5
$\begingroup$

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)

$1-z\ln[(1+z)/z]$.

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:

enter image description here

However, the pictures shows another branch cut between -2 and -1. How come?

$\endgroup$
1
  • $\begingroup$ @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. $\endgroup$
    – user67023
    Commented Aug 18, 2019 at 13:18

1 Answer 1

6
$\begingroup$

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}]
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.