# Level curves of $|e^{1/z}|= s$ for $s \in \mathbb{R}$

In order to find the behavior of a complex-valued function at an essential singularity, I want to plot $$|e^{1/z}|= s$$ for $$s \in \mathbb{R}$$, say for $$s= 1, 1/2$$. Following is the code but it doesn't show an output.

ContourPlot[{
Abs[Exp[1/z]] == 1,
Abs[Exp[1/z]] == 1/2},
{z, -3 - 3 I, 3 + 3 I},
ContourStyle -> {Blue, Red},
FrameLabel -> {"Re(z)", "Im(z)"},
PlotLegends -> {"|e^(1/z)| = 1", "|e^(1/z)| = 1/2"}
]


Any help in getting the desired output is much appreciated.

Edit : I tried the suggested answer but it still doesn't show an output.

## Edit

For the old version (before 12.1) we can use

Clear["Global*"];
Block[{z = x + I*y},
ContourPlot[
Abs[Exp[1/z]] == {1, 1/2, 1/3, 1/4, 1/5} // Thread //
Evaluate, {x, -3, 3}, {y, -3, 3}, ContourStyle -> Automatic]]

• For complex functions ,we using ComplexContourPlot.
ComplexContourPlot[
Abs[Exp[1/z]] == {1, 1/2, 1/3, 1/4, 1/5} // Thread //
Evaluate, {z, -3 - 3 I, 3 + 3 I}, ContourStyle -> Automatic]


• thank you very much for the prompt response. I've edited the question as I still didn't get an output. Oct 11, 2023 at 0:01
• @Eureka see the updated. Oct 11, 2023 at 7:47
• many thanks ! Will it be possible to put a label corresponding to each curve with the respective $s$ value ? Oct 11, 2023 at 20:29
• @Eureka Maybe Block[{z = x + I*y}, ContourPlot[Abs[Exp[1/z]], {x, -2, 2}, {y, -2, 2}, Contours -> {1, 1/2, 1/3, 1/4, 1/5}, ContourStyle -> ColorData[97] /@ Range[5], ContourLabels -> All, ContourShading -> None]]` Oct 12, 2023 at 2:25