Deriving parametric plot of a branch cut
- To: mathgroup at smc.vnet.net
- Subject: [mg78804] Deriving parametric plot of a branch cut
- From: chuck009 <dmilioto at comcast.com>
- Date: Tue, 10 Jul 2007 06:26:37 -0400 (EDT)
I've been working on an interesting problem of late: f[z_]:=Log[1+Exp[-z]/Sqrt[z]] posted in here earlier. I'm unable to plot the Riemann surface of this function even with the nice code written by Michael Trott (his code fails to handle the branch cuts).
This is a contour plot of Im[f] showing three branch cuts and a circle of radius 3:
Show[{ContourPlot[Evaluate[Im[wExample] /. z -> x + I*y],
{x, -5, 5}, {y, -5, 5}, PlotPoints -> 200, Contours -> 30,
ColorFunction -> (Hue[0.7*Im] & ), ContourStyle ->
{{Thickness[0.00001]}}, PlotLabel -> Im],
Graphics[{Blue, Thickness[0.01], Circle[{0, 0}, 3]}]}]
Note the two dark lines which are two cuts as well as the negative real axis which is also a cut. The cuts occur when:
1 + Exp[-z]/(r^(1/2)*Exp[(I/2)*Arg[z]]) < 0
That is, whenever the expression in the log function is pure real and negative. For example, on the circle of radius 3, I can find the point by using FindRoot:
In[88]:=
Plot[Im[1 + Exp[-3*Exp[I*t]]/(3^(1/2)*Exp[(I/2)*Arg[3*Exp[I*t]]])],
{t, 0, 2*Pi}, AspectRatio -> 1, PlotRange ->
{{0, 2*Pi}, {-20, 20}}]
FindRoot[
Im[1 + Exp[-3*Exp[I*t]]/(3^(1/2)*Exp[(I/2)*Arg[3*Exp[I*t]]])] ==
0, {t, 24/10}, AccuracyGoal -> 10, WorkingPrecision -> 20]
Out[88]=
â??Graphicsâ??
Out[89]=
{t\[Rule]2.4481765384891584965}
Check:
In[90]:=
N[1 + Exp[-3*Exp[I*t]]/(3^(1/2)*Exp[(I/2)*Arg[3*Exp[I*t]]]) /.
t -> 2.448176538489158]
Out[90]=
-4.800214850397471 + 7.017140493687672*^-15*I
I'd like to be able to plot the cuts parametrically but I can't figure out how to do that. Can anybody suggest a way to do so?
Thanks,
Chuck
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- From: Daniel Lichtblau <danl@wolfram.com>
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