Visualizing Riemann surface (two branches) of logarithmTo plot branch cut of logarithmDomain ColoringRiemann surface for cubic rootHow do I plot a line delineating a subset of values on a 3D surface plot?How to calculate residues using different branches of the logarithmPlotting Riemann Surface of $w(z)=sqrt1-z^2$Visualizing the Riemann zeta functionDrawing a hyperelliptic Riemann surfaceOn visualizing Riemann surfaces of algebraic functionsVisualizing the complex logarithmVisualising a Riemann surface
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Visualizing Riemann surface (two branches) of logarithm
To plot branch cut of logarithmDomain ColoringRiemann surface for cubic rootHow do I plot a line delineating a subset of values on a 3D surface plot?How to calculate residues using different branches of the logarithmPlotting Riemann Surface of $w(z)=sqrt1-z^2$Visualizing the Riemann zeta functionDrawing a hyperelliptic Riemann surfaceOn visualizing Riemann surfaces of algebraic functionsVisualizing the complex logarithmVisualising a Riemann surface
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I'm trying to plot two branches of the complex multi-valued function: (In a previous post - linked above-, Mathematica found the branch cut of the following function between -1 and 0)
$1-zln[(1+z)/z]$.
So, to plot the real parts of the two sheets, we have:
Plot3D[Re[1 - (x + I y) (Log[1 + x + I y] - Log[x + I y])],
Re[1 - (x + I y) (Log[1 + x + I y] - Log[x + I y] -
I 2 Pi)], x, -2, 2, y, -3, 3, BoxRatios -> 1, 1, 1.5,
PlotRange -> All, PlotPoints -> 50, Mesh -> 30,
MeshFunctions -> Im[Log[#1 + I #2]] &, Re[Log[#1 + I #2]] &,
ImageSize -> Large, ColorFunction -> mycolor]
which as expected the two sheets are connected through the branch cut between -1 and 0 on the x-axis:
However, the pictures shows another branch cut between -2 and -1. How come?
plotting calculus-and-analysis complex
asked 12 hours ago
Call me potato.Call me potato.
354 bronze badges
New contributor
Call me potato. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
$begingroup$
I'm trying to plot two branches of the complex multi-valued function: (In a previous post - linked above-, Mathematica found the branch cut of the following function between -1 and 0)
$1-zln[(1+z)/z]$.
So, to plot the real parts of the two sheets, we have:
Plot3D[Re[1 - (x + I y) (Log[1 + x + I y] - Log[x + I y])],
Re[1 - (x + I y) (Log[1 + x + I y] - Log[x + I y] -
I 2 Pi)], x, -2, 2, y, -3, 3, BoxRatios -> 1, 1, 1.5,
PlotRange -> All, PlotPoints -> 50, Mesh -> 30,
MeshFunctions -> Im[Log[#1 + I #2]] &, Re[Log[#1 + I #2]] &,
ImageSize -> Large, ColorFunction -> mycolor]
which as expected the two sheets are connected through the branch cut between -1 and 0 on the x-axis:
However, the pictures shows another branch cut between -2 and -1. How come?
plotting calculus-and-analysis complex
asked 12 hours ago
Call me potato.Call me potato.
354 bronze badges
New contributor
Call me potato. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
@MariuszIwaniuk In that question, I needed to know if the Mathematica understands the branch cut of the logarithm. Here, I'm interested in its Riemann surface.
$endgroup$
– Call me potato.
11 hours ago
add a comment |
$begingroup$
I'm trying to plot two branches of the complex multi-valued function: (In a previous post - linked above-, Mathematica found the branch cut of the following function between -1 and 0)
$1-zln[(1+z)/z]$.
So, to plot the real parts of the two sheets, we have:
Plot3D[Re[1 - (x + I y) (Log[1 + x + I y] - Log[x + I y])],
Re[1 - (x + I y) (Log[1 + x + I y] - Log[x + I y] -
I 2 Pi)], x, -2, 2, y, -3, 3, BoxRatios -> 1, 1, 1.5,
PlotRange -> All, PlotPoints -> 50, Mesh -> 30,
MeshFunctions -> Im[Log[#1 + I #2]] &, Re[Log[#1 + I #2]] &,
ImageSize -> Large, ColorFunction -> mycolor]
which as expected the two sheets are connected through the branch cut between -1 and 0 on the x-axis:
However, the pictures shows another branch cut between -2 and -1. How come?
plotting calculus-and-analysis complex
asked 12 hours ago
Call me potato.Call me potato.
