Plot the solution complex transcendental equation

I'm trying to plot an implicit function satisfying a complex transcendental equation. The code is the following:

ContourPlot[(e (-1 + e^2) l (-3 + 2 e^2 - l^2) Cosh[2 Sqrt[1 - e^2]] +
(-1 + e^2) (-1 + l^2) (e l Cos[2] - I (1 + l^2) Sin[2]) + Sqrt[1 - e^2]
(I - I l^4 - 2 e (-2 + e^2) l Sqrt[-e^2 + l^2]) Sinh[2 Sqrt[1 - e^2]]) == 0,
{l, 1, 2}, {e, -2, 2}]


I need a real solution of e as a function of l and also e<=Min{1,l}. I don't know why it's empty in the plot. I don't think it's because the solution is out of the range since I tried l=1, and there are several solutions, one of which is e = 0. Thus I think there must be something wrong with the plot.

Your expression is not real, but complex. Think about how you want to turn your expression real. The documentation states

At positions where f does not evaluate to a real number, holes are left so that the background to the contour plot shows through.

So what you probably want is to test Abs[f[..]]==0 which gives you the lines where the whole left side vanishes. Unfortunately, they are not in your plot region. Therefore, take a look at this and hover with the mouse over the region borders to see the iso-value.

f[l_, e_] := (e (-1 + e^2) l (-3 + 2 e^2 - l^2) Cosh[
2 Sqrt[1 - e^2]] + (-1 + e^2) (-1 + l^2) (e l Cos[2] -
I (1 + l^2) Sin[2]) +
Sqrt[1 - e^2] (I - I l^4 - 2 e (-2 + e^2) l Sqrt[-e^2 + l^2]) Sinh[
2 Sqrt[1 - e^2]]);

ContourPlot[Abs[f[l, e]], {l, 1, 2}, {e, -2, 2}]


• Thanks for answering. I don't quite understand why only the real part of function is zero. The equation is complex does not mean the solution of e has to be complex, right? Oct 18, 2013 at 4:18
• I see what you mean, but this does not equivalent to the original equation anymore. I try Im[f]==0, the solution is different. Seems like my equation is problematic.. Oct 18, 2013 at 4:24
• I just wanted to point out where the issue lies. You don't have to (and probably shouldn't) use Re. What you want is probably Abs[f[l,e]] but this is for you to decide. Oct 18, 2013 at 4:26
• Thanks a lot, I see where I got wrong. And it seems to me that ContourPlot ignores isolated solution, doesn't it? Oct 18, 2013 at 6:52