# Plot complex path (essentially a 2D path) expressed in one real parameter

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.)

• 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 9.0.1.0 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

f[u_, a_] := 1./(Exp[I u] + a);

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