354 bronze badges
New contributor
Call me potato. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I'm trying to plot two branches of the complex multi-valued function: (In a previous post - linked above-, Mathematica found the branch cut of the following function between -1 and 0)
$1-zln[(1+z)/z]$.
So, to plot the real parts of the two sheets, we have:
Plot3D[Re[1 - (x + I y) (Log[1 + x + I y] - Log[x + I y])],
Re[1 - (x + I y) (Log[1 + x + I y] - Log[x + I y] -
I 2 Pi)], x, -2, 2, y, -3, 3, BoxRatios -> 1, 1, 1.5,
PlotRange -> All, PlotPoints -> 50, Mesh -> 30,
MeshFunctions -> Im[Log[#1 + I #2]] &, Re[Log[#1 + I #2]] &,
ImageSize -> Large, ColorFunction -> mycolor]
which as expected the two sheets are connected through the branch cut between -1 and 0 on the x-axis:
However, the pictures shows another branch cut between -2 and -1. How come?
plotting calculus-and-analysis complex
plotting calculus-and-analysis complex
asked 12 hours ago
Call me potato.Call me potato.
354 bronze badges
New contributor
Call me potato. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 12 hours ago
Call me potato.Call me potato.
354 bronze badges
New contributor
Call me potato. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 9 hours ago
Call me potato.
asked 12 hours ago
Call me potato.Call me potato.
354 bronze badges
New contributor
Call me potato. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 12 hours ago
Call me potato.Call me potato.
354 bronze badges
asked 12 hours ago
Call me potato.Call me potato.
354 bronze badges
354 bronze badges
New contributor
Call me potato. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Call me potato. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
@MariuszIwaniuk In that question, I needed to know if the Mathematica understands the branch cut of the logarithm. Here, I'm interested in its Riemann surface.
$endgroup$
– Call me potato.
11 hours ago
add a comment |
$begingroup$
@MariuszIwaniuk In that question, I needed to know if the Mathematica understands the branch cut of the logarithm. Here, I'm interested in its Riemann surface.
$endgroup$
– Call me potato.
11 hours ago
$begingroup$
@MariuszIwaniuk In that question, I needed to know if the Mathematica understands the branch cut of the logarithm. Here, I'm interested in its Riemann surface.
$endgroup$
– Call me potato.
11 hours ago
$begingroup$
@MariuszIwaniuk In that question, I needed to know if the Mathematica understands the branch cut of the logarithm. Here, I'm interested in its Riemann surface.
$endgroup$
– Call me potato.
11 hours ago
add a comment |
1 Answer
1
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votes
$begingroup$
There's no branch-cut from -1 to -2. Whenever plotting multi-valued functions, I'd recommend you use ParametricPlot3D so that you have better control over the Arg function and can run it through -Pi to Pi. When using Plot3D, you'll sometimes get rough branch cuts because it's in rectilinear form. This is what I'd use for your function:
f[z_, n_] := 1 - z (Log[Abs[(1 + z) z]] + I (2 n Pi + Arg[(1 + z)/z]))
p1 = ParametricPlot3D[Re[z], Im[z], Re[f[z, 0]] /.
z -> r Exp[I t], r, 0, 2, t, -Pi, Pi, PlotStyle -> Red];
p2 = ParametricPlot3D[Re[z], Im[z], Re[f[z, 1]] /.
z -> r Exp[I t], r, 0, 2, t, -Pi, Pi, PlotStyle -> Blue];
Show[p1, p2, PlotRange -> All, BoxRatios -> 1, 1, 1]
answered 7 hours ago
DominicDominic
7024 silver badges9 bronze badges
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1 Answer
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1 Answer
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$begingroup$
There's no branch-cut from -1 to -2. Whenever plotting multi-valued functions, I'd recommend you use ParametricPlot3D so that you have better control over the Arg function and can run it through -Pi to Pi. When using Plot3D, you'll sometimes get rough branch cuts because it's in rectilinear form. This is what I'd use for your function:
f[z_, n_] := 1 - z (Log[Abs[(1 + z) z]] + I (2 n Pi + Arg[(1 + z)/z]))
p1 = ParametricPlot3D[Re[z], Im[z], Re[f[z, 0]] /.
