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Consider a complex path expressed in one real parameter, such as $$f(\alpha)=\frac{1}{e^{i \alpha} + 1},$$ which is a Möbius transformation of the unit circle. I know I can expand it all out and use the parametric plot, but it is cumbersome and not very easy (sometimes practically impossible) to do so.

Therefore, I want to know if there is some plotting function specifically designed for this. (I searched for it in the documentation but did not find it; sorry if it's all obvious.)

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The first argument of ParametricPlot should be {Re@#, Im@#} &@f[x] -- not so cumbersome, I think. – C. E. Sep 29 '13 at 22:47
@Anon Thank you for the answer, I never thought of that nice solution. Anyway, I find it hard to get the code working. I've tried a couple of possibilities like ParametricPlot[{Re@#, Im@#} & (1 / (Cos[u] + I Sin[u] + 1)), {u, 0, 2 Pi}], ParametricPlot[{Re@#, Im@#} & (1 / (Exp[I u] + 1)), {u, 0, 2 Pi}] and ParametricPlot[{Re[1/(Cos[u] + I Sin[u] + 1)], Im[1/(Cos[u] + I Sin[u] + 1)]}, {u, 0, 2 Pi}]. None of them worked. The only thing that produced a result was ParametricPlot[{Re[Cos[u] + I Sin[u]], Im[Cos[u] + I Sin[u]]}, {u, 0, 2 Pi}]. – 4ae1e1 Sep 29 '13 at 23:07
It's because you've omitted @. – C. E. Sep 29 '13 at 23:14
@Anon Oops sorry, those are typos. Inserting @ does not seem to help though, and note there's simply no @ ParametricPlot[{Re[1/(Cos[u] + I Sin[u] + 1)], Im[1/(Cos[u] + I Sin[u] + 1)]}, {u, 0, 2 Pi}], it is explicit. Maybe you could open up a Mathematica session and give them a try? At least they don't work in my student edition. – 4ae1e1 Sep 29 '13 at 23:41
I mean, the problem is with the complexity of the function, not the syntax. Since the very simple ParametricPlot[{Re[Cos[u] + I Sin[u]], Im[Cos[u] + I Sin[u]]}, {u, 0, 2 Pi}] does work. – 4ae1e1 Sep 29 '13 at 23:43
up vote 2 down vote accepted

The problem is not with the complexity of the function, but with the fact that your function diverges. Consider:

f[u_, a_] := 1./(Exp[I u] + a);
Manipulate[ParametricPlot[{Re@#, Im@#} &@f[u, a], {u, 0, 2 Pi}, PlotRange -> All], 
          {a, 0, 2}]

As long as you stay away from values of $a$ near 1, the plot is reasonable. As $a$ approaches 1, the function diverges to +/-$\infty$.

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So I was trapped by my stupid example, into which I didn't give much thinking... Okay. – 4ae1e1 Sep 29 '13 at 23:57
Well the real part is a constant 1/2 so you get a vertical line that goes to infinity at Pi or -Pi. Otherwise Presentations has a ComplexCurve[f[t], {tmin, tmax}] routine for plotting complex curves, which you would have to email me to obtain more information. – David Park Sep 30 '13 at 2:03

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