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Question on branch cuts and branch points
Updating Wagon's FindAllCrossings2D[] functionpresenting a real number as real instead of imaginarySqrt — how to get negative branch?About how Mathematica understands the branchcuts of the complex logarithm [Part 3]Differential Equation in Complex Plane and Parametric PlotContour Integration along a contour containing two branch pointsHow to declare a function as real in order to get rid of Conjugate in front of the expressions?Branch cuts of sqrtTricky inverse Laplace transformHow do I convert parentheses of a function into brackets?Constuct a function that plus equations
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Is it possible to determine branch cuts and branch points for complicated functions using mathematica
Iam trying to determine the brnach cuts and branch points of this complicated function
$$sqrt{(tanh(z) -tanh(2z))^2 +(tanh(z)*tanh(2z)+1)^2-1-2tanh(z)^2 tanh(2z)^2}$$
I have tried in mathematica but it's not obvious for me where are the branch cuts ?
ContourPlot[Im[Sqrt[(Tanh[x + I*y] - Tanh[2 x + I*2 y])^2 + (Tanh[x + I*y]
Tanh[2 x + I*2 y] + 1)^2-1 - 2 ((Tanh[x + I*2 y])^2)((Tanh[x + I*y])^2) ]],
{x, -10, 10}, {y, -10, 10}, AxesLabel -> Automatic,ContourShading -> Automatic,
ColorFunction -> "Rainbow", Contours -> 20]
ContourPlot[Re[Sqrt[(Tanh[x + I*y] - Tanh[2 x + I*2 y])^2 + (Tanh[x + I*y]Tanh[2 x + I*2 y] + 1)^2 - 1 - 2 ((Tanh[x + I*2 y])^2) ((Tanh[x + I*y])^2) ]],
{x, -10, 10}, {y, -10, 10}, AxesLabel -> Automatic,
ContourShading -> Automatic, ColorFunction -> "Rainbow", Contours -> 20]
functions complex
New contributor
$endgroup$
add a comment |
$begingroup$
Is it possible to determine branch cuts and branch points for complicated functions using mathematica
Iam trying to determine the brnach cuts and branch points of this complicated function
$$sqrt{(tanh(z) -tanh(2z))^2 +(tanh(z)*tanh(2z)+1)^2-1-2tanh(z)^2 tanh(2z)^2}$$
I have tried in mathematica but it's not obvious for me where are the branch cuts ?
ContourPlot[Im[Sqrt[(Tanh[x + I*y] - Tanh[2 x + I*2 y])^2 + (Tanh[x + I*y]
Tanh[2 x + I*2 y] + 1)^2-1 - 2 ((Tanh[x + I*2 y])^2)((Tanh[x + I*y])^2) ]],
{x, -10, 10}, {y, -10, 10}, AxesLabel -> Automatic,ContourShading -> Automatic,
ColorFunction -> "Rainbow", Contours -> 20]
ContourPlot[Re[Sqrt[(Tanh[x + I*y] - Tanh[2 x + I*2 y])^2 + (Tanh[x + I*y]Tanh[2 x + I*2 y] + 1)^2 - 1 - 2 ((Tanh[x + I*2 y])^2) ((Tanh[x + I*y])^2) ]],
{x, -10, 10}, {y, -10, 10}, AxesLabel -> Automatic,
ContourShading -> Automatic, ColorFunction -> "Rainbow", Contours -> 20]
functions complex
New contributor
$endgroup$
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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
12 hours ago
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
12 hours ago
$begingroup$
Ok, Thank you . I have just edited my post .
$endgroup$
– topspin
11 hours ago
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
11 hours ago
add a comment |
$begingroup$
Is it possible to determine branch cuts and branch points for complicated functions using mathematica
Iam trying to determine the brnach cuts and branch points of this complicated function
$$sqrt{(tanh(z) -tanh(2z))^2 +(tanh(z)*tanh(2z)+1)^2-1-2tanh(z)^2 tanh(2z)^2}$$
I have tried in mathematica but it's not obvious for me where are the branch cuts ?
ContourPlot[Im[Sqrt[(Tanh[x + I*y] - Tanh[2 x + I*2 y])^2 + (Tanh[x + I*y]
Tanh[2 x + I*2 y] + 1)^2-1 - 2 ((Tanh[x + I*2 y])^2)((Tanh[x + I*y])^2) ]],
{x, -10, 10}, {y, -10, 10}, AxesLabel -> Automatic,ContourShading -> Automatic,
ColorFunction -> "Rainbow", Contours -> 20]
ContourPlot[Re[Sqrt[(Tanh[x + I*y] - Tanh[2 x + I*2 y])^2 + (Tanh[x + I*y]Tanh[2 x + I*2 y] + 1)^2 - 1 - 2 ((Tanh[x + I*2 y])^2) ((Tanh[x + I*y])^2) ]],
{x, -10, 10}, {y, -10, 10}, AxesLabel -> Automatic,
ContourShading -> Automatic, ColorFunction -> "Rainbow", Contours -> 20]
functions complex
New contributor
$endgroup$
Is it possible to determine branch cuts and branch points for complicated functions using mathematica
Iam trying to determine the brnach cuts and branch points of this complicated function
$$sqrt{(tanh(z) -tanh(2z))^2 +(tanh(z)*tanh(2z)+1)^2-1-2tanh(z)^2 tanh(2z)^2}$$
I have tried in mathematica but it's not obvious for me where are the branch cuts ?
ContourPlot[Im[Sqrt[(Tanh[x + I*y] - Tanh[2 x + I*2 y])^2 + (Tanh[x + I*y]
Tanh[2 x + I*2 y] + 1)^2-1 - 2 ((Tanh[x + I*2 y])^2)((Tanh[x + I*y])^2) ]],
{x, -10, 10}, {y, -10, 10}, AxesLabel -> Automatic,ContourShading -> Automatic,
ColorFunction -> "Rainbow", Contours -> 20]
ContourPlot[Re[Sqrt[(Tanh[x + I*y] - Tanh[2 x + I*2 y])^2 + (Tanh[x + I*y]Tanh[2 x + I*2 y] + 1)^2 - 1 - 2 ((Tanh[x + I*2 y])^2) ((Tanh[x + I*y])^2) ]],
{x, -10, 10}, {y, -10, 10}, AxesLabel -> Automatic,
ContourShading -> Automatic, ColorFunction -> "Rainbow", Contours -> 20]
functions complex
functions complex
New contributor
New contributor
edited 11 hours ago
topspin
New contributor
asked 13 hours ago
topspintopspin
1313
1313
New contributor
New contributor
$begingroup$
The first step might be to find all the zeros of the function under the square root. Perhaps this might help.
$endgroup$
– Hugh
12 hours ago
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
12 hours ago
$begingroup$
Ok, Thank you . I have just edited my post .
$endgroup$
– topspin
11 hours ago
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
11 hours ago
add a comment |
$begingroup$
The first step might be to find all the zeros of the function under the square root. Perhaps this might help.
$endgroup$
– Hugh
12 hours ago
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
12 hours ago
$begingroup$
Ok, Thank you . I have just edited my post .
$endgroup$
– topspin
11 hours ago
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
11 hours ago
$begingroup$
The first step might be to find all the zeros of the function under the square root. Perhaps this might help.
$endgroup$
– Hugh
12 hours ago
$begingroup$
The first step might be to find all the zeros of the function under the square root. Perhaps this might help.
$endgroup$
– Hugh
12 hours ago
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
12 hours ago
$begingroup$
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
12 hours ago
$begingroup$
Ok, Thank you . I have just edited my post .
$endgroup$
– topspin
11 hours ago
$begingroup$
Ok, Thank you . I have just edited my post .
$endgroup$
– topspin
11 hours ago
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
11 hours ago
$begingroup$
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
11 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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])))
$endgroup$
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Thank you very much .
$endgroup$
– topspin
8 hours ago
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Is there any way to visualize those branch points and the branch cuts in Mathematica Instead of ContourPlot ?
$endgroup$
– topspin
8 hours ago
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Or visualizing the branch points and the branch cuts using ContourPlot .
$endgroup$
– topspin
7 hours ago
add a comment |
Your Answer
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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])))
$endgroup$
$begingroup$
Thank you very much .
$endgroup$
– topspin
8 hours ago
$begingroup$
Is there any way to visualize those branch points and the branch cuts in Mathematica Instead of ContourPlot ?
$endgroup$
– topspin
8 hours ago
$begingroup$
Or visualizing the branch points and the branch cuts using ContourPlot .
$endgroup$
– topspin
7 hours ago
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])))
$endgroup$
$begingroup$
Thank you very much .
$endgroup$
– topspin
8 hours ago
$begingroup$
Is there any way to visualize those branch points and the branch cuts in Mathematica Instead of ContourPlot ?
$endgroup$
– topspin
8 hours ago
$begingroup$
Or visualizing the branch points and the branch cuts using ContourPlot .
$endgroup$
– topspin
7 hours ago
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])))
$endgroup$
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 10 hours ago
Carl WollCarl Woll
72.8k396188
72.8k396188
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Thank you very much .
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– topspin
8 hours ago
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Is there any way to visualize those branch points and the branch cuts in Mathematica Instead of ContourPlot ?
$endgroup$
– topspin
8 hours ago
$begingroup$
Or visualizing the branch points and the branch cuts using ContourPlot .
$endgroup$
– topspin
7 hours ago
add a comment |
$begingroup$
Thank you very much .
$endgroup$
– topspin
8 hours ago
$begingroup$
Is there any way to visualize those branch points and the branch cuts in Mathematica Instead of ContourPlot ?
$endgroup$
– topspin
8 hours ago
$begingroup$
Or visualizing the branch points and the branch cuts using ContourPlot .
$endgroup$
– topspin
7 hours ago
$begingroup$
Thank you very much .
$endgroup$
– topspin
8 hours ago
$begingroup$
Thank you very much .
$endgroup$
– topspin
8 hours ago
$begingroup$
Is there any way to visualize those branch points and the branch cuts in Mathematica Instead of ContourPlot ?
$endgroup$
– topspin
8 hours ago
$begingroup$
Is there any way to visualize those branch points and the branch cuts in Mathematica Instead of ContourPlot ?
$endgroup$
– topspin
8 hours ago
$begingroup$
Or visualizing the branch points and the branch cuts using ContourPlot .
$endgroup$
– topspin
7 hours ago
$begingroup$
Or visualizing the branch points and the branch cuts using ContourPlot .
$endgroup$
– topspin
7 hours ago
add a comment |
topspin is a new contributor. Be nice, and check out our Code of Conduct.
topspin is a new contributor. Be nice, and check out our Code of Conduct.
topspin is a new contributor. Be nice, and check out our Code of Conduct.
topspin is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
The first step might be to find all the zeros of the function under the square root. Perhaps this might help.
$endgroup$
– Hugh
12 hours ago
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
12 hours ago
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Ok, Thank you . I have just edited my post .
$endgroup$
– topspin
11 hours ago
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
11 hours ago