z -> r Exp[I t], r, 0, 2, t, -Pi, Pi, PlotStyle -> Red];
p2 = ParametricPlot3D[Re[z], Im[z], Re[f[z, 1]] /.
z -> r Exp[I t], r, 0, 2, t, -Pi, Pi, PlotStyle -> Blue];
Show[p1, p2, PlotRange -> All, BoxRatios -> 1, 1, 1]
answered 7 hours ago
DominicDominic
7024 silver badges9 bronze badges
$endgroup$
add a comment |
$begingroup$
There's no branch-cut from -1 to -2. Whenever plotting multi-valued functions, I'd recommend you use ParametricPlot3D so that you have better control over the Arg function and can run it through -Pi to Pi. When using Plot3D, you'll sometimes get rough branch cuts because it's in rectilinear form. This is what I'd use for your function:
f[z_, n_] := 1 - z (Log[Abs[(1 + z) z]] + I (2 n Pi + Arg[(1 + z)/z]))
p1 = ParametricPlot3D[Re[z], Im[z], Re[f[z, 0]] /.
z -> r Exp[I t], r, 0, 2, t, -Pi, Pi, PlotStyle -> Red];
p2 = ParametricPlot3D[Re[z], Im[z], Re[f[z, 1]] /.
z -> r Exp[I t], r, 0, 2, t, -Pi, Pi, PlotStyle -> Blue];
Show[p1, p2, PlotRange -> All, BoxRatios -> 1, 1, 1]
answered 7 hours ago
DominicDominic
7024 silver badges9 bronze badges
$endgroup$
add a comment |
$begingroup$
There's no branch-cut from -1 to -2. Whenever plotting multi-valued functions, I'd recommend you use ParametricPlot3D so that you have better control over the Arg function and can run it through -Pi to Pi. When using Plot3D, you'll sometimes get rough branch cuts because it's in rectilinear form. This is what I'd use for your function:
f[z_, n_] := 1 - z (Log[Abs[(1 + z) z]] + I (2 n Pi + Arg[(1 + z)/z]))
p1 = ParametricPlot3D[Re[z], Im[z], Re[f[z, 0]] /.
z -> r Exp[I t], r, 0, 2, t, -Pi, Pi, PlotStyle -> Red];
p2 = ParametricPlot3D[Re[z], Im[z], Re[f[z, 1]] /.
z -> r Exp[I t], r, 0, 2, t, -Pi, Pi, PlotStyle -> Blue];
Show[p1, p2, PlotRange -> All, BoxRatios -> 1, 1, 1]
answered 7 hours ago
DominicDominic
7024 silver badges9 bronze badges
$endgroup$
There's no branch-cut from -1 to -2. Whenever plotting multi-valued functions, I'd recommend you use ParametricPlot3D so that you have better control over the Arg function and can run it through -Pi to Pi. When using Plot3D, you'll sometimes get rough branch cuts because it's in rectilinear form. This is what I'd use for your function:
f[z_, n_] := 1 - z (Log[Abs[(1 + z) z]] + I (2 n Pi + Arg[(1 + z)/z]))
p1 = ParametricPlot3D[Re[z], Im[z], Re[f[z, 0]] /.
z -> r Exp[I t], r, 0, 2, t, -Pi, Pi, PlotStyle -> Red];
p2 = ParametricPlot3D[Re[z], Im[z], Re[f[z, 1]] /.
z -> r Exp[I t], r, 0, 2, t, -Pi, Pi, PlotStyle -> Blue];
Show[p1, p2, PlotRange -> All, BoxRatios -> 1, 1, 1]
answered 7 hours ago
DominicDominic
7024 silver badges9 bronze badges
answered 7 hours ago
DominicDominic
7024 silver badges9 bronze badges
answered 7 hours ago
DominicDominic
7024 silver badges9 bronze badges
answered 7 hours ago
DominicDominic
7024 silver badges9 bronze badges
7024 silver badges9 bronze badges
add a comment |
add a comment |
Call me potato. is a new contributor. Be nice, and check out our Code of Conduct.
Call me potato. is a new contributor. Be nice, and check out our Code of Conduct.
Call me potato. is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
@MariuszIwaniuk In that question, I needed to know if the Mathematica understands the branch cut of the logarithm. Here, I'm interested in its Riemann surface.
$endgroup$
– Call me potato.
11 hours ago