts and ComplexAnalysisBranchPoints. 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 = ComplexAnalysisBranchPoints[expr, z]


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

The above can be simplified a bit with:

Simplify[pts, C ∈ Integers]


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

Similarly, the branch cuts can be found with:

ComplexAnalysisBranchCuts[expr, z]


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

• Thank you very much . – Mahmoud Hassan Apr 5 '19 at 17:26
• Is there any way to visualize those branch points and the branch cuts in Mathematica Instead of ContourPlot ? – Mahmoud Hassan Apr 5 '19 at 17:28
• Or visualizing the branch points and the branch cuts using ContourPlot . – Mahmoud Hassan Apr 5 '19 at 17:51
• Just to compare: the Maple's result dropbox.com/s/zh7mq932rlvb1uk/branch_cuts.pdf?dl=0 seems to be different. – user64494 Apr 6 '19 at 6:08

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: 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]

• Thank you very much . – Mahmoud Hassan Apr 14 '19 at 21:10