Question on branch cuts and branch points
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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
asked 17 hours ago
topspintopspin
1313
New contributor
topspin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
$begingroup$
Is it possible to determine branch cuts and branch points for complicated functions using mathematica
Iam trying to determine the brnach cuts and branch points of this complicated function
$$sqrt{(tanh(z) -tanh(2z))^2 +(tanh(z)*tanh(2z)+1)^2-1-2tanh(z)^2 tanh(2z)^2}$$
I have tried in mathematica but it's not obvious for me where are the branch cuts ?
ContourPlot[Im[Sqrt[(Tanh[x + I*y] - Tanh[2 x + I*2 y])^2 + (Tanh[x + I*y]
Tanh[2 x + I*2 y] + 1)^2-1 - 2 ((Tanh[x + I*2 y])^2)((Tanh[x + I*y])^2) ]],
{x, -10, 10}, {y, -10, 10}, AxesLabel -> Automatic,ContourShading -> Automatic,
ColorFunction -> "Rainbow", Contours -> 20]
ContourPlot[Re[Sqrt[(Tanh[x + I*y] - Tanh[2 x + I*2 y])^2 + (Tanh[x + I*y]Tanh[2 x + I*2 y] + 1)^2 - 1 - 2 ((Tanh[x + I*2 y])^2) ((Tanh[x + I*y])^2) ]],
{x, -10, 10}, {y, -10, 10}, AxesLabel -> Automatic,
ContourShading -> Automatic, ColorFunction -> "Rainbow", Contours -> 20]
functions complex
asked 17 hours ago
topspintopspin
1313
New contributor
topspin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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The first step might be to find all the zeros of the function under the square root. Perhaps this might help.
$endgroup$
– Hugh
16 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
16 hours ago
$begingroup$
Ok, Thank you . I have just edited my post .
$endgroup$
– topspin
15 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
14 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
asked 17 hours ago
topspintopspin
1313
New contributor
topspin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Is it possible to determine branch cuts and branch points for complicated functions using mathematica
Iam trying to determine the brnach cuts and branch points of this complicated function
$$sqrt{(tanh(z) -tanh(2z))^2 +(tanh(z)*tanh(2z)+1)^2-1-2tanh(z)^2 tanh(2z)^2}$$
I have tried in mathematica but it's not obvious for me where are the branch cuts ?
ContourPlot[Im[Sqrt[(Tanh[x + I*y] - Tanh[2 x + I*2 y])^2 + (Tanh[x + I*y]
Tanh[2 x + I*2 y] + 1)^2-1 - 2 ((Tanh[x + I*2 y])^2)((Tanh[x + I*y])^2) ]],
{x, -10, 10}, {y, -10, 10}, AxesLabel -> Automatic,ContourShading -> Automatic,
ColorFunction -> "Rainbow", Contours -> 20]
ContourPlot[Re[Sqrt[(Tanh[x + I*y] - Tanh[2 x + I*2 y])^2 + (Tanh[x + I*y]Tanh[2 x + I*2 y] + 1)^2 - 1 - 2 ((Tanh[x + I*2 y])^2) ((Tanh[x + I*y])^2) ]],
{x, -10, 10}, {y, -10, 10}, AxesLabel -> Automatic,
ContourShading -> Automatic, ColorFunction -> "Rainbow", Contours -> 20]
functions complex
functions complex
asked 17 hours ago
topspintopspin
1313
New contributor
topspin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 17 hours ago
topspintopspin
1313
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topspin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 15 hours ago
topspin
asked 17 hours ago
topspintopspin
1313
New contributor
topspin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 17 hours ago
topspintopspin
1313
asked 17 hours ago
topspintopspin
1313
1313
New contributor
topspin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
topspin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
topspin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
The first step might be to find all the zeros of the function under the square root. Perhaps this might help.
$endgroup$
– Hugh
16 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
16 hours ago
$begingroup$
Ok, Thank you . I have just edited my post .
$endgroup$
– topspin
15 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
14 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
16 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
16 hours ago
$begingroup$
Ok, Thank you . I have just edited my post .
$endgroup$
– topspin
15 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
14 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
16 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
16 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
16 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
16 hours ago
$begingroup$
Ok, Thank you . I have just edited my post .
$endgroup$
– topspin
15 hours ago
$begingroup$
Ok, Thank you . I have just edited my post .
$endgroup$
– topspin
15 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
14 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
14 hours ago
add a comment |
1 Answer
1
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 14 hours ago
Carl WollCarl Woll
72.8k396188
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Thank you very much .
$endgroup$
– topspin
12 hours ago
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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
12 hours ago
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Or visualizing the branch points and the branch cuts using ContourPlot .
$endgroup$
– topspin
11 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])))
answered 14 hours ago
Carl WollCarl Woll
72.8k396188
$endgroup$
$begingroup$
Thank you very much .
$endgroup$
– topspin
12 hours ago
$begingroup$
Is there any way to visualize those branch points and the branch cuts in Mathematica Instead of ContourPlot ?
$endgroup$
– topspin
12 hours ago
$begingroup$
Or visualizing the branch points and the branch cuts using ContourPlot .
$endgroup$
– topspin
11 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])))
answered 14 hours ago
Carl WollCarl Woll
72.8k396188
$endgroup$
$begingroup$
Thank you very much .
$endgroup$
– topspin
12 hours ago
$begingroup$
Is there any way to visualize those branch points and the branch cuts in Mathematica Instead of ContourPlot ?
$endgroup$
– topspin
12 hours ago
$begingroup$
Or visualizing the branch points and the branch cuts using ContourPlot .
$endgroup$
– topspin
11 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])))
answered 14 hours ago
Carl WollCarl Woll
72.8k396188
$endgroup$
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 14 hours ago
Carl WollCarl Woll
72.8k396188
answered 14 hours ago
Carl WollCarl Woll
72.8k396188
answered 14 hours ago
Carl WollCarl Woll
72.8k396188
answered 14 hours ago
Carl WollCarl Woll
72.8k396188
72.8k396188
$begingroup$
Thank you very much .
$endgroup$
– topspin
12 hours ago
$begingroup$
Is there any way to visualize those branch points and the branch cuts in Mathematica Instead of ContourPlot ?
$endgroup$
– topspin
12 hours ago
$begingroup$
Or visualizing the branch points and the branch cuts using ContourPlot .
$endgroup$
– topspin
11 hours ago
add a comment |
$begingroup$
Thank you very much .
$endgroup$
– topspin
12 hours ago
$begingroup$
Is there any way to visualize those branch points and the branch cuts in Mathematica Instead of ContourPlot ?
$endgroup$
– topspin
12 hours ago
$begingroup$
Or visualizing the branch points and the branch cuts using ContourPlot .
$endgroup$
– topspin
11 hours ago
$begingroup$
Thank you very much .
$endgroup$
– topspin
12 hours ago
$begingroup$
Thank you very much .
$endgroup$
– topspin
12 hours ago
$begingroup$
Is there any way to visualize those branch points and the branch cuts in Mathematica Instead of ContourPlot ?
$endgroup$
– topspin
12 hours ago
$begingroup$
Is there any way to visualize those branch points and the branch cuts in Mathematica Instead of ContourPlot ?
$endgroup$
– topspin
12 hours ago
$begingroup$
Or visualizing the branch points and the branch cuts using ContourPlot .
$endgroup$
– topspin
11 hours ago
$begingroup$
Or visualizing the branch points and the branch cuts using ContourPlot .
$endgroup$
– topspin
11 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
16 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
16 hours ago
$begingroup$
Ok, Thank you . I have just edited my post .
$endgroup$
– topspin
15 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
14 hours